2. Some notation

Trees are finite except when explicitly said to be infinite. Trees may be rooted or unrooted; in a rooted tree, the root is denoted o. The rooted trees may be ordered or not. The unrooted trees will always be labelled; we do not consider unrooted unlabelled trees in the present paper.

If T is a tree, then its number of vertices is denoted by $|T|$ ; this is called the order or the size of T. (Unlike some authors, we do not distinguish between order and size.) The notation $v\in T$ means that v is a vertex in T.

The degree of a vertex $v\in T$ is denoted d(v). In a rooted tree, we also define the outdegree $d^+(v)$ as the number of children of v; thus,

(5) \begin{align} d^+(v)= \begin{cases} d(v)-1, & v\neq o,\\d(v), & v=o. \end{cases}\end{align}

A leaf in an unrooted tree is a vertex v with $d(v)=1$ . In a rooted tree, we instead require $d^+(v)=0$ ; this may make a difference only for the root.

A fringe subtree in a rooted tree is a subtree consisting of some vertex v and all its descendants. We regard v as the root of the fringe tree. The branches of a rooted tree are the fringe trees rooted at the children of the root. The number of branches thus equals the degree d(o) of the root.

Let $\mathfrak{T}_n$ be the set of all ordered rooted trees of order n, and let $\mathfrak{T}\,:\!=\,\bigcup_1^\infty\mathfrak{T}_n$ . Note that $\mathfrak{T}_n$ is a finite set; we may identify the vertices of an ordered rooted tree by finite strings of positive integers, such that the root is the empty string and the children of v are vi, $i=1,\dots,d(v)$ . (Thus, an ordered rooted tree is regarded as a subtree of the infinite Ulam–Harris tree.) In fact, it is well known that $|\mathfrak{T}_n|=\frac{1}{n}\binom{2n-2}{n-1}$ , the Catalan number $C_{n-1}$ .

Let $\mathfrak{L}_n$ be the set of all unrooted trees of order n, with the labels $1,\dots,n$ ; thus $\mathfrak{L}_n$ is the set of all trees on $[n]\,:\!=\,{\{{1,\dots,n}\}}$ . $\mathfrak{L}_n$ is evidently finite and by Cayley’s formula $|\mathfrak{L}_n|=n^{n-2}$ . Let $\mathfrak{L}\,:\!=\,\bigcup_1^\infty\mathfrak{L}_n$ .

A probability sequence is the same as a probability distribution on $\mathbb N_0\,:\!=\,{\{{0,1,2,\dots}\}}$ , i.e., a sequence $\textbf{p}=(p_k)_0^\infty$ with $p_k\geqslant0$ and $\sum_{k=0}^\infty p_k=1$ . The mean $\mu(\textbf{p})$ and variance $\sigma^2(\textbf{p})$ of a probability sequence are defined to be the mean and variance of a random variable with distribution $\textbf{p}$ , i.e.,

(6) \begin{align} \mu(\textbf{p})\,:\!=\,\sum_{k=0}^\infty kp_k, &&&\sigma^2(\textbf{p})\,:\!=\,\sum_{k=0}^\infty k^2p_k-\mu(\textbf{p})^2.\end{align}

We use $\overset{\textrm{d}}{\longrightarrow}$ and $\overset{\textrm{p}}{\longrightarrow}$ for convergence in distribution and in probability, respectively, for a sequence of random variables in some metric space; see e.g. [Reference Billingsley11]. Also, $\overset{\textrm{d}}{=}$ means convergence in distribution.

The total variation distance between two random variables X and Y in a metric space (or rather between their distributions) is defined by

(7) \begin{align} d_{\textrm{TV}}(X,Y)\,:\!=\,\sup_A\bigl\lvert{{\mathbb P{}}(X\in A)-{\mathbb P{}}(Y\in A) }\bigr\rvert,\end{align}

taking the supremum over all measurable subsets A. It is well known that in a complete separable metric space, there exists a coupling of X and Y (i.e., a joint distribution with the given marginal distributions) such that

(8) \begin{align} {\mathbb P{}}(X\neq Y) = d_{\textrm{TV}}(X,Y),\end{align}

and this is best possible.

$O_{\textrm{p}}(1)$ denotes a sequence of real-valued random variables $(X_n)_n$ that is stochastically bounded, i.e., for every $\varepsilon>0$ , there exists C such that ${\mathbb P{}}(|X_n|>C)\leqslant\varepsilon$ . This is equivalent to $(X_n)_n$ being tight. For tightness in more general metric spaces, see e.g. [Reference Billingsley11].

Let f be a real-valued function defined on an interval $I\subseteq\mathbb R$ . The modulus of continuity of f is the function $[0,\infty)\to[0,\infty]$ defined by, for $\delta\geqslant0$ ,

(9) \begin{align} \omega(\delta;\,f)=\omega(\delta;\,f;\,I)\,:\!=\,\sup\bigl({|f(s)-f(t)|\,:\, s,t\in I, |s-t|\leqslant\delta}\bigr).\end{align}

If x and y are real numbers, $x\land y\,:\!=\,\min{\{{x,y}\}}$ and $x\lor y\,:\!=\,\max{\{{x,y}\}}$ .

C denotes unspecified constants that may vary from one occurrence to the next. They may depend on parameters such as weight sequences or offspring distributions, but they never depend on the size of the trees. Sometimes we write, e.g., $C_r$ to emphasize that the constant depends on the parameter r.

Unspecified limits are as ${{n\to\infty}}$ .

2.1. Profiles

For two vertices v and w in a tree T, let $\textsf{d}(x,y)=\textsf{d}_T(x,y)$ denote the distance between v and w, i.e., the number of edges in the unique path joining v and w. In particular, in a rooted tree, $\textsf{d}(v,o)$ is the distance to the root, often called the depth (or sometimes height) of v.Footnote 1

For a rooted tree T, the height of T is $H(T)\,:\!=\,\max_{v\in T}\textsf{d}(v,o)$ , i.e., the maximum depth. The diameter of a tree T, rooted or not, is $\textrm{diam}(T)\,:\!=\,\max_{v,w\in T}\textsf{d}(v,w)$ .

The profile of a rooted tree is the function $L=L_T\,:\,\mathbb R\to[0,\infty)$ defined by

(10) \begin{align} L(i)\,:\!=\,\bigl\lvert{{\{{v\in T\,:\,\textsf{d}(v,o)=i}\}}}\bigr\rvert,\end{align}

for integers i, extended by linear interpolation to all real x. (We are mainly interested in $x\geqslant0$ , and trivially $L(x)=0$ for $x\leqslant -1$ , but it will be convenient to allow negative x.) The linear interpolation can be written as

(11) \begin{align} L(x)\,:\!=\,\sum_{i=0}^\infty L(i)\tau(x-i),\end{align}

where $\tau$ is the triangular function $\tau(x)\,:\!=\,(1-|x|)\lor0$ .

Note that $L(0)=1$ , and that L is a continuous function with compact support $[\!-1,H(T)+1]$ . Furthermore, since $\int\tau(x)\,\textrm{d} x=1$ ,

(12) \begin{align} \int_{-1}^\infty L(x)\,\textrm{d} x = \sum_{i=0}^\infty L(i) = |T|,\end{align}

where we integrate from $-1$ because of the linear interpolation; we have $ \int_0^\infty L(x)\,\textrm{d} x = \sum_{i=0}^\infty L(i) -\frac12 = |T|-\frac12$ .

The width of T is defined as

(13) \begin{align} W(T)\,:\!=\,\max_{i\in\mathbb N_0} L(i)=\max_{x\in\mathbb R} L(x).\end{align}

Similarly, in any tree T, rooted or unrooted, we define the distance profile as the function $\Lambda=\Lambda_T\,:\,[0,\infty)\to[0,\infty)$ defined by

(14) \begin{align}\Lambda(i)\,:\!=\,\bigl\lvert{{\{{(v,w)\in T\,:\,\textsf{d}(v,w)=i}\}}}\bigr\rvert\end{align}

for integers i, again extended by linear interpolaton to all real $x\geqslant0$ . For definiteness, we count ordered pairs in (14), and we include the case $v=w$ , so $\Lambda(0)=|T|$ . $\Lambda$ is a continuous function on $[0,\infty)$ with support $[\!-1,\textrm{diam}(T)+1]$ . We have, similarly to (12),

(15) \begin{align} \int_{-1}^\infty \Lambda(t)\,\textrm{d} t = \sum_{i=0}^\infty \Lambda(i)= |T|^2.\end{align}

If T is an unrooted tree, let T(v) denote the rooted tree obtained by declaring v as the root, for $v\in T$ . Then, as a consequence of (10) and (14),

(16) \begin{align} \Lambda_T(x)=\sum_{v\in T} L_{T(v)}(x).\end{align}

Hence, the distance profile can be regarded as the sum (or, after normalisation, average) of the profiles for all possible choices of a root.

Remark 2.1. Alternatively, one might extend L to a step function by $L(x)\,:\!=\,L(\lfloor x\rfloor)$ , and similarly for $\Lambda$ . The asymptotic results are the same (and equivalent by simple arguments), with L and $\Lambda$ elements of $D[0,\infty]$ instead of $C[0,\infty]$ and limit theorems taking place in that space. This has some advantages, but for technical reasons (e.g. simpler tightness criteria), we prefer to work in the space $C[0,\infty]$ of continuous functions.

Remark 2.2. Another version of $\Lambda$ would count unordered pairs of distinct vertices. The two versions are obviously equivalent and our results hold for the alternative version too, mutatis mutandis.

2.2. Brownian excursion and its local time

The standard Brownian excursion $\textbf{e}$ is a random continuous function ${[0,1]}\to[0,\infty)$ such that $\textbf{e}(0)=\textbf{e}(1)=0$ and $\textbf{e}(t)>0$ for $t\in(0,1)$ . Informally, $\textbf{e}$ can be regarded as a Brownian motion conditioned on these properties; this can be formalised as an appropriate limit [Reference Durrett, Iglehart and Miller23]. There are several other, quite different but equivalent, definitions, see e.g. [Reference Revuz and Yor49, XII.(2.13)], [Reference Blumenthal13, Example II.1d)], and [Reference Drmota17, Section 4.1.3].

The local time $L_\textbf{e}$ of $\textbf{e}$ is a continuous random function that is defined (almost surely) as a functional of $\textbf{e}$ satisfying

(17) \begin{align} \int_0^\infty f(x)L_\textbf{e}(x)\,\textrm{d} x = \int_0^1 f\bigl({\textbf{e}(t)}\bigr)\,\textrm{d} t,\end{align}

for every bounded (or non-negative) measurable function $f\,:\,[0,\infty)\to\mathbb R$ . In particular, (17) yields, for any $x\geqslant0$ and $\varepsilon>0$ ,

(18) \begin{align} \int_x^{x+\varepsilon}L_\textbf{e}(y)\,\textrm{d} y = \int_0^1 \boldsymbol1\{{{\textbf{e}(t)\in[x,x+\varepsilon)}}\}\,\textrm{d} t\end{align}

and thus

(19) \begin{align}L_\textbf{e}(x) = \lim_{\varepsilon\to0} \frac{1}{\varepsilon}\int_0^1 \boldsymbol1\{{{\textbf{e}(t)\in[x,x+\varepsilon)}}\}\,\textrm{d} t.\end{align}

Hence, $L_\textbf{e}(x)$ can be regarded as the occupation density of $\textbf{e}$ at the value x.

Note that the existence (almost surely) of a function $L_\textbf{e}(x)$ satisfying (17)–(19) is far from obvious; this is part of the general theory of local times for semimartingales, see e.g. [Reference Revuz and Yor49, Chapter VI]. The existence also follows from (some of) the proofs of Theorem 1.2.

2.3. Brownian continuum random tree

Given a continuous function $g\,:\,{[0,1]}\to[0,\infty)$ with $g(0)=g(1)=0$ , one can define a pseudometric $\textsf{d}$ on ${[0,1]}$ by

(20) \begin{align}\textsf{d}(s,t)= \textsf{d}(s,t;\,g)\,:\!=\,g(s)+g(t)-2\min_{u\in [s,t]}g(u),\qquad 0\leqslant s\leqslant t\leqslant 1.\end{align}

By identifying points with distance 0, we obtain a metric space $T_g$ , which is a compact real tree, see e.g. Le Gall [Reference Le Gall39, Theorem 2.2]. We denote the natural quotient map ${[0,1]}\to T_g$ by $\rho_g$ , and let $T_g$ be rooted at $\rho_{g}(0)$ . The Brownian continuum random tree defined by Aldous[Reference Aldous2–Reference Aldous4] can be defined as the random real tree $T_{\textbf{e}}$ constructed in this way from the random Brownian excursion $\textbf{e}$ , see [Reference Le Gall39, Section 2.3]. (Aldous [Reference Aldous2–Reference Aldous4] used another definition, and another scaling corresponding to $T_{2\textbf{e}}$ .) Note that using (20), (4) can be written

(21) \begin{align}\int_0^\infty f(x) \Lambda_\textbf{e}(x) \,\textrm{d} x=\iint_{s,t\in{[0,1]}} f\bigl({\textsf{d}(s,t;\,\textbf{e})}\bigr)\,\textrm{d} s\,\textrm{d} t,\end{align}

for any bounded (or non-negative) measurable function f. This means that $\Lambda_\textbf{e}$ is the density of the distance in $T_\textbf{e}$ between two random points, chosen independently with the probability measure on $T_\textbf{e}$ induced by the uniform measure on ${[0,1]}$ . This justifies the equivalence of the two definitions of $\Lambda_\textbf{e}$ stated in Theorem 1.4. As for the local time $L_\textbf{e}$ , the existence (almost surely) of a continuous function $\Lambda_\textbf{e}$ satisfying (21) is far from trivial; this will be a consequence of our proof.

An important feature of the Brownian continuum random tree is its re-rooting invariance property. More precisely, fix $s \in [0,1]$ and set

(22) \begin{align} \textbf{e}^{[s]}(t)= \begin{cases}\textsf{d}(s,s+t;\,\textbf{e}), & 0\leqslant t < 1-s\\\textsf{d}(s,s+t-1;\,\textbf{e}), &1-s \leqslant t \leqslant 1. \end{cases}\end{align}

Note that $\textbf{e}^{[s]}$ is a random continuous function ${[0,1]}\to[0,\infty)$ such that $\textbf{e}^{[s]}(0)=\textbf{e}^{[s]}(1)=0$ and a.s. $\textbf{e}^{[s]}(t)>0$ for $t\in(0,1)$ ; clearly, $\textbf{e}^{[0]} = \textbf{e}$ . By Duquesne and Le Gall [Reference Duquesne and Le Gall21, Lemma 2.2], the compact real tree $T_{\textbf{e}^{[s]}}$ is then canonically identified with the $T_{\textbf{e}}$ tree re-rooted at the vertex $\rho_{\textbf{e}}(s)$ . Marckert and Mokkadem [Reference Marckert and Mokkadem40, Proposition 4.9] (see also Duquesne and Le Gall [Reference Duquesne and Le Gall22, Theorem 2.2]) have shown that for every fixed $s \in [0,1]$ ,

(23) \begin{align} \textbf{e}^{[s]}\overset{\textrm{d}}{=} \textbf{e} \quad \text{and} \quad T_{\textbf{e}^{[s]}} =T_{\textbf{e}},\end{align}

in distribution. Thus, the re-rooted tree $T_{\textbf{e}^{[s]}}$ is a version of the Brownian continuum random tree.

Remark 2.3. Indeed, Aldous [Reference Aldous3, (20)] already observed that the Brownian continuum random tree is invariant under uniform re-rooting and that this property corresponds to the invariance of the law of the Brownian excursion under the path transformation (22) if $s = U$ is uniformly random on [0, 1] and independent of $\textbf{e}$ .

As a consequence of the previous re-rooting invariance property, we deduce the following explicit expression for the continuous function $\Lambda_\textbf{e}$ . For every fixed $s \in [0,1]$ , let $L_{\textbf{e}^{[s]}}$ denote the local time of $\textbf{e}^{[s]}$ , which is perfectly defined thanks to (23). It follows from (20), (21) and (22) that

\begin{align*}\int_0^\infty f(x) \Lambda_\textbf{e}(x) \,\textrm{d} x=\int_0^1\int_0^1 f\bigl({\textbf{e}^{[s]}(t)}\bigr)\,\textrm{d} s \,\textrm{d} t=\int_{0}^{1} \int_0^\infty f(x) L_{\textbf{e}^{[s]}}(x) \,\textrm{d} x\,\textrm{d} s,\end{align*}

for any bounded (or non-negative) measurable function f, or equivalently,

(24) \begin{align} \Lambda_\textbf{e}(x) =\int_{0}^{1} L_{\textbf{e}^{[s]}}(x) \,\textrm{d} s,\quad x \geqslant 0.\end{align}

In accordance with the discrete analogue of $\Lambda_\textbf{e}$ in (16), the identity (24) shows that $\Lambda_\textbf{e}$ can be regarded as the average of the profiles for all possible choices of a root in $T_{\textbf{e}}$ .

3. Rooted simply generated trees

As a background, we recall first the definition of random rooted simply generated trees and the almost equivalent conditioned Galton–Watson trees, see e.g. [Reference Drmota17] or [Reference Janson30] for further details, and [Reference Athreya and Ney6] for more on Galton–Watson processes.

3.1. Simply generated trees

Let $\boldsymbol{\phi}=(\phi_k)_0^\infty$ be a given sequence of non-negative weights, with $\phi_0>0$ and $\phi_k>0$ for at least one $k\geqslant2$ . (The latter conditions exclude only trivial cases when the random tree ${\mathcal T}^{{\boldsymbol{\phi}}}_n$ defined below either does not exist or is a deterministic path.)

For any rooted ordered tree $T\in\mathfrak{T}$ , define the weight of T as

(26) \begin{align} \phi(T)\,:\!=\,\prod_{v\in T} \phi_{d^+(v)}.\end{align}

For a given n, we define the random rooted simply generated tree ${\mathcal T}^{{\boldsymbol{\phi}}}_n$ of order n as a random tree in $\mathfrak{T}_n$ with probability proportional to its weight; i.e.,

(27) \begin{align} {\mathbb P{}}({\mathcal T}^{{\boldsymbol{\phi}}}_n=T) \,:\!=\,\frac{\phi(T)}{\sum_{T'\in\mathfrak{T}_n}\phi(T')},\qquad T\in\mathfrak{T}_n.\end{align}

We consider only n such that at least one tree T with $\phi(T)>0$ exists.

A weight sequence $\boldsymbol{\phi}'=(\phi^{\prime}_k)_0^\infty$ with

(28) \begin{align} \phi^{\prime}_k=a b^k\phi_k,\qquad k\geqslant0,\end{align}

for some $a,b>0$ is said to be equivalent to $(\phi_k)_0^\infty$ . It is easily seen that equivalent weight sequences define the same random tree ${\mathcal T}^{{\boldsymbol{\phi}}}_n$ , i.e., ${\mathcal T}^{{\boldsymbol{\phi}'}}_n\overset{\textrm{d}}{=}{\mathcal T}^{{\boldsymbol{\phi}}}_n$ .

3.3. Equivalence

A random simply generated tree ${\mathcal T}^{{\boldsymbol{\phi}}}_n$ with a weight sequence $(\phi_k)_0^\infty$ that is a probability sequence equals, as just said, the conditioned Galton–Watson tree ${\mathcal T}^{{\boldsymbol{\phi}}}_n$ . Much more generally, any weight sequence $\boldsymbol{\phi}$ such that its generating function

(29) \begin{align} \Phi(z)\,:\!=\,\sum_{k=0}^\infty \phi_k z^k\end{align}

has positive radius of convergence is equivalent to some probability weight sequence; hence, ${\mathcal T}^{{\boldsymbol{\phi}}}_n$ can be regarded as a conditioned Galton–Watson tree in this case too. Moreover, in many cases, we can choose an equivalent probability weight sequence that has mean 1 and finite variance; see e.g. [Reference Janson30, Section 4].

We will use this to switch between simply generated trees and conditioned Galton–Watson trees without comment in the sequel; we will use the name that seems best and most natural in different contexts.

3.4. Simply generated forests

The Galton–Watson process above starts with one individual. More generally, we may start with m individuals, which we may assume are numbered $1,\dots,m$ ; this yields a Galton–Watson forest consisting of m independent copies of ${\mathcal T}^{{\textbf{p}}}$ . Conditioning on the total size being $n\geqslant m$ , we obtain a conditioned Galton–Watson forest ${\mathcal T}^{{\textbf{p}}}_{n,m}$ , which thus consists of m random trees ${\mathcal T}^{{\textbf{p}}}_{n,m;\,1}, \dots,{\mathcal T}^{{\textbf{p}}}_{n,m;\,m}$ with $|{\mathcal T}^{{\textbf{p}}}_{n,m;\,1}|+\dots+|{\mathcal T}^{{\textbf{p}}}_{n,m;\,{m}}|=n$ . Conditioned on the sizes $|{\mathcal T}^{{\textbf{p}}}_{n,m;\,1}|,\dots,|{\mathcal T}^{{\textbf{p}}}_{n,m;\,m}|$ , the trees are independent conditioned Galton–Watson trees with the given sizes.

More generally, given any weight sequence $\boldsymbol{\phi}$ , a random simply generated forest ${\mathcal T}^{{\boldsymbol{\phi}}}_{n,m}$ is a random forest with m rooted trees and total size n, chosen with probability proportional to its weight, defined as in (26). Again, conditioned on their sizes, the trees are independent simply generated trees.

Thus, the distribution of the sizes of the trees in the forest is of major importance. Consider the Galton–Watson case, and let ${\mathcal T}^{{\textbf{p}}}_{n,m;\,(1)}, \dots,{\mathcal T}^{{\textbf{p}}}_{n,m;\,(m)}$ denote the trees arranged in decreasing order: $|{\mathcal T}^{{\textbf{p}}}_{n,m;\,(1)}|\geqslant\dots\geqslant|{\mathcal T}^{{\textbf{p}}}_{n,m;\,(m)}|$ . (Ties are resolved randomly, say; this applies tacitly to all similar situations.) We have the following general result, which was proved by Marzouk [38, Lemma 5.7(iii)] under an additional regularity hypothesis.

Lemma 3.1. Let $m\geqslant1$ be fixed, and consider the conditioned Galton–Watson forest ${\mathcal T}^{{{\textbf{p}}}}_{n,m}$ as ${{n\to\infty}}$ . Then

(30) \begin{align}|{\mathcal T}^{{{\textbf{p}}}}_{n,m;\,(i)}|= \begin{cases}n-O_{\textrm{p}}(1), & i=1\\O_{\textrm{p}}(1), & i=2,\dots,m. \end{cases}\end{align}

Proof. Suppose first that $\mu(\textbf{p})=1$ . Suppose also, for simplicity, that $p_m>0$ . Consider the conditioned Galton–Watson tree ${\mathcal T}^{{\textbf{p}}}_{n+1}$ and condition on the event $\mathcal E_m$ that the root degree is m. Conditioned on $\mathcal E_m$ , there are m branches, which form a conditioned Galton–Watson forest ${\mathcal T}^{{\textbf{p}}}_{n,m}$ .

As ${{n\to\infty}}$ , the random tree ${\mathcal T}^{{\textbf{p}}}_{n+1}$ converges in distribution to an infinite random tree $\widehat{{\mathcal T}}$ (the size-biased Galton–Watson tree defined by Kesten [Reference Kesten35]), see [Reference Janson30, Theorem 7.1]. Moreover, ${\mathbb P{}}(\mathcal E_m)\to mp_m>0$ by [Reference Janson30, Theorem 7.10]. Hence, ${\mathcal T}^{{\textbf{p}}}_{n+1}$ conditioned on $\mathcal E_m$ converges in distribution to $\widehat{{\mathcal T}}$ conditioned on $\mathcal E_m$ . In other words, the forest ${\mathcal T}^{{\textbf{p}}}_{n,m}$ converges in distribution to the branches of $\widehat{{\mathcal T}}$ conditioned on having exactly m branches; denote this random limit by $({\mathcal T}_1,\dots,{\mathcal T}_m)$ . By the Skorohod coupling theorem [Reference Kallenberg32, Theorem 4.30], we may (for simplicity) assume that this convergence is a.s. The convergence here is in the local topology used in [Reference Janson30], which means [Reference Janson30, Lemma 6.2] that for any fixed $\ell\geqslant1$ , if $T^{[\ell]}$ denotes the tree T truncated at height $\ell$ , then a.s., for sufficiently large n, ${\mathcal T}^{{\textbf{p},[\ell]}}_{n,m;\,i}={\mathcal T}_i^{[\ell]}$ .

The infinite tree $\widehat{{\mathcal T}}$ has exactly one infinite branch; thus, there exists a (random) $j\leqslant m$ such that ${\mathcal T}_j$ is infinite but ${\mathcal T}_i$ is finite for $i\neq j$ . Truncating the trees at an $\ell$ chosen larger than the heights $H({\mathcal T}_i)$ for all $i\neq j$ , we see that for large n, ${\mathcal T}^{{\textbf{p}}}_{n,m;\,i}={\mathcal T}_i$ . Thus, $|{\mathcal T}^{{\textbf{p}}}_{n,m;\,i}|=O(1)$ for $i\neq j$ , and necessarily the remaining branch ${\mathcal T}^{{\textbf{p}}}_{n,m;\,j}$ has size $n-O(1)$ . Hence, for large enough n, ${\mathcal T}^{{\textbf{p}}}_{n,m;\,(1)}={\mathcal T}^{{\textbf{p}}}_{n,m;\,j}$ .

Consequently, ${\mathcal T}^{{\textbf{p}}}_{n,m;\,(2)},\dots,{\mathcal T}^{{\textbf{p}}}_{n,m;\,(m)}$ converge a.s., and thus in distribution, to the $m-1$ finite branches of $\widehat{{\mathcal T}}$ , arranged in decreasing order and conditioned on $\mathcal E_m$ . In particular, their sizes converge in distribution and are thus $O_{\textrm{p}}(1)$ .

We assumed for simplicity $p_m>0$ . In general, we may select a rooted tree T with $\geqslant m$ leaves, such that $p_{d^+(v)}>0$ for every $v\in T$ . Fix m leaves $v_1,\dots,v_m$ in T, and consider the conditioned Galton–Watson tree ${\mathcal T}^{{\textbf{p}}}_{n+|T|-m}$ conditioned on the event $\mathcal E_T$ that it consists of T with subtrees added at $v_1,\dots,v_m$ . Then these subtrees form a conditioned Galton–Watson forest ${\mathcal T}^{{\textbf{p}}}_{n,m}$ , and we can argue as above, conditioning $\widehat{{\mathcal T}}$ on $\mathcal E_T$ .

This completes the proof when $\mu(\textbf{p})=1$ . If $\mu(\textbf{p})>1$ , there always exists an equivalent probability weight $\tilde{\textbf{p}}$ with $\mu(\tilde{\textbf{p}})=1$ , and the result follows. If $\mu(\textbf{p})<1$ , the same may hold, and if it does not hold, then there is a similar infinite limit tree $\widehat{{\mathcal T}}$ [Reference Janson30, Theorem 7.1]; in this case, $\widehat{{\mathcal T}}$ has one vertex of infinite degree, but the proof above holds with minor modifications.

Remark 3.2. The proof shows that in the case $\mu(\textbf{p})=1$ , the small trees ${\mathcal T}^{{\textbf{p}}}_{n,m;\,(2)},\dots,{\mathcal T}^{{\textbf{p}}}_{n,m;\,(m)}$ in the forest converge in distribution to $m-1$ independent copies of the unconditioned Galton–Watson tree ${\mathcal T}^{{\textbf{p}}}$ , arranged in decreasing order. More generally, the small trees converge in distribution to independent Galton–Watson trees for a probability distribution equivalent to $\textbf{p}$ . (This too was shown in [38, Lemma 5.7(iii)] under stronger assumptions.)

Remark 3.3. In the standard case $\mu(\textbf{p})=1$ , $\sigma^2(\textbf{p})<\infty$ , it is also easy to show Lemma 3.1 using the fact ${\mathbb P{}}(|{\mathcal T}^{{\textbf{p}}}|=n)\sim c n^{-3/2}$ , for some $c>0$ , which is a well-known consequence of the local limit theorem, cf. (36)–(37).

Problem 3.4. A simply generated forest ${\mathcal T}^{{\boldsymbol{\phi}}}_{n,m}$ is covered by Lemma 3.1 when the generating function (29) has positive radius of convergence, since then it is equivalent to a conditioned Galton–Watson forest. We conjecture that Lemma 3.1 holds for simply generated forests also in the case when the generating function has radius of convergence 0, but we leave this as an open problem.

4. Modified simply generated trees

One frequently meets random trees where the root has a special distribution, see, for example [Reference Kortchemski and Marzouk38, Reference Marckert and Panholzer41]. Thus, let $\boldsymbol{\phi}$ and $\boldsymbol{\phi}^0$ be two weight sequences, where $\boldsymbol{\phi}$ is as above, and $\boldsymbol{\phi}^0=(\phi^0_k)_0^\infty$ satisfies $\phi^0_k\geqslant0$ , with strict inequality for at least one k. We modify (26) and now define the weight of a tree $T\in\mathfrak{T}$ as

(31) \begin{align} \phi^*(T)\,:\!=\,\phi^0_{d^+(o)}\prod_{v\neq o} \phi_{d^+(v)}.\end{align}

The random modified simply generated tree ${\mathcal T}^{{\boldsymbol{\phi}},{\boldsymbol{\phi}^0}}_n$ is defined as in (27), using the modified weight (31).

We say that a pair $(\boldsymbol{\phi}',\boldsymbol{\phi}^{0\prime})$ is equivalent to $(\boldsymbol{\phi},\boldsymbol{\phi}^0)$ if (28) holds and similarly

(32) \begin{align} \phi^{0\prime}_k=a_0 b^k\phi^0_k,\qquad k\geqslant0.\end{align}

It is important that the same b is used in (28) and (32), while a and $a_0$ may be different. It is easy to see that equivalent pairs of weight sequences define the same modified simply generated tree.

Similarly, given two probability sequences $\textbf{p}=(p_k)_0^\infty$ and $\textbf{p}^0=({p}^0_k)_0^\infty$ , we define the modified Galton–Watson tree ${\mathcal T}^{{\textbf{p},\textbf{p}^0}}$ and conditioned modified Galton–Watson tree ${\mathcal T}^{{\textbf{p},\textbf{p}^0}}_n$ as in Section 3.2, but now giving children to the root with distribution $\textbf{p}^0$ , and to everyone else with distribution $\textbf{p}$ .

Again, as indicated by our notation, we have an equality: the conditioned modified Galton–Watson tree ${\mathcal T}^{{\textbf{p},\textbf{p}^0}}_n$ equals the modified simply generated tree with weight sequences $\textbf{p}$ and $\textbf{p}^0$ . Conversely, if two weight sequences $\boldsymbol{\phi}$ and $\boldsymbol{\phi}^0$ both have positive radius of convergence, then it is possible (by taking b small enough) to find equivalent weight sequences $\boldsymbol{\phi}'$ and $\boldsymbol{\phi}^{0\prime}$ that are probability sequences, and thus ${\mathcal T}^{{\boldsymbol{\phi},\boldsymbol{\phi}^0}}_n= {\mathcal T}^{{\boldsymbol{\phi}',\boldsymbol{\phi}^{0\prime}}}_n$ can be interpreted as a conditioned modified Galton–Watson tree.

Proof. Exercise.

Note that Lemma 4.1 applies also to the simply generated tree ${\mathcal T}^{{\boldsymbol{\phi}}}_n$ (by taking $\boldsymbol{\phi}^0=\boldsymbol{\phi}$ ). Thus, conditioned on the root degree the branches have the same distribution for ${\mathcal T}^{{\boldsymbol{\phi}}}_n$ and ${\mathcal T}^{{\boldsymbol{\phi},\boldsymbol{\phi}^0}}_n$ . Hence, the distribution of the root degree is of central importance. The following lemma is partly shown by [Reference Kortchemski and Marzouk38, Proposition 5.6] in greater generality (the stable case), although we add the estimate (35).

Lemma 4.2. (mainly Kortchemski and Marzouk [Reference Kortchemski and Marzouk38]) Suppose that ${\textbf{p}}$ is a probability sequence with mean $\mu({\textbf{p}})=1$ and variance $\sigma^2({\textbf{p}})\in(0,\infty)$ and that ${\textbf{p}}^0$ is a probability sequence with finite mean $\mu({\textbf{p}}^0)$ . Then the root degree d(o) in the conditioned modified Galton–Watson tree ${\mathcal T}_n^{{{\textbf{p}},{\textbf{p}}^0}}$ converges in distribution to a random variable $\widetilde D$ with distribution

(33) \begin{align} {\mathbb P{}}(\widetilde D=k)= \frac{k{p}^0_k}{\sum_{j=1}^\infty j{p}^0_j}= \frac{k{p}^0_k}{\mu({\textbf{p}}^0)}.\end{align}

In other words, for every fixed $k\geqslant0$ ,

(34) \begin{align} {\mathbb P{}}(d(o)=k)\to{\mathbb P{}}(\widetilde D=k),\qquad{\rm{as}}\,n \to \infty.\end{align}

Moreover, if n is large enough, we have uniformly

(35) \begin{align} {\mathbb P{}}(d(o)=k)\leqslant 2{\mathbb P{}}(\widetilde D=k),\qquad k\geqslant1.\end{align}

As a consequence, ${\mathbb E{}}\widetilde D<\infty$ if and only if $\sigma^2(\textbf{p}^0)<\infty$ .

Proof. This uses well-known standard arguments, but we give a full proof for completeness; see also [Reference Kortchemski and Marzouk38]. Let D be the root degree in the modified Galton–Watson tree ${\mathcal T}^{{\textbf{p},\textbf{p}^0}}$ . If $D=k$ , then the rest of the tree consists of k independent copies of ${\mathcal T}^{{\textbf{p}}}$ . Thus, the conditional probability ${\mathbb P{}}\bigl({|{\mathcal T}^{{\textbf{p},\textbf{p}^0}}|=n\mid D=k}\bigr)$ equals the probability that a Galton–Watson process started with k individuals has in total $n-1$ individuals; hence, by a formula by Dwass [Reference Dwass24], see e.g. [Reference Janson30, Section 15] and the further references there,

(36) \begin{align} {\mathbb P{}}\bigl({|{\mathcal T}^{{\textbf{p},\textbf{p}^0}}|=n\mid D=k}\bigr)=\frac{k}{n-1}{\mathbb P{}}\bigl({S_{n-1}=n-k-1}\bigr),\end{align}

where $S_{n-1}$ denotes the sum of $n-1$ independent random variables with distribution $\textbf{p}$ .

Suppose for simplicity that the distribution $\textbf{p}$ is aperiodic, i.e., not supported on any subgroup $d\mathbb N$ . (The general case follows similarly using standard modifications.) It then follows by the local limit theorem, see e.g. [Reference Kolchin36, Theorem 1.4.2] or [Reference Petrov46, Theorem VII.1], that, as ${{n\to\infty}}$ ,

(37) \begin{align} {\mathbb P{}}\bigl({S_{n-1}=n-k-1}\bigr)= \frac{1}{\sqrt{2\pi\sigma^2 n}}\bigl({e^{-k^2/(2n\sigma^2)}+o(1)}\bigr),\end{align}

uniformly in k. Consequently, combining (36) and (37) with ${\mathbb P{}}(D=k)={p}^0_k$ ,

(38) \begin{align} {\mathbb P{}}\bigl({|{\mathcal T}^{{\textbf{p},\textbf{p}^0}}|=n \text{ and } D=k}\bigr)&=\frac{k{p}^0_k}{n-1}{\mathbb P{}}\bigl({S_{n-1}=n-k-1}\bigr)\notag\\&= c\frac{k{p}^0_k}{n^{3/2}}\bigl({e^{-k^2/(2n\sigma^2)}+o(1)}\bigr),\end{align}

uniformly in k, where $c\,:\!=\,(2\pi\sigma^2)^{-1/2}$ .

Summing over k we find as ${{n\to\infty}}$ , using $\sum k{p}^0_k<\infty$ and monotone convergence,

(39) \begin{align} {\mathbb P{}}\bigl({|{\mathcal T}^{{\textbf{p},\textbf{p}^0}}|=n}\bigr)&=\frac{c}{n^{3/2}}\Bigl({\sum_{k=1}^\infty {k{p}^0_k}e^{-k^2/(2n\sigma^2)}+o(1)}\Bigr)\sim \frac{c}{n^{3/2}}\sum_{k=1}^\infty {k{p}^0_k}.\end{align}

Thus, combining (38) and (39), for any fixed $k\geqslant1$ , as ${{n\to\infty}}$ ,

(40) \begin{align} {\mathbb P{}}\bigl({D=k\mid|{\mathcal T}^{{\textbf{p},\textbf{p}^0}}|=n}\bigr)&=\frac{ {\mathbb P{}}\bigl({|{\mathcal T}^{{\textbf{p},\textbf{p}^0}}|=n\text{ and } D=k}\bigr)}{ {\mathbb P{}}\bigl({|{\mathcal T}^{{\textbf{p},\textbf{p}^0}}|=n}\bigr)}\notag\\&\to \frac{k{p}^0_k}{\sum_{j=1}^\infty j{p}^0_j}.\end{align}

The limits on the right-hand side sum to 1, and thus the result (33) follows.

Moreover, (38) and (39) also yield

(41) \begin{align} {\mathbb P{}}\bigl({D=k\mid|{\mathcal T}^{{\textbf{p},\textbf{p}^0}}|=n}\bigr)&=\frac{ {\mathbb P{}}\bigl({|{\mathcal T}^{{\textbf{p},\textbf{p}^0}}|=n\text{ and } D=k}\bigr)}{ {\mathbb P{}}\bigl({|{\mathcal T}^{{\textbf{p},\textbf{p}^0}}|=n}\bigr)}\notag\\&\leqslant \frac{k{p}^0_k}{\sum_{j=1}^\infty j{p}^0_j}\bigl({1+o(1)}\bigr),\end{align}

uniformly in k. In particular, (35) holds for all k if n is large enough.

Finally, by (33), ${\mathbb E{}}\widetilde D=\sum k{\mathbb P{}}(\widetilde D=k)<\infty$ if and only if $\sum_k k^2{p}^0_k<\infty$ .

It follows from Lemma 4.2 that the tree is overwhelmingly dominated by one branch. (Again, this was shown by Kortchemski and Marzouk [Reference Kortchemski and Marzouk38] in greater generality.)

Lemma 4.3. (essentially Kortchemski and Marzouk [38, Proposition 5.6]) Suppose that ${\textbf{p}}$ is a probability sequence with mean $\mu({\textbf{p}})=1$ and variance $\sigma^2({\textbf{p}})\in(0,\infty)$ and that ${\textbf{p}}^0$ is a probability sequence with finite mean $\mu({\textbf{p}}^0)$ . Let ${\mathcal T}_{n,(1)},\dots,{\mathcal T}_{n,({d(o)})}$ be the branches of ${\mathcal T}^{{{\textbf{p}},{\textbf{p}}^0}}_n$ arranged in decreasing order. Then

(42) \begin{align} |{\mathcal T}_{n,(1)}|=n-O_{{p}}(1).\end{align}

Proof. Let $D_n=d(o)$ and condition on $D_n=m$ , for a fixed m. Then, by Lemmas 4.1 and 3.1, (42) holds. In other words, for every $\varepsilon>0$ , there exists $C_{m,\varepsilon}$ such that

(43) \begin{align} {\mathbb P{}}\bigl({n-|{\mathcal T}_{n,(1)}|>C_{m,\varepsilon}\mid D_n=m}\bigr) <\varepsilon.\end{align}

By Lemma 4.2, $D_n\overset{\textrm{d}}{\longrightarrow}\widetilde D$ , and thus $(D_n)_n$ is tight, i.e., $O_{\textrm{p}}(1)$ , so there exists M such that ${\mathbb P{}}(D_n>M)<\varepsilon$ for all n. Consequently, if $C_{\varepsilon}\,:\!=\,\max_{m\leqslant M}C_{m,\varepsilon}$ ,

(44) \begin{align} {\mathbb P{}}\bigl({n-|{\mathcal T}_{n,(1)}|>C_{\varepsilon}}\bigr)&={\mathbb E{}} {\mathbb P{}}\bigl({n-|{\mathcal T}_{n,(1)}|>C_{\varepsilon}\mid D_n}\bigr)\leqslant \varepsilon + {\mathbb P{}}(D_n>M)\notag\\&\leqslant 2\varepsilon,\end{align}

which completes the proof.

Lemmas 4.1–4.3 make it possible to transfer many results that are known for simply generated trees (conditioned Galton–Watson trees) to the modified version. See Section 9 for a few examples.

We sketch a proof. First, from (38) and Fatou’s lemma (for sums),

(45) \begin{align} \liminf_{{{n\to\infty}}}n^{3/2} {\mathbb P{}}\bigl({|{\mathcal T}^{{\textbf{p},\textbf{p}^0}}|=n}\bigr)&\geqslant\sum_{k=0}^\infty \liminf_{{{n\to\infty}}}n^{3/2} {\mathbb P{}}\bigl({|{\mathcal T}^{{\textbf{p},\textbf{p}^0}}|=n \text{ and } D=k}\bigr)\notag\\&=\sum_{k=0}^\infty c k{p}^0_k=\infty.\end{align}

In other words, $ n^{3/2} {\mathbb P{}}\bigl({|{\mathcal T}^{{\textbf{p},\textbf{p}^0}}|=n}\bigr)\to \infty$ . Then, (38) and (45) yield, for any fixed $k\geqslant0$ ,

(46) \begin{align} {\mathbb P{}}\bigl({D =k \mid|{\mathcal T}^{{\textbf{p},\textbf{p}^0}}|=n}\bigr)&=\frac{ {\mathbb P{}}\bigl({|{\mathcal T}^{{\textbf{p},\textbf{p}^0}}|=n\text{ and } D=k}\bigr)}{ {\mathbb P{}}\bigl({|{\mathcal T}^{{\textbf{p},\textbf{p}^0}}|=n}\bigr)} \to 0.\end{align}

This proves our claim.

5. Unrooted simply generated trees

We make definitions corresponding to Section 3 for unrooted trees. In this case, we consider labelled trees, so that we can distinguish the vertices. (This is not needed for ordered trees, since their vertices can be labelled canonically as described in Section 2.) Of course, we may then ignore the labelling when we want.

Let $(w_k)_0^\infty$ be a given sequence of non-negative weights, with $w_1>0$ and $w_k>0$ for some $k\geqslant3$ . (The weight $w_0$ is needed only for the trivial case $n=1$ and might be ignored. We may take $w_0=0$ without essential loss of generality.)

For any labelled tree $T\in\mathfrak{L}_n$ , now define the weight of T as

(47) \begin{align} w(T)\,:\!=\,\prod_{v\in T} w_{d(v)}.\end{align}

Given $n\geqslant1$ , we define the random unrooted simply generated tree ${\mathcal T}^\circ_{n}={\mathcal T}^{\textbf{w},\circ}_{n}$ as a labelled tree in ${\mathcal L}_n$ , chosen randomly with probability proportional to the weight (47). (We consider only n such that at least one tree of positive weight exists.)

In the following sections, we give three (related but different) relations with the more well-known rooted simply generated trees.

6. Mark a vertex

Let ${\mathcal T}^{\textbf{w},\circ}_{n}$ be a random unrooted simply generated tree as in Section 5 and mark one of its n vertices, chosen uniformly at random. Regard the marked vertex as a root and denote the resulting rooted tree by ${\mathcal T}^{\textbf{w},\bullet}_n$ .

Thus, ${\mathcal T}^{\textbf{w},\bullet}_n$ is a random unordered rooted tree, where an unordered rooted tree T has probability proportional to its weight given by (47).

We make ${\mathcal T}^{\textbf{w},\bullet}_n$ ordered by ordering the children of each vertex uniformly at random; denote the resulting random labelled ordered rooted tree by ${\mathcal T}^{\textbf{w},*}_n$ . Since each vertex v has $d^+(v)!$ possible orders, the probability that ${\mathcal T}^{\textbf{w},*}_n$ equals a given ordered tree T is proportional to the weight

(48) \begin{align} w^{*}(T) \,:\!=\,\frac{w(T)}{\prod_{v\in T}d^+(v)!}=\frac{w_{d(o)}}{d(o)!}\prod_{v\neq o}\frac{w_{d(v)}}{d^+(v)!}=\frac{w_{d(o)}}{d(o)!}\prod_{v\neq o}\frac{w_{d^+(v)+1}}{d^+(v)!}.\end{align}

The tree ${\mathcal T}^{\textbf{w},*}_n$ is constructed as a labelled tree, but each ordered rooted tree $T\in\mathfrak{T}_n$ has the same number $n!$ of labellings, and they have the same weight (48) and thus appear with the same probability. Hence, we may forget the labelling and regard ${\mathcal T}^{\textbf{w},*}_n$ as a random ordered tree in $\mathfrak{T}_n$ , with probabilities proportional to the weight (48). This is the same as the weight (31) with

(49) \begin{align}\phi_{k}&\,:\!=\,\frac{w_{k+1}}{k!},\qquad k\geqslant0,\end{align}

(50) \begin{align} \phi^0_{k}&\,:\!=\,\frac{w_{k}}{k!},\qquad k\geqslant0.\end{align}

Thus, ${\mathcal T}^{\textbf{w},*}_n={\mathcal T}^{{\boldsymbol{\phi},\boldsymbol{\phi}^0}}_n$ , the modified simply generated tree defined in Section 4.

We recover ${\mathcal T}^{\textbf{w},\circ}_{n}$ from ${\mathcal T}^{*}_n={\mathcal T}^{{\boldsymbol{\phi},\boldsymbol{\phi}^0}}_n$ by ignoring the root (and adding a uniformly random labelling). This yields thus a method to construct ${\mathcal T}^{\textbf{w},\circ}_{n}$ .

Example 6.1. Marckert and Panholzer [Reference Marckert and Panholzer41] studied uniformly random non-crossing trees of a given size n and found that if they are regarded as ordered rooted trees, then they have the same distribution as the conditioned modified Galton–Watson tree ${\mathcal T}^{{\textbf{p},\textbf{p}^0}}_n$ , where

(51) \begin{align} p_k&=4(k+1)3^{-k-2}, k\geqslant0,\end{align}

(52) \begin{align} {p}^0_k&=2\cdot3^{-k}, k\geqslant1.\end{align}

These weights are equivalent to $\phi_k=k+1$ and $\phi^0_k=1$ , which are given by (49)–(50) with $w_k=k!$ . We may thus reformulate the result by Marckert and Panholzer [Reference Marckert and Panholzer41] as: A uniformly random non-crossing tree is the same as a random unrooted simply generated tree with weights $w_k=k!$ .

More generally, Kortchemski and Marzouk [Reference Kortchemski and Marzouk38] studied simply generated non-crossing trees, which are random non-crossing trees with probability proportional to the weight (47) for some weight sequence $\textbf{w}=(w_k)_k$ , and showed that they (under a condition) are equivalent to conditioned modified Galton–Watson trees. In fact, for any weight sequence $\textbf{w}$ , the proofs in [Reference Marckert and Panholzer41, in particular Lemma 2] and [Reference Kortchemski and Marzouk38, in particular Proposition 2.1] show that the simply generated non-crossing tree, regarded as an ordered rooted tree, is the same as ${\mathcal T}^{{\boldsymbol{\phi},\boldsymbol{\phi}^0}}_n$ with

(53) \begin{align} \phi_k&\,:\!=\,(k+1)w_{k+1},\quad k\geqslant0,\end{align}

(54) \begin{align} \phi^0_k&\,:\!=\,w_k,\quad k\geqslant0.\end{align}

Thus, comparing with (49)–(50), it follows that the simply generated non-crossing tree is an unordered simply generated tree, with weight sequence $\overline{w}_k\,:\!=\,w_kk!$ .

Note that non-crossing trees are naturally defined as unrooted trees. A root is introduced in [Reference Kortchemski and Marzouk38, Reference Marckert and Panholzer41] for the analysis, which as said above makes the trees conditioned modified Galton–Watson trees (or, more generally, modified simply generated trees). This is precisely the marking of an unrooted simply generated tree discussed in the present section.

7. Mark an edge

In the random unrooted tree ${\mathcal T}^{\textbf{w},\circ}_{n}$ , mark a (uniformly) random edge, and give it a direction; i.e., mark two adjacent vertices, say $o_+$ and $o_-$ . Since each tree ${\mathcal T}^{\textbf{w},\circ}_{n}$ has the same number $n-1$ of edges, the resulting marked tree ${\mathcal T}^{\textbf{w},\bullet\bullet}_n$ is distributed over all labelled trees on [n] with a marked and directed edge with probabilities proportional to the weight (47).

Now ignore the marked edge, and regard the tree ${\mathcal T}^{\textbf{w},\bullet\bullet}_n$ as two rooted trees ${\mathcal T}_{n,1}$ and ${\mathcal T}_{n,2}$ with roots $o_+$ and $o_-$ , respectively. Furthermore, order randomly the children of each vertex in each of these rooted trees; this makes ${\mathcal T}_{n,1}$ and ${\mathcal T}_{n,2}$ a pair of ordered trees, and each pair $(T_+,T_-)$ of labelled ordered rooted trees with $|T_+|+|T_-|=n$ and the labels $1,\dots,n$ appears with probability proportional to

(55) \begin{align} \widehat{w}(T_+,T_-)&\,:\!=\,\widehat{w}(T_+)\widehat{w}(T_-).\end{align}

where, for a rooted tree T,

(56) \begin{align}\widehat{w}(T)&\,:\!=\,\prod_{v\in T}\frac{w_{d^+(v)+1}}{d^+(v)!}.\end{align}

Using again the definition (49), we have by (26),

(57) \begin{align} \widehat{w}(T)=\prod_{v\in T}\phi_{d^+(v)}=\phi(T).\end{align}

Moreover, since we now have ordered rooted trees, the vertices are distinguishable, and each pair $(T_+,T_-)$ of ordered trees with $|T_+|+|T_-|=n$ has the same number $n!$ of labellings. Hence, we may ignore the labelling and regard the marked tree ${\mathcal T}^{\textbf{w},\bullet\bullet}_n$ as a pair of ordered trees $({\mathcal T}_{n,1},{\mathcal T}_{n,2})$ with $|{\mathcal T}_{n,1}|+|{\mathcal T}_{n,2}|=n$ and probabilities proportional to the weight given by (55) and (57). This means that ${\mathcal T}_{n,1}$ and ${\mathcal T}_{n,2}$ , conditioned on their sizes, are two independent random rooted simply generated trees, with the weight sequence $\boldsymbol{\phi}$ given by (49); in other words, $({\mathcal T}_{n,1},{\mathcal T}_{n,2})$ is a simply generated forest ${\mathcal T}^{{\boldsymbol{\phi}}}_{n,2}$ .

Consequently, we can construct the random unrooted simply generated tree ${\mathcal T}^{\textbf{w},\circ}_{n}$ by taking two random rooted simply generated trees ${\mathcal T}_{n,1}$ and ${\mathcal T}_{n,2}$ constructed in this way, with the right distribution of their sizes, and joining their roots.

Note that $|{\mathcal T}_{n,1}|=n-|{\mathcal T}_{n,2}|$ is random, with a distribution given by the construction above. More precisely, if $a_n$ is the total weight (57) summed over all ordered trees of order n, then

(58) \begin{align} {\mathbb P{}}\bigl({|{\mathcal T}_{n,1}|=m}\bigr)=\frac{a_ma_{n-m}}{\sum_{k=1}^{n-1}a_ka_{n-k}}.\end{align}

If $\phi_2=0$ , we can instead take any $k>2$ with $\phi_k>0$ , and take a random rooted simply generated tree ${\mathcal T}^{{\boldsymbol{\phi}}}_{n+k-1}$ of order $n+k-1$ , conditioned on the event that the root degree is k, and the $k-2$ last children of the root are leaves; we remove the root and these children, and join the first two children.

Consider a Galton–Watson process with offspring distribution $(\phi_k)_0^\infty$ , starting with two individuals, and conditioned on the total progeny being n. This creates a forest with two trees; join their roots to obtain ${\mathcal T}^{\textbf{w},\circ}_{n}$ .

Note that it follows from the arguments above that if we mark the edge joining the two roots, then the marked edge will be distributed uniformly over all edges in the tree ${\mathcal T}^{\textbf{w},\circ}_{n}$ .

In the construction above, ${\mathcal T}_{n,1}$ and ${\mathcal T}_{n,2}$ have the same distribution by symmetry. Now define ${\mathcal T}_{n,+}$ as the largest and ${\mathcal T}_{n,-}$ as the smallest of ${\mathcal T}_{n,1}$ and ${\mathcal T}_{n,2}$ . The next lemma shows that (at least under a weak condition), ${\mathcal T}_{n,-}$ is stochastically bounded, so ${\mathcal T}^{\textbf{w},\circ}_{n}$ is dominated by the subtree ${\mathcal T}_{n,+}$ .

Proof. This is a special case of Lemma 3.1, see also Remark 3.2.

As remarked in Problem 3.4, we conjecture that $|{\mathcal T}_{n,-}|=O_{\textrm{p}}(1)$ also when the generating function has radius of convergence 0, but we leave this as an open problem.

8. Mark a leaf

This differs from the preceding two sections in that we do not recover the distribution of ${\mathcal T}^{\textbf{w},\circ}_{n}$ exactly, but only asymptotically.

Let $N_0(T)$ be the number of leaves in an unrooted tree T. Let $\widehat{{\mathcal T}}^{\textbf{w},\circ}_{n}$ be a random unrooted labelled tree with probability proportional to $N_0(T)w(T)$ ; in other words, we bias the distribution of ${\mathcal T}^{\textbf{w},\circ}_{n}$ by the factor $N_0(T)$ .

Let $\widehat{{\mathcal T}}^{\textbf{w},\bullet}_{n}$ be the random rooted tree obtained by marking a uniformly random leaf in $\widehat{{\mathcal T}}^{\textbf{w},\circ}_{n}$ , regarding the marked leaf as the root. Then, any pair (T, o) with $T\in\mathfrak{L}_n$ and $o\in T$ with $d(o)=1$ will be chosen as $\widehat{{\mathcal T}}^{\textbf{w},\bullet}_{n}$ and its root with probability proportional to the weight (47). We order the children of each vertex at random as in Sections 6 and 7, and obtain an ordered rooted tree $\widehat{{\mathcal T}}^{\textbf{w},*}_{n}$ . Then each tree with root degree 1 appears with probability proportional to (48).

Consequently, if we ignore the labelling, $\widehat{{\mathcal T}}^{\textbf{w},*}_{n}={\mathcal T}^{{\boldsymbol{\phi}},{\boldsymbol{\phi}^0}}_n$ , where $\boldsymbol{\phi}$ is given by (49), and $\phi^0_k\,:\!=\,\delta_{k1}$ (with a Kronecker delta). Equivalently, $\widehat{{\mathcal T}}^{\textbf{w},*}_{n}$ has a root of degree 1, and its single branch is a ${\mathcal T}^{{\boldsymbol{\phi}}}_{n-1}$ .

Conversely, we may obtain $\widehat{{\mathcal T}}^{\textbf{w},\circ}_{n}$ from ${\mathcal T}^{{\boldsymbol{\phi}}}_{n-1}$ by adding a new root under the old one and then adding a random labelling.

As said above, $\widehat{{\mathcal T}}^{\textbf{w},\circ}_{n}$ does not have the distribution of ${\mathcal T}^{\textbf{w},\circ}_{n}$ , but it is not far from it.

Proof. We may construct ${\mathcal T}^{\textbf{w},\circ}_{n}$ as in Section 7 from two random ordered trees ${\mathcal T}_{n,+}$ and ${\mathcal T}_{n,-}$ , where $|{\mathcal T}_{n,-}|=O_{\textrm{p}}(1)$ . Conditioned on $|{\mathcal T}_{n,-}|=\ell$ , for any fixed $\ell$ , we have ${\mathcal T}_{n,+}\overset{\textrm{d}}{=} {\mathcal T}^{{\boldsymbol{\phi}}}_{n-\ell}$ , where $\boldsymbol{\phi}$ is given by (49). Thus, by [Reference Janson30, Theorem 7.11] (see comments there for earlier references to special cases, and to further results), as ${{n\to\infty}}$ , conditioned on $|{\mathcal T}_{n,-}|=\ell$ for any fixed $\ell$ ,

(59) \begin{align} \frac{N_0({\mathcal T}_{n,+})}{n}\overset{\textrm{d}}{=}\frac{N_o({\mathcal T}^{{\boldsymbol{\phi}}}_{n-\ell})}{n}\overset{\textrm{p}}{\longrightarrow} \pi_0,\end{align}

for some constant $\pi_0>0$ . (If $\boldsymbol{\phi}$ is a probability sequence, then $\pi_0=\phi_0.)$ Furthermore, $N_0({\mathcal T}^{\textbf{w},\circ}_{n})=N_0({\mathcal T}_{n,+})+N_0({\mathcal T}_{n,-})=N_0({\mathcal T}_{n,+})+O(1)$ , since $N_0({\mathcal T}_{n,-})\leqslant|{\mathcal T}_{n,-}|=\ell$ . Consequently, still conditioned on $|{\mathcal T}_{n,-}|=\ell$ for any fixed $\ell$ ,

(60) \begin{align} \frac{N_0({\mathcal T}^{\textbf{w},\circ}_{n})}{n}\overset{\textrm{p}}{\longrightarrow} \pi_0>0.\end{align}

Since $ |{\mathcal T}_{n,-}|=O_{\textrm{p}}(1)$ , it follows that (60) holds also unconditionally.

Since ${N_0({\mathcal T}^{\textbf{w},\circ}_{n})}/{n}\leqslant1$ , dominated convergence yields

(61) \begin{align} \frac{{\mathbb E{}} N_0({\mathcal T}^{\textbf{w},\circ}_{n})}{n}={\mathbb E{}} \frac{N_0({\mathcal T}^{\textbf{w},\circ}_{n})}{n}\to \pi_0.\end{align}

By (60) and (61),

(62) \begin{align}\frac{N_0({\mathcal T}^{\textbf{w},\circ}_{n})}{{\mathbb E{}} N_0({\mathcal T}^{\textbf{w},\circ}_{n})}\overset{\textrm{p}}{\longrightarrow}1,\end{align}

and thus, by dominated convergence again,

(63) \begin{align}{\mathbb E{}}\Bigl\lvert{ \frac{N_0({\mathcal T}^{\textbf{w},\circ}_{n})}{{\mathbb E{}} N_0({\mathcal T}^{\textbf{w},\circ}_{n})}-1}\Bigr\rvert\to0.\end{align}

The definition of $\widehat{{\mathcal T}}^{\textbf{w},\circ}_{n}$ by biasing means that for any bounded (or non-negative) function $f\,:\,\mathfrak{L}_n\to\mathbb R$ ,

(64) \begin{align} {\mathbb E{}} f(\widehat{{\mathcal T}}^{\textbf{w},\circ}_{n}) =\frac{{\mathbb E{}}\bigl[{f({\mathcal T}^{\textbf{w},\circ}_{n})N_0({\mathcal T}^{\textbf{w},\circ}_{n})}\bigr]}{{\mathbb E{}} N_0({\mathcal T}^{\textbf{w},\circ}_{n})}.\end{align}

and thus, for any indicator function f,

(65) \begin{align}\bigl\lvert{{\mathbb E{}} f(\widehat{{\mathcal T}}^{\textbf{w},\circ}_{n})-{\mathbb E{}} f({\mathcal T}^{\textbf{w},\circ}_{n})}\bigr\rvert&=\Bigl\lvert{{\mathbb E{}} \Bigl[{f({\mathcal T}^{\textbf{w},\circ}_{n})\Bigl({\frac{{N_0({\mathcal T}^{\textbf{w},\circ}_{n})}}{{\mathbb E{}} N_0({\mathcal T}^{\textbf{w},\circ}_{n})}-1}\Bigr)}\Bigr]}\Bigr\rvert\notag\\&\leqslant{\mathbb E{}} \Bigl\lvert{\frac{{N_0({\mathcal T}^{\textbf{w},\circ}_{n})}}{{\mathbb E{}} N_0({\mathcal T}^{\textbf{w},\circ}_{n})}-1}\Bigr\rvert.\end{align}

Hence, taking the supremum over all $f=\boldsymbol1_{A}$ ,

(66) \begin{align}d_{\textrm{TV}}(\widehat{{\mathcal T}}^{\textbf{w},\circ}_{n},{\mathcal T}^{\textbf{w},\circ}_{n})\leqslant{\mathbb E{}} \Bigl\lvert{\frac{{N_0({\mathcal T}^{\textbf{w},\circ}_{n})}}{{\mathbb E{}} N_0({\mathcal T}^{\textbf{w},\circ}_{n})}-1}\Bigr\rvert,\end{align}

and the result follows by (63).

Lemma 8.2 implies that any result on convergence in probability or distribution for one of $\ {\widehat{{\mathcal T}}^{\textbf{w},\circ}_{n}}$ and ${\mathcal T}^{\textbf{w},\circ}_{n}$ also hold for the other.

9. Profile of conditioned modified Galton–Watson trees

We will use the following extension of Theorem 1.2 to conditioned modified Galton–Watson trees.

Theorem 9.1. Let $L_n$ be the profile of a conditioned modified Galton–Watson tree ${\mathcal T}^{{{\textbf{p}},{\textbf{p}}^0}}_n$ of order n and assume that $\mu({\textbf{p}})=1$ , $\sigma^2({\textbf{p}})<\infty$ and $\mu({\textbf{p}}^0)<\infty$ . Then, as ${{n\to\infty}}$ ,

(67) \begin{align} n^{-1/2} L_n(x n^{1/2}) \overset{{d}}{\longrightarrow} \frac{\sigma}2L_{{\textbf{e}}}\Bigl({\frac{\sigma}2 x}\Bigr),\end{align}

in the space $C[0,\infty]$ , where $L_{{\textbf{e}}}$ is, as in Theorem 1.2, the local time of a standard Brownian excursion ${\textbf{e}}$ .

Proof. Denote the branches of ${\mathcal T}^{{\textbf{p},\textbf{p}^0}}_n$ by ${\mathcal T}_1,\dots,{\mathcal T}_{d(o)}$ and let ${\mathcal T}_0$ be a single root. Then, regarding the branches as rooted trees, which means that their vertices have their depths shifted by 1 from the original tree,

(68) \begin{align}L_n(x)=\sum_{i=1}^{d(o)} L_{{\mathcal T}_i}(x-1)+L_{{\mathcal T}_0}(x).\end{align}

Let ${\mathcal T}_{(1)},\dots,{\mathcal T}_{({d(o)})}$ be the branches arranged in decreasing order. Lemma 4.3 shows that $|{\mathcal T}_{(1)}|=n-O_{\textrm{p}}(1)$ . Hence, (68) and the trivial estimate $0\leqslant L_T(x)\leqslant |T|$ for any T and x yield

(69) \begin{align}\bigl\lvert{L_n(x)-L_{{\mathcal T}_{(1)}}(x-1)}\bigr\rvert \leqslant\sum_{i=2}^{d(o)}|{\mathcal T}_{({i})}|+1=n-|{\mathcal T}_{(1)}|=O_{\textrm{p}}(1).\end{align}

Furthermore, conditioned on $|{\mathcal T}_{(1)}|=n-\ell$ , for any fixed $\ell$ , ${\mathcal T}_{(1)}$ has the same distribution as ${\mathcal T}^{{\textbf{p}}}_{n-\ell}$ , and thus Theorem 1.2 shows that

(70) \begin{align} (n-\ell)^{-1/2} L_{{\mathcal T}_{(1)}}(x (n-\ell)^{1/2})\overset{\textrm{d}}{\longrightarrow} \frac{\sigma}2L_\textbf{e}\Bigl({\frac{\sigma}2 x}\Bigr),\qquad\text{in }C[0,\infty],\end{align}

and it follows easily that, still conditioned,

(71) \begin{align} n^{-1/2} L_{{\mathcal T}_{(1)}}(x n^{1/2} -1)\overset{\textrm{d}}{\longrightarrow} \frac{\sigma}2L_\textbf{e}\Bigl({\frac{\sigma}2 x}\Bigr),\qquad\text{in }C[0,\infty].\end{align}

Together with (69), this shows that for every fixed $\ell$ ,

(72) \begin{align} \bigl({L_n(x)\mid |{\mathcal T}_{(1)}|=n-\ell}\bigr)\overset{\textrm{d}}{\longrightarrow} \frac{\sigma}2L_\textbf{e}\Bigl({\frac{\sigma}2 x}\Bigr),\qquad\text{in }C[0,\infty]. \end{align}

It follows that (72) holds also if we condition on $n-|{\mathcal T}_{(1)}|\leqslant K$ , for any fixed K, and then (67) follows easily from $n-|{\mathcal T}_{(1)}|=O_{\textrm{p}}(1)$ .

Recall that for conditioned Galton–Watson trees ${\mathcal T}^{{\textbf{p}}}_n$ with $\mu(\textbf{p})=1$ and $\sigma^2(\textbf{p})<\infty$ , the width divided by $\sqrt n$ converges in distribution: we have

(73) \begin{align} n^{-1/2} W({\mathcal T}^{{\textbf{p}}}_n) \overset{\textrm{d}}{\longrightarrow} \sigma W,\end{align}

for some random variable W (not depending on $\textbf{p}$ ). In fact, as noted by Drmota and Gittenberger [Reference Drmota and Gittenberger18], this is an immediate consequence of (13) and (1), with

(74) \begin{align} W\,:\!=\,\tfrac12\max_{x\geqslant0} L_\textbf{e}(x).\end{align}

It is also known that all moments converge, see [Reference Drmota and Gittenberger19] (assuming an exponential moment) and [Reference Addario-Berry, Devroye and Janson1] (in general).

The next theorem records that (73) extends to conditioned modified Galton–Watson trees, together with two partial results on moments.

Theorem 9.2. Consider a conditioned modified Galton–Watson tree ${\mathcal T}^{{{\textbf{p}},{\textbf{p}}^0}}_n$ where $\mu({\textbf{p}})=1$ , $\sigma^2({\textbf{p}})<\infty$ and $\sigma^2({\textbf{p}}^0)<\infty$ . Then, as ${{n\to\infty}}$ ,

(75) \begin{align} n^{-1/2} W({\mathcal T}^{{{\textbf{p}},{\textbf{p}}^0}}_n) &\overset{\textrm{d}}{\longrightarrow} \sigma W,\end{align}

(76) \begin{align} n^{-1/2} {\mathbb E{}} W({\mathcal T}^{{{\textbf{p}},{\textbf{p}}^0}}_n) &\to \sigma {\mathbb E{}} W =\sigma\sqrt{\pi/2},\end{align}

(77) \begin{align}{\mathbb E{}} \bigl[{W({\mathcal T}^{{{\textbf{p}},{\textbf{p}}^0}}_n)^{2}}\bigr]& = O(n). \end{align}

Proof. First, (75) follows as in [Reference Drmota and Gittenberger18]: $f\to\sup f$ is a continuous functional on $C[0,\infty]$ , and thus (75) follows from (67), (13) and (74).

We next prove (77). Denote the branches of ${\mathcal T}^{{\textbf{p},\textbf{p}^0}}_n$ by ${\mathcal T}_1,\dots,{\mathcal T}_{d(o)}$ . Assume $n>1$ , then the width is attained above the root, and we have, for every $i\leqslant d(o)$ ,

(78) \begin{align} W({\mathcal T}_i) \leqslant W({\mathcal T}^{{\textbf{p},\textbf{p}^0}}_n) \leqslant \sum_{i=1}^{d(o)}W({\mathcal T}_i).\end{align}

Condition on d(o) and $|{\mathcal T}_1|,\dots,|{\mathcal T}_{d(o)}|$ as in Lemma 4.1. For a random variable X, denote its conditioned $L^2$ norm by

(79) \begin{align}\lVert{{X}}\rVert^{\prime}_2\,:\!=\,\bigl({{\mathbb E{}}\bigl[{ X^2\mid d(o),|{\mathcal T}_1|,\dots,|{\mathcal T}_{d(o)}|}\bigr]}\bigr)^{1/2}.\end{align}

By (78) and Minkowski’s inequality, we have

(80) \begin{align} \lVert{{W({\mathcal T}^{{\textbf{p},\textbf{p}^0}}_n)}}\rVert^{\prime}_2\leqslant \sum_{i=1}^{d(o)}\lVert{W({\mathcal T}_i)}\rVert^{\prime}_2.\end{align}

Furthermore, by Lemma 4.1 and [Reference Addario-Berry, Devroye and Janson1, Corollary 1.3], if $|{\mathcal T}_i|=n_i$ ,

(81) \begin{align} {\mathbb E{}} \bigl({W({\mathcal T}_i)^2 \mid d(o),|{\mathcal T}_1|,\dots,|{\mathcal T}_{d(o)}|}\bigr) = {\mathbb E{}} \bigl[{W({\mathcal T}^{{\textbf{p}}}_{n_i})^2}\bigr]\leqslant C n_i,\end{align}

and thus $\lVert{{W({\mathcal T}_i)}}\rVert^{\prime}_2 \leqslant C n_i^{1/2} = C|{\mathcal T}_i|^{1/2}$ . Hence, by (80),

(82) \begin{align} \lVert{{W({\mathcal T}^{{\textbf{p},\textbf{p}^0}}_n)}}\rVert^{\prime}_2\leqslant \sum_{i=1}^{d(o)} C |{\mathcal T}_i|^{1/2}\end{align}

and thus, by the Cauchy–Schwarz inequality,

(83) \begin{align}&{{\mathbb E{}}\bigl[{ W({\mathcal T}^{{\textbf{p},\textbf{p}^0}}_n)^2\mid d(o),|{\mathcal T}_1|,\dots,|{\mathcal T}_{d(o)}|}\bigr]}=\bigl({ \lVert{{W({\mathcal T}^{{\textbf{p},\textbf{p}^0}}_n)}}\rVert^{\prime}_2}\bigr)^2\notag\\&\qquad\leqslant C\Bigl({ \sum_{i=1}^{d(o)} |{\mathcal T}_i|^{1/2}}\Bigr)^2\leqslant C d(o) \sum_{i=1}^{d(o)} |{\mathcal T}_i|\leqslant C d(o) n.\end{align}

Taking the expectation yields

(84) \begin{align}{\mathbb E{}}\bigl[{ W({\mathcal T}^{{\textbf{p},\textbf{p}^0}}_n)^2}\bigr]\leqslant C n {\mathbb E{}}\bigl[{ d(o)}\bigr].\end{align}

Furthermore, (35) implies that for large n, ${\mathbb E{}}[{ d(o)}] \leqslant 2{\mathbb E{}}{ \widetilde D}$ , where ${\mathbb E{}}\widetilde D<\infty $ by Lemma 4.2. Thus, ${\mathbb E{}} [{d(o)}]\leqslant C$ , and (84) yields

(85) \begin{align}{\mathbb E{}}\bigl[{ W({\mathcal T}^{{\textbf{p},\textbf{p}^0}}_n)^2}\bigr]\leqslant C n,\end{align}

showing (77).

Finally, (77) implies that the variables on the left-hand side of (75) are uniformly integrable [Reference Gut26, Theorem 5.4.2], and thus (76) follows from (75). ${\mathbb E{}} W=\sqrt{\pi/2}$ is well known, see e.g. [Reference Biane, Pitman and Yor12].

10. Distance profile, first step

We now turn to distance profiles. We begin with a weak version of Theorem 1.4; recall the pseudometric $\textsf{d}$ defined in (20), and (21).

Lemma 10.1. Consider a conditioned Galton–Watson tree ${\mathcal T}^{{{\textbf{p}}}}_n$ where $\mu({\textbf{p}})=1$ and $\sigma^2=\sigma^2({\textbf{p}})<\infty$ . Then, as ${{n\to\infty}}$ , for any continuous function with compact support $f\,:\,[0,\infty)\to\mathbb R$ ,

(86) \begin{align}\int_0^\infty n^{-3/2} \Lambda_{{\mathcal T}^{{{\textbf{p}}}}_n}\bigl({x n^{1/2}}\bigr)f(x) \,{d} x\overset{{d}}{\longrightarrow}\int_0^1\int_0^1 f\Bigl({\frac{2}{\sigma}{\textsf{d}}\bigl({s,t;\,{\textbf{e}}}\bigr)}\Bigr)\,{d} s\,{d} t. \end{align}

Proof. The function f is bounded, and also uniformly continuous, i.e., its modulus of continuity $\omega(\delta;\,f)$ , defined in (9), satisfies $\omega(\delta;\,f)\to0$ as $\delta\to0$ . Thus, for any rooted tree $T\in\mathfrak{T}_n$ , noting that $\Lambda_T(x)\leqslant n$ on $[\!-\!1,0]$ and using the analogue of (11) for $\Lambda$ ,

(87) \begin{align}\hskip2em&\hskip-2em\int_0^\infty n^{-3/2} \Lambda_T\bigl({x n^{1/2}}\bigr)f(x) \,\textrm{d} x= n^{-2}\int_0^\infty \Lambda_T({x})f\bigl({n^{-1/2} x}\bigr) \,\textrm{d} x\notag\\&= n^{-2}\int_{-1}^\infty f\bigl({n^{-1/2} x}\bigr)\Lambda_T({x}) \,\textrm{d} x+O\bigl({n^{-1}}\bigr)\notag\\&= n^{-2}\sum_{i=0}^\infty \int_{i-1}^{i+1}f\bigl({n^{-1/2} x}\bigr)\Lambda_T(i)\tau(x-i) \,\textrm{d} x +O\bigl({n^{-1}}\bigr)\notag\\&= n^{-2}\sum_{i=0}^\infty f\bigl({n^{-1/2} i}\bigr)\Lambda_T(i)+ O\bigl({\omega(n^{-1/2};\,f)}\bigr)+O\bigl({n^{-1}}\bigr)\notag\\&= n^{-2}\sum_{v,w\in T} f\bigl({n^{-1/2}\textsf{d}(v,w)}\bigr)+ o(1),\end{align}

where (as throughout the proof) o(1) tends to 0 as ${{n\to\infty}}$ , uniformly in $T\in\mathfrak{T}_n$ . Recall that the contour process $C_T(x)$ of T is a continuous function $C_T\,:\,[0,2n-2]\to[0,\infty)$ that describes the distance from the root to a particle that travels with speed 1 on the ‘outside’ of the tree. (Equivalently, it performs a depth first walk at integer times $0,1,\dots,2n-2$ .) For each vertex $v\neq o$ , the particle travels through the edge leading from v towards the root during two time intervals of unit length (once in each direction). Thus, as is well known,

(88) \begin{align} \int_0^{2n-2} f\bigl({n^{-1/2} C_T(x)}\bigr)\,\textrm{d} x= 2\sum_{v\neq o} f\bigl({n^{-1/2} \textsf{d}(v,o)}\bigr)+O\bigl({n\omega(n^{-1/2};\, f)}\bigr).\end{align}

We will use a bivariate version of this. It is also well known that if v(i) is the vertex visited by the particle at time i, then, for any integers $i,j\in[0,2n-2]$ ,

(89) \begin{align} \textsf{d}\bigl({v(i),v(j)}\bigr)=\textsf{d}(i,j;\,C_T),\end{align}

where the first $\textsf{d}$ is the graph distance in T, and the second is the pseudometric defined by (20) (now on the interval $[0,2n-2]$ ). Hence, the argument yielding (88) also yields

(90) \begin{align}\int_0^{2n-2}\int_0^{2n-2}f\bigl({n^{-1/2}\textsf{d}(x,y;\,C_T)}\bigr)\,\textrm{d} x\,\textrm{d} y =4 \sum_{v,w\neq o} f\bigl({n^{-1/2}\textsf{d}(v,w)}\bigr) + O\bigl({n^2\omega(n^{-1/2};\,f)}\bigr).\end{align}

We use the standard rescaling of the contour process

(91) \begin{align} \widetilde C_T(t)\,:\!=\,n^{-1/2} C_T\bigl({(2n-2)t}\bigr), \qquad t \in [0,1],\end{align}

and note that for any $g\,:\,[0,1] \rightarrow [0, \infty)$ with $g(0) = g(1) = 0$ and $c>0$ ,

(92) \begin{align} \textsf{d}(s,t;\,cg)=c\textsf{d}(s,t;\,g), \qquad s,t \in [0,1].\end{align}

Thus, by (90) and a change of variables,

(93) \begin{align}\hskip4em&\hskip-4em\int_0^{1}\int_0^{1}f\bigl({\textsf{d}(s,t;\,\widetilde C_T)}\bigr)\,\textrm{d} s\,\textrm{d} t\notag\\&=\frac{1}{(2n-2)^2}\int_0^{2n-2}\int_0^{2n-2}f\bigl({n^{-1/2}\textsf{d}(x,y;\,C_T)}\bigr)\,\textrm{d} x\,\textrm{d} y\notag\\&=\frac{1}{(n-1)^2} \sum_{v,w\neq o} f\bigl({n^{-1/2}\textsf{d}(v,w)}\bigr)+ O\bigl({\omega(n^{-1/2};\,f)}\bigr).\notag\\&=\frac{1}{n^2} \sum_{v,w\neq o} f\bigl({n^{-1/2}\textsf{d}(v,w)}\bigr) +o(1).\end{align}

Combining (87) and (93), we find

(94) \begin{align}\hskip2em&\hskip-2em\int_0^\infty n^{-3/2} \Lambda_T\bigl({x n^{1/2}}\bigr)f(x) \,\textrm{d} x=\int_0^{1}\int_0^{1}f\bigl({\textsf{d}(s,t;\,\widetilde C_T)}\bigr)\,\textrm{d} s\,\textrm{d} t+o(1).\end{align}

We apply this to $T={\mathcal T}^{{\textbf{p}}}_n$ and use the result by Aldous [Reference Aldous3, Reference Aldous4],

(95) \begin{align} \widetilde C_{{\mathcal T}^{{\textbf{p}}}_n}(t)\overset{\textrm{d}}{\longrightarrow} \frac{2}{\sigma}\textbf{e}(t),\qquad \text{in $C{[0,1]}$}.\end{align}

The functional $g\to\iint f\bigl({d(s,t;\,g)}\bigr)\,\textrm{d} s\,\textrm{d} t$ is continuous on $C{[0,1]}$ , and the result (86) follows from (94) and (95) by the continuous mapping theorem, using also (92).

11. Distance profile of unrooted trees

We continue with the distance profile, now turning to unrooted simply generated trees for a while. Throughout this section, we assume that $\textbf{w}$ is a weight sequence and that $\boldsymbol{\phi}$ and $\boldsymbol{\phi}^0$ are the weight sequences given by (49) and (50). We assume that the exponential generating function of $\textbf{w}$ has positive radius of convergence; this means that the generating function $\Phi(z)$ in (29) has positive radius of convergence, which in turn implies that there exists a probability weight sequence $\textbf{p}$ equivalent to $\boldsymbol{\phi}$ . We assume furthermore that it is possible to choose $\textbf{p}$ such that $\mu(\textbf{p})=1$ ; $\textbf{p}$ will denote this choice. (For algebraic conditions on $\Phi$ for such a $\textbf{p}$ to exist, see e.g. [Reference Janson30].)

We note that by (49)–(50), $\phi^0_k\leqslant\phi_{k-1}$ , $k\geqslant1$ . Hence, if $p_k=ab^k\phi_k$ , then $\sum_k b^k\phi^0_k<\infty$ , and it is possible to find $a_0>0$ such that $\textbf{p}^0\,:\!=\,\boldsymbol{\phi}^{0\prime}$ given by (32) also is a probability sequence; hence ${\mathcal T}^{{\boldsymbol{\phi},\boldsymbol{\phi}^0}}_n={\mathcal T}^{{\textbf{p},\textbf{p}^0}}_n$ is a modified Galton–Watson tree. Furthermore, ${p}^0_k\leqslant (a_0/a)p_k$ , and thus if $\sigma^2(\textbf{p})<\infty$ , then $\sigma^2(\textbf{p}^0)<\infty$ .

We begin with an unrooted version of Lemma 10.1.

Lemma 11.1. Let ${\textbf{w}}$ , $\boldsymbol{\phi}$ and ${\textbf{p}}$ be as above and assume $\sigma^2\,:\!=\,\sigma^2({\textbf{p}})<\infty$ . Let $\Lambda_n$ be the distance profile of the unrooted simply generated tree ${\mathcal T}^{{\textbf{w}},\circ}_{n}$ . Then, as ${{n\to\infty}}$ , for any continuous function with compact support $f\,:\,[0,\infty)\to\mathbb R$ ,

(96) \begin{align}\int_0^\infty n^{-3/2} \Lambda_{n}\bigl({x n^{1/2}}\bigr)f(x) \,{d} x\overset{{d}}{\longrightarrow}\int_0^1\int_0^1 f\Bigl({\frac{2}{\sigma}{\textsf{d}}\bigl({s,t;\,{\textbf{e}}}\bigr)}\Bigr)\,{d} s\,{d} t. \end{align}

Proof. Consider the leaf-biased random tree $\widehat{{\mathcal T}}^{\textbf{w},\circ}_{n}$ defined in Section 8. By Lemma 8.2, we may assume ${\mathbb P{}}(\widehat{{\mathcal T}}^{\textbf{w},\circ}_{n}\neq{\mathcal T}^{\textbf{w},\circ}_{n})\to0$ and thus it suffices to show (96) with $\Lambda_{\widehat{{\mathcal T}}^{\textbf{w},\circ}_{n}}$ instead of $\Lambda_n$ . If ${\mathcal T}_{n,+}$ denotes the unique branch of $\widehat{{\mathcal T}}^{\textbf{w},\circ}_{n}$ , then, trivially,

(97) \begin{align}0\leqslant \Lambda_{\widehat{{\mathcal T}}^{\textbf{w},\circ}_{n}}(x)-\Lambda_{{\mathcal T}_{n,+}}(x) \leqslant 2n-1, \qquad x\geqslant0,\end{align}

and thus we may further reduce and replace $\Lambda_n$ in (96) by $\Lambda_{{\mathcal T}_{n,+}}$ . As shown in Section 8, ${\mathcal T}_{n,+}\overset{\textrm{d}}{=}{\mathcal T}^{{\boldsymbol{\phi}}}_{n-1}={\mathcal T}^{{\textbf{p}}}_{n-1}$ , and the result now follows from Lemma 10.1, replacing n there by $n-1$ and x by $x=(n/(n-1))^{1/2} x$ , noting that $\sup_x|f(x)-f\bigl({(n/(n-1))^{1/2} x}\bigr)|\to0$ as ${{n\to\infty}}$ .

Theorem 11.2. Let ${\textbf{w}}$ , $\boldsymbol{\phi}$ and ${\textbf{p}}$ be as above, and assume $\sigma^2\,:\!=\,\sigma^2({\textbf{p}})<\infty$ . Let $\Lambda_n$ be the distance profile of the unrooted simply generated tree ${\mathcal T}^{{\textbf{w}},\circ}_{n}$ . Then, as ${{n\to\infty}}$ ,

(98) \begin{align} n^{-3/2}\Lambda_n\bigl({x n^{1/2}}\bigr)\overset{{d}}{\longrightarrow}\frac{\sigma}2\Lambda_{{\textbf{e}}}\Bigl({\frac{\sigma}2 x}\Bigr),\end{align}

in the space $C[0,\infty]$ , where $\Lambda_{{\textbf{e}}}(x)$ is as in Theorem 1.4.

Proof. Let

(99) \begin{align} Y_n(x)\,:\!=\,n^{-3/2}\Lambda_n\bigl({x n^{1/2}}\bigr)=n^{-3/2} \Lambda_{{\mathcal T}^{\textbf{w},\circ}_{n}}\bigl({xn^{1/2}}\bigr).\end{align}

Regard $Y_n$ as a random element of $C[0,\infty]$ . Define also the mapping $\psi\,:\,C[0,\infty]\to\mathcal M([0,\infty))$ , the space of all locally finite Borel measures on $[0,\infty)$ , defined by $\psi(h)\,:\!=\,h(x)\,\textrm{d} x$ ; i.e., for $h\in C[0,\infty]$ and $f\in C[0,\infty)$ with compact support,

(100) \begin{align} \int_0^\infty f(x) \,\textrm{d} \psi(h)\,:\!=\,\int_0^\infty f(x)h(x)\,\textrm{d} x.\end{align}

In other word, $\psi(h)$ has density h.

We give $\mathcal M([0,\infty))$ the vague topology, i.e., $\nu_n\to\nu$ in $\mathcal M([0,\infty))$ if $\int f\,\textrm{d}\nu_n\to\int f\,\textrm{d}\nu$ for every $f\in C[0,\infty)$ with compact support, and note that $\mathcal M([0,\infty))$ is a Polish space, see e.g. [Reference Kallenberg32, Theorem A2.3]. Clearly, the separable Banach space $C[0,\infty]$ is also a Polish space. (Recall that a Polish space has a topology that can be defined by a complete separable metric.) It follows from the definition (100) that $\psi$ is continuous $C[0,\infty]\to\mathcal M([0,\infty))$ . Furthermore, $\psi$ is injective, since the density of a measure is a.e. uniquely determined.

We will use the alternative method of proof in [Reference Drmota17, p. 123–125], and show the following two properties:

It then follows from [Reference Bousquet-Mélou and Janson14, Lemma 7.1] (see also [Reference Drmota17, Theorem 4.17]) that

(101) \begin{align} Y_n\overset{\textrm{d}}{\longrightarrow} Z,\qquad \text{in }C[0,\infty],\end{align}

for some random $Z\in C[0,\infty]$ such that

(102) \begin{align}\psi(Z)\overset{\textrm{d}}{=}\zeta.\end{align}

It will then be easy to complete the proof.

Proof of Claim 1: For $i=1,\dots,n$ , let ${\mathcal T}(i)$ be ${\mathcal T}^{\textbf{w},\circ}_{n}$ rooted at i. By symmetry, all ${\mathcal T}(i)$ have the same distribution; moreover, they equal in distribution $\widehat{{\mathcal T}}^{\textbf{w},\bullet}_{n}$ defined in Section 6 (which has a random root). Hence, if we order each ${\mathcal T}(i)$ randomly, we have by Section 6

(103) \begin{align} {\mathcal T}(i)\overset{\textrm{d}}{=} {\mathcal T}^{\textbf{w},*}_n ={\mathcal T}^{{\boldsymbol{\phi},\boldsymbol{\phi}^0}}_n={\mathcal T}^{{\textbf{p},\textbf{p}^0}}_n.\end{align}

By (16),

(104) \begin{align}Y_n(x)=n^{-3/2} \Lambda_{{\mathcal T}^{\textbf{w},\circ}_{n}}\bigl({xn^{1/2}}\bigr)=\frac{1}{n}\sum_{i=1}^n n^{-1/2} L_{T(i)}\bigl({xn^{1/2}}\bigr).\end{align}

Since the sequence $n^{-1/2} L_{{\mathcal T}^{{\textbf{p},\textbf{p}^0}}_n}\bigl({xn^{1/2}}\bigr)$ converges in $C[0,\infty]$ by Theorem 9.1, it is tight in $C[0,\infty]$ . Furthermore,

(105) \begin{align} \sup_x\bigl\lvert{ n^{-1/2} L_{{\mathcal T}^{{\textbf{p},\textbf{p}^0}}_n}\bigl({xn^{1/2}}\bigr)}\bigr\rvert=n^{-1/2} W({\mathcal T}^{{\textbf{p},\textbf{p}^0}}_n),\end{align}

which are uniformly integrable by (77) in Theorem 9.2. Hence, by Lemma A.2 (which we state and prove in Appendix A) and Remark A.3, (104) and (103) imply that the sequence $Y_n$ is tight in $C[0,\infty]$ , proving Claim 1.

Proof of Claim 2: Let $f\in C[0,\infty)$ have compact support. Then, Lemma 11.1 shows that

(106) \begin{align} \int_0^\infty f(x)\,\textrm{d}\psi(Y_n)&=\int_0^\infty f(x)Y_n(x)\,\textrm{d} x\overset{\textrm{d}}{\longrightarrow}\int_0^1\int_0^1 f\Bigl({\frac{2}{\sigma}\textsf{d}\bigl({s,t;\,\textbf{e}}\bigr)}\Bigr)\,\textrm{d} s\,\textrm{d} t\notag\\&=\int_0^\infty f(x)\,\textrm{d}\zeta(x),\end{align}

where $\zeta$ is the (random) probability measure on $[0,\infty)$ defined as the push-forward of Lebesgue measure on ${[0,1]}\times{[0,1]}$ by the map $(s,t)\to(2/\sigma)\textsf{d}(s,t;\,\textbf{e})$ ; in other words, $\zeta$ is the conditional distribution, given $\textbf{e}$ , of $(2/\sigma)\textsf{d}(U_1,U_2;\,\textbf{e})$ where $U_1$ and $U_2$ are independent uniform $U{[0,1]}$ random variables. This convergence in distribution for each f with compact support is equivalent to convergence in $\mathcal M([0,\infty))$ , see [Reference Kallenberg32, Theorem 16.16]. Thus, $\psi(Y_n)\overset{\textrm{d}}{\longrightarrow} \zeta$ in $\mathcal M([0,\infty))$ , proving Claim 2.

As said above, the claims imply (101)–(102). Thus, by the definition of $\zeta$ above, see (106), for any bounded measurable $f\,:\,[0,\infty)\to\mathbb R$ ,

(107) \begin{align}\int_0^\infty f(x) Z(x)\,\textrm{d} x&=\int_0^\infty f(x)\,\textrm{d}\psi(Z)=\int_0^\infty f(x)\,\textrm{d}\zeta(x)\notag\\&=\int_0^1\int_0^1 f\Bigl({\frac{2}{\sigma}\textsf{d}\bigl({s,t;\,\textbf{e}}\bigr)}\Bigr)\,\textrm{d} s\,\textrm{d} t.\end{align}

Define

(108) \begin{align} \Lambda_\textbf{e}(x)\,:\!=\,\frac{2}{\sigma} Z\Bigl({\frac{2}{\sigma}x}\Bigr).\end{align}

Then, (101) is the same as (98). Furthermore, replacing f(x) by $f(\sigma x/2)$ in (107) yields after a simple change of variables (21) and thus (4). (In particular, $\Lambda_\textbf{e}$ does not depend on the weight sequence $\textbf{w}$ .)

12. Distance profile of rooted trees

Finally, we can prove Theorem 1.4 as a simple consequence of the corresponding result Theorem 11.2 for unrooted trees.

Proof of Theorem 1.4. Let $\boldsymbol{\phi}=\textbf{p}$ , and use (49) to define the weight sequence $\textbf{w}$ . Then, Theorem 11.2 applies to ${\mathcal T}^{\textbf{w},\circ}_{n}$ . Consider, as in the proof of Lemma 11.1, the leaf-biased unrooted random tree $\widehat{{\mathcal T}}^{\textbf{w},\circ}_{n}$ . By Lemma 8.2, Theorem 11.2 holds also for $\widehat{{\mathcal T}}^{\textbf{w},\circ}_{n}$ .

Let again ${\mathcal T}_{n,+}\overset{\textrm{d}}{=} {\mathcal T}^{{\boldsymbol{\phi}}}_{n-1}={\mathcal T}^{{\textbf{p}}}_{n-1}$ denote the unique branch of $\widehat{{\mathcal T}}^{\textbf{w},\circ}_{n}$ . By (97), Theorem 11.2 holds also for ${\mathcal T}_{n,+}$ and thus for ${\mathcal T}^{{\textbf{p}}}_{n-1}$ . Replace n by $n+1$ , then a change of variables shows that Theorem 11.2 holds for ${\mathcal T}^{{\textbf{p}}}_n$ too, which is Theorem 1.4.

As part of the proof, we have shown the corresponding result Theorem 11.2 for unrooted trees. The result also extends easily to conditioned modified Galton–Watson trees, using the method by Marckert and Panholzer [Reference Marckert and Panholzer41] and Kortchemski and Marzouk [Reference Kortchemski and Marzouk38].

Proof. This is a simple consequence of Theorem 1.4 and Lemma 4.3. It follows from (42) that

(109) \begin{align}\sup_x\bigl\lvert{\Lambda_{{\mathcal T}^{{\textbf{p},\textbf{p}^0}}_n}(x)-\Lambda_{{\mathcal T}_{n,(1)}}(x)}\bigr\rvert\leqslant 2n \bigl({n-|{\mathcal T}_{n,(1)}|}\bigr) = O_{\textrm{p}}(n), \end{align}

and thus (3) for ${\mathcal T}^{{\textbf{p},\textbf{p}^0}}_n$ follows from the same result for ${\mathcal T}_{n,(1)}$ , which follows from (42) and Theorem 1.4 by conditioning on $|{\mathcal T}_{n,(1)}|$ . We omit the details.

13. Wiener index

Recall that the Wiener index of a tree T is defined as

(110) \begin{align} {Wie}(T)\,:\!=\,\frac12\sum_{v,w\in T} \textsf{d}(v,w),\end{align}

where $\textsf{d}$ is the graph distance in T. Thus,

(111) \begin{align} {Wie}(T)=\frac12\sum_{i=1}^\infty i \Lambda_T(i)= \frac12\int_{-1}^\infty x\Lambda_T(x)\,\textrm{d} x.\end{align}

Since the integrand x in (111) is unbounded, convergence of the Wiener index does not follow immediately from convergence of the profile, but only a simple extra truncation argument is needed. For a conditioned Galton–Watson tree ${\mathcal T}^{{\textbf{p}}}_n$ , with $\mu(\textbf{p})=1$ and $\sigma^2(\textbf{p})<\infty$ , convergence is more easily proved directly from Aldous’s result (95) [Reference Aldous3, Reference Aldous4], see [Reference Janson28], but as an application of the results above, we show the corresponding result for unrooted trees.

Theorem 13.1. Let ${\textbf{w}}$ and ${\textbf{p}}$ be as in Theorem 11.2. Then

(112) \begin{align} n^{-5/2}{Wie}\bigl({{\mathcal T}^{{\textbf{w}},\circ}_{n}}\bigr)\overset{{d}}{\longrightarrow}\frac{1}{\sigma}\int_0^\infty x\Lambda_{{\textbf{e}}}(x)\,{d} x=\frac{1}{\sigma}\int_0^1\int_0^1 d(s,t;\, {\textbf{e}})\,{d} s\,{d} t.\end{align}

Proof. Let $T\in\mathfrak{T}_n$ , and define a modified version by

(113) \begin{align} {Wie}'(T)\,:\!=\, \frac12\int_0^\infty x\Lambda_T(x)\,{d} x={Wie}(T)+O(n),\end{align}

using (111) and noting that $\Lambda_T(x)\leqslant n$ for $x\in[\!-1,0]$ . It suffices to prove (112) for ${Wie}'({\mathcal T}^{\textbf{w},\circ}_{n})$ . By (113),

(114) \begin{align}n^{-5/2}{Wie}'(T)&=\frac{n^{-5/2}}2\int_0^\infty x\Lambda_T(x)\,\textrm{d} x=\frac12\int_0^\infty xn^{-3/2}\Lambda_T(n^{1/2} x)\,\textrm{d} x. \end{align}

Define a truncated version by, for $m \geq 0$ ,

(115) \begin{align}{Wie}_m(T)\,:\!=\,\frac{n^{5/2}}2\int_0^\infty (x\land m)n^{-3/2}\Lambda_T(n^{1/2} x)\,\textrm{d} x. \end{align}

Then, since the support of $\Lambda_T$ is $[-1,\textrm{diam}(T)+1]$ ,

(116) \begin{align} {\mathbb P{}}\bigl({\textrm{Wie}'({\mathcal T}^{\textbf{w},\circ}_{n})\neq\textrm{Wie}_m({\mathcal T}^{\textbf{w},\circ}_{n})}\bigr)\leqslant{\mathbb P{}}\bigl({\textrm{diam}({\mathcal T}^{\textbf{w},\circ}_{n})> n^{1/2} m-1}\bigr).\end{align}

Since $\textrm{diam}({\mathcal T}^{\textbf{w},\circ}_{n}) = O_{\textrm{p}}(n^{1/2})$ , as an easy consequence of any of the constructions in Sections 6–8 and known results on the height of rooted Galton–Watson trees, see e.g. [Reference Aldous3, Reference Kolchin36] or [Reference Addario-Berry, Devroye and Janson1, Theorem 1.2], it follows that

(117) \begin{align}\sup_n {\mathbb P{}}\bigl({\textrm{Wie}'({\mathcal T}^{\textbf{w},\circ}_{n})\neq\textrm{Wie}_m({\mathcal T}^{\textbf{w},\circ}_{n})}\bigr)\to0,\qquad{\rm{as}}\,m \to \infty.\end{align}

Furthermore, for each fixed m, (115) and Theorem 11.2 imply, as ${{n\to\infty}}$ ,

(118) \begin{align}{n^{-5/2}}\textrm{Wie}_m({\mathcal T}^{\textbf{w},\circ}_{n})&\overset{\textrm{d}}{\longrightarrow}\frac{1}2\int_0^\infty (x\land m)\frac{\sigma}2\Lambda_\textbf{e}\Bigl({\frac{\sigma}2 x}\Bigr)\,\textrm{d} x\notag\\&=\frac{1}\sigma\int_0^\infty \bigl({x\land \frac{\sigma m}2}\bigr)\Lambda_\textbf{e}({x})\,\textrm{d} x.\end{align}

The convergence in (112) follows by (117) and (118), see [11, Theorem 4.2]. The equality in (112) holds by (21).

Of course, the limit in (112) agrees with the limit in [Reference Janson28] for the rooted case.

14. Moments of the distance profile

In this section, we prove the following estimates on moments of the distance profile for a conditioned Galton–Watson tree; we use again the simplified notation $L_n$ and $\Lambda_n$ for the profile and distance profile as in Theorems 1.2 and 1.4. Throughout the section, C and c denote some positive constants that may depend on the offspring distribution $\textbf{p}$ only; $C_r$ denotes constants depending on $\textbf{p}$ and the parameter r only. (As always, these may change from one occurrence to the next.)

1. (i) Let $r\geqslant 1$ be a real number. Then, for all $i, n \geqslant 1$ ,

   (119) \begin{align} {\mathbb E{}}[ \Lambda_n(i)^{r} ]\leqslant C_r n^{3r/2}e^{-ci^{2}/n}.\end{align}

2. (ii) Let $r\geqslant1$ be an integer, and suppose that $\textbf{p}$ has a finite $(r+1)th$ moment:

   (120) \begin{align} \sum_{k} k^{1+r}p_{k} < \infty.\end{align}
   Then, for all $i, n \geqslant 1$ ,
   (121) \begin{align} {\mathbb E{}}[ \Lambda_n(i)^{r} ]\leqslant C_r i^{r} n^{r}.\end{align}
   Furthermore, we may in this case combine (119) and (121) to
   (122) \begin{align} {\mathbb E{}}[ \Lambda_n(i)^{r} ]\leqslant C_r i^r n^{r}e^{-ci^{2}/n},\end{align}
   for all $i, n \geqslant 1$ .

The proof is given later. Note that (122) trivially implies (121) and (changing $C_r$ and c) (119); conversely (119) and (121) imply (122) by considering $i\leqslant n^{1/2}$ and $i>n^{1/2}$ separately.

The special case $r=1$ of (121), i.e.,

(123) \begin{align}{\mathbb E{}} \Lambda_n(i)\leqslant C in,\end{align}

was proved in [16, Theorem 1.3]; note that when $r=1$ , (120) holds automatically by our assumption $\sigma^2(\textbf{p})<\infty$ .

Remark 14.2. The estimates above are natural analogues of estimates for the profile $L_n$ . First, as proved in [Reference Addario-Berry, Devroye and Janson1, Theorem 1.6], under the conditions in Theorem 14.1(i),

(124) \begin{align} {\mathbb E{}}[ L_n(i)^{r} ]&\leqslant C_r n^{r/2}e^{-ci^{2}/n},\end{align}

for all $i,n\geqslant1$ . Second, as proved in [Reference Janson29, Theorem 1.13], under the conditions in Theorem 14.1(ii),

(125) \begin{align}{\mathbb E{}}[L_n(i)^{r}] &\leqslant C_r i^{r},\end{align}

for all $i,n\geqslant1$ . The estimates (124)–(125) are used in our proof of Theorem 14.1.

As an immediate consequence of Theorems 1.4 and 14.1, we obtain the corresponding results for the asymptotic profile $\Lambda_\textbf{e}$ .

Theorem 14.6. For any $r \geqslant 1$ and all $x\geqslant0$ ,

(126) \begin{align} {\mathbb E{}}[\Lambda_{{\textbf{e}}}(x)^r]\leqslant C_r \min\bigl({x^r, e^{-c x^2}}\bigr).\end{align}

Proof. Fix an offspring distribution $\textbf{p}$ with $\mu(\textbf{p})=1$ and all moments finite. (For example, we may choose a well-known example such as ${Po}(1)$ or ${Bi}(2,\frac12)$ .) Let $x\in(0,\infty)$ . Define $i_n\,:\!=\,\lceil{2\sigma^{-1} xn^{1/2}}\rceil$ . Then $i_n/n^{1/2}\to 2x/\sigma$ as ${{n\to\infty}}$ , and (3) implies

(127) \begin{align} n^{-3/2}\Lambda_n(i_n)\overset{\textrm{d}}{\longrightarrow} \frac{\sigma}2\Lambda_\textbf{e}(x).\end{align}

Hence, Fatou’s lemma and (119) yield, for any $r\geqslant1$ ,

(128) \begin{align} {\mathbb E{}}[\Lambda_\textbf{e}(x)^r] \leqslant(2/\sigma)^r \liminf_{{n\to\infty}} {\mathbb E{}}\bigl[{ n^{-3r/2}\Lambda_n(i_n)^r}\bigr]\leqslant C_r e^{-cx^2}.\end{align}

Similarly, Fatou’s lemma and (121) yield

(129) \begin{align} {\mathbb E{}}[\Lambda_\textbf{e}(x)^r]\leqslant C_r x^r\end{align}

for integer $r\geqslant1$ ; equivalently, the $L^r$ norm is estimated by $\lVert{\Lambda_\textbf{e}(x)}\rVert_r \leqslant C_r x$ . This estimate extends to all real $r\geqslant1$ by Lyapounov’s inequality. Hence, (128) and (129) both hold for all $r\geqslant1$ , which yields (126) for $x>0$ .

Finally, the result (126) is trivial for $x=0$ , since $\Lambda_\textbf{e}(0)=0$ a.s.

Remark 14.7. The same argument shows that Theorem 1.2 and (124)–(125) imply

(130) \begin{align} {\mathbb E{}}[L_\textbf{e}(x)^r]\leqslant C_r \min\bigl({x^r, e^{-c x^2}}\bigr),\end{align}

for any $r \geqslant 1$ and all $x\geqslant0$ .

The proof of Theorem 14.1 relies on an invariance property of the law of Galton–Watson trees under random re-rooting proved by Bertoin and Miermont [Reference Bertoin and Miermont9, Proposition 2]. A pointed tree is a pair (T, v), where T is an ordered rooted tree (also called planar rooted tree) and v is a vertex of T. We endow the space of pointed trees with the $\sigma$ -finite measure ${\mathbb P{}}^{\bullet}$ defined by

(131) \begin{align} {\mathbb P{}}^{\bullet}((T,v)) = {\mathbb P{}}( {\mathcal T}^{{\textbf{p}}} = T),\end{align}

where ${\mathcal T}^{{\textbf{p}}}$ is a Galton–Watson tree with offspring distribution $\textbf{p}$ such that $\mu(\textbf{p})=1$ . We let ${\mathbb E{}}^{\bullet}$ denote the expectation under this ‘law’. In particular, the conditional law ${\mathbb P{}}^{\bullet}(\, \cdot \, \mid |T|= n)$ on the space of pointed trees with n vertices is well defined and equals the distribution of $({\mathcal T}^{{\textbf{p}}}_{n}, v)$ where given the conditioned Galton–Watson tree ${\mathcal T}^{{\textbf{p}}}_{n}$ of order n, v is a uniform random vertex of ${\mathcal T}^{{\textbf{p}}}_{n}$ .

Let us now describe the transformation of pointed trees of Bertoin and Miermont [Reference Bertoin and Miermont9, Section 4]. They work with planted planar trees; the base in the planted tree is useful since it implicitly specifies the ordering of the transformed tree. However, we ignore this detail and formulate their transformation for rooted trees; our formulation is easily seen to be equivalent to theirs. For any rooted planar tree T and vertex v of T, let $T_{v}$ be the fringe subtree of T rooted at v, and let $T^{v}$ be the subtree of T obtained by deleting all the strict descendants of v in T. We define a new pointed tree $(\hat{T},{\hat{v}})$ in the following way. If v is the root, we do nothing, and let $\bigl({\hat{T},\hat{v}}\bigr)=(T,v)$ . Otherwise, first remove the edge e(v) between v and its parent pr(v) in T, and instead connect v to the root of T. We then re-root the resulting tree at pr(v) and obtain the new rooted tree $\hat{T}$ , which we point at $\hat v = v$ . Note that $\hat T_{\hat v}=T_v$ , and that $\hat{T}^{\hat{v}}\setminus{\{{\hat v}\}}$ equals $T^v\setminus{\{{v}\}}=T\setminus T_v$ rerooted at pr(v).

Bertoin and Miermont [Reference Bertoin and Miermont9, Proposition 2 and its proof] and its proof establish that this transformation preserves the measure ${\mathbb P{}}^\bullet$ ; this includes the following.

Proof of Theorem 14.1. For a rooted plane tree T and $i \geqslant 0$ an integer, it will be convenient to write $Z_{i}(T)\,:\!=\,L_T(i)$ for the number of vertices at distance i from the root of T. For a vertex v in T, note that the number of vertices at distance $i \geqslant 1$ from v is less than or equal to $Z_{i-1}(\hat{T}^{\hat{v}}) +Z_{i}(T_{v})$ . (Strict inequality may occur because of the extra vertex $\hat v$ added in $\hat{T}^{\hat{v}}$ .) From (16), we thus obtain that if $\Lambda_{T}$ is the distance profile of T defined in (14), then

(132) \begin{align} \Lambda_{T}(i)\leqslant \sum_{v \in T} \bigl({ Z_{i-1}(\hat{T}^{\hat{v}}) + Z_{i}(T_{v})}\bigr),\qquad i \geq 1.\end{align}

If $r\geqslant1$ and $|T|=n$ , then (132) and Jensen’s inequality yield

(133) \begin{align} \bigl({n^{-1}\Lambda_{T}(i)}\bigr)^r\leqslant \frac{1}{n}\sum_{v \in T} \bigl({ Z_{i-1}(\hat{T}^{\hat{v}}) + Z_{i}(T_{v})}\bigr)^r.\end{align}

Consequently, using (131),

(134) \begin{align} {\mathbb E{}}[ \Lambda_n(i)^{r} ] & \leqslant n^{r} {\mathbb E{}}^{\bullet} \bigl[{ \bigl({ Z_{i-1}(\hat{T}^{\hat{v}}) + Z_{i}(T_{v})}\bigr)^{r} \, \big | \, |T| = n }\bigr] \notag \\& \leqslant 2^{r} n^{r} {\mathbb E{}}^{\bullet} \bigl[{ Z_{i-1}(\hat{T}^{\hat{v}})^{r} + Z_{i}(T_{v})^{r} \, \big | \, |T| = n }\bigr].\end{align}

By Proposition 14.8, we have on the one hand that

(135) \begin{align} {\mathbb E{}}^{\bullet} \bigl[{ Z_{i-1}(\hat{T}^{\hat{v}})^{r} \, \big | \, |T| = n }\bigr] & = {\mathbb E{}}^{\bullet} \bigl[{ Z_{i-1}(T^{v})^{r} \, \big | \, |T| = n }\bigr] \notag \\& \leqslant {\mathbb E{}}^{\bullet} \bigl[{ Z_{i-1}(T)^{r} \, \big | \, |T| = n }\bigr] \notag \\& = {\mathbb E{}} [ Z_{i-1}( {\mathcal T}^{{\textbf{p}}}_{n})^{r} ].\end{align}

On the other hand, since $|T| = |T^{v}| + |T_{v}| -1$ , we see from Proposition 14.8 that

(136) \begin{align}{\mathbb E{}}^{\bullet} \bigl[{ Z_{i}(T_{v})^{r} \, \big | \, |T| = n }\bigr] & = \sum_{m=1}^{n} {\mathbb E{}}^{\bullet} \bigl[{ Z_{i}(T_{v})^{r} \, \big | \, |T_{v}| = m, |T^{v}| = n-m+1 }\bigr] \notag \\& \times {\mathbb P{}}^{\bullet} \bigl[{ |T_{v}| = m \, \big | \, |T| = n }\bigr] \notag \\& = \sum_{m=1}^{n} {\mathbb E{}} \bigl[{ Z_{i}({\mathcal T}^{{\textbf{p}}}_{m})^{r} }\bigr] {\mathbb P{}}^{\bullet} \bigl[{ |T_{v}| = m \, \big | \, |T| = n }\bigr] \notag \\& \leqslant \sup_{1 \leqslant m \leqslant n} {\mathbb E{}} \bigl[{ Z_{i}({\mathcal T}^{{\textbf{p}}}_{m})^{r} }\bigr] .\end{align}

Combining (134), (135) and (136), we have that

(137) \begin{align} {\mathbb E{}}[ \Lambda_n(i)^{r} ]&\leqslant 2^{r} n^{r} \bigl({{\mathbb E{}} \bigl[{ Z_{i-1}( {\mathcal T}^{{\textbf{p}}}_{n})^{r} }\bigr]+ \sup_{1 \leqslant m \leqslant n} {\mathbb E{}} \bigl[{ Z_{i}({\mathcal T}^{{\textbf{p}}}_{m})^{r} }\bigr]}\bigr)\notag\\&= 2^{r} n^{r} \bigl({{\mathbb E{}} \bigl[{L_n(i-1)^{r} }\bigr]+ \sup_{1 \leqslant m \leqslant n} {\mathbb E{}} \bigl[{ L_m(i)^{r} }\bigr]}\bigr).\end{align}

Therefore, (119) and (121) follow from (124) and (125), proved in [Reference Addario-Berry, Devroye and Janson1, Theorem 1.6] and [Reference Janson29, Theorem 1.13], respectively.

Finally (122) follows from (119) and (121) as noted above.

15. Hölder continuity

We now discuss Hölder continuity properties of the continuous random functions $L_\textbf{e}$ and $\Lambda_\textbf{e}$ . We begin with $L_\textbf{e}$ , the local time of $\textbf{e}$ . The results are to a large extent known, although we do not know any reference to the form of them stated here; nevertheless, we treat also $L_\textbf{e}$ in detail, as a background to and preparation for the discussion of $\Lambda_\textbf{e}$ below.

It is well known that the local time of Brownian motion at some fixed time a.s. is Hölder continuous of order $\alpha$ for any $\alpha <1/2$ , see [Reference Revuz and Yor49, VI.(1.8)]. It is also known that this Hölder continuity extends by standard arguments to the local times of Brownian bridge and Brownian excursion, and thus (see Theorem 1.2) to $L_\textbf{e}$ . We need a quantitative version of this.

For $\alpha>0$ and an interval $I\subseteq\mathbb R$ , let the Hölder space $\mathcal H_\alpha=\mathcal H_\alpha(I)$ be the space of functions $f\,:\,I\to\mathbb R$ such that

(138) \begin{align} \lVert{{f}}\rVert_{\mathcal H_\alpha}\,:\!=\,\sup\Bigl\{{\frac{|f(x)-f(y)|}{|x-y|^\alpha}\,:\, x,y\in I, x\neq y}\Bigr\}<\infty\end{align}

This is a semi-norm. We may regard $\mathcal H_\alpha$ as a space of functions modulo constants, and then $\lVert{\cdot}\rVert_{\mathcal H_\alpha}$ is a norm and $\mathcal H_\alpha$ a Banach space.

Theorem 15.1. Let $0<\alpha<\frac12$ and let $A<\infty$ . Then

(139) \begin{align} {\mathbb E{}}\, \lVert{{L_\textbf{e}}}\rVert_{\mathcal H_\alpha[0,A]} <\infty.\end{align}

In particular, $L_\textbf{e}\in\mathcal H_\alpha[0,\infty)$ a.s.

Proof. We let $\alpha$ and A be fixed. Constants C below may depend on $\alpha$ and A.

Recall that $L_\textbf{e}(x)=L^x_1$ , where $L^x_t$ is the local time of a (standard) Brownian excursion $\textbf{e}$ . The proof will actually show the result for $L^x_t$ for any fixed $t\in{[0,1]}$ .

We first note that if $\beta$ is a 3-dimensional Bessel process started from 0 (see e.g. [Reference Revuz and Yor49, Section VI.3]), then a standard Brownian excursion $\textbf{e}$ is given by $\textbf{e}(1)=0$ and

(140) \begin{align} \textbf{e}(t) = (1- t) \beta \Bigl({ \frac{t}{1-t} }\Bigr), 0 \leq t < 1;\end{align}

see [Reference Blumenthal13, II.(1.5)]. One then can deduce from [Reference Revuz and Yor49, VI.(3.3)] and an application of the Itô integration by parts formula that $\textbf{e}$ satisfies

(141) \begin{align} \textbf{e}(t)= \int_{0}^{t} \Bigl({ \frac{1}{\textbf{e}(s)} - \frac{\textbf{e}(s)}{1-s} }\Bigr) \,\textrm{d} s + B(t),\qquad 0 \leqslant t \leqslant 1,\end{align}

where B is a standard Brownian motion. In particular, $\textbf{e}$ is a continuous semi-martingale.

We now follow the proof of [Reference Revuz and Yor49, VI.(1.7), see also VI.(1.32)], but avoid localising at the cost of further calculations. Let

(142) \begin{align} \textbf{v}(t)\,:\!=\,\frac{1}{\textbf{e}(t)} - \frac{\textbf{e}(t)}{1-t} ,\qquad 0 \leqslant t <1,\end{align}

so (141) can be written

(143) \begin{align} \textbf{e}(t)= B(t) +\int_{0}^{t} \textbf{v}(s) \,\textrm{d} s.\end{align}

Then, by Tanaka’s formula [Reference Revuz and Yor49, VI.(1.2)], writing $x^+\,:\!=\,x\lor 0$ ,

(144) \begin{align} L^x_t = 2 \bigl({\textbf{e}(t)-x}\bigr)^+- 2 \bigl({-x}\bigr)^+&-2\int_0^t \boldsymbol1\{{{\textbf{e}(s)>x}}\}\,\textrm{d} B(s)\notag\\&\qquad-2\int_0^t \boldsymbol1\{{{\textbf{e}(t)>x}}\} \textbf{v}(s)\,\textrm{d} s.\end{align}

Fix $t\in{[0,1]}$ and denote the four random functions of x on the right-hand side of (144) by $F_1(x),\dots,F_4(x)$ . Trivially, the first two are in $\mathcal H_1$ (= Lipschitz), with norm at most 2. Hence,

(145) \begin{align}\lVert{{F_1}}\rVert_{\mathcal H_\alpha[0,A]}+\lVert{{F_2}}\rVert_{\mathcal H_\alpha[0,A]}\leqslant C.\end{align}

For each x, $F_3(x)$ is a continuous martingale in t. Thus, if $0\leqslant x<y\leqslant A$ , the Burkholder–Davis–Gundy inequality [Reference Revuz and Yor49, IV.(4.1)] and the occupation times formula [Reference Revuz and Yor49, VI.(1.9)] yield, for any $p\geqslant2$ ,

(146) \begin{align} {\mathbb E{}} |F_3(x)-F_3(y)|^p&\leqslant C_p {\mathbb E{}} \Bigl[{\Bigl({\int_0^t \boldsymbol1\{{{x<\textbf{e}(s)\leqslant y}}\}\,\textrm{d} s}\Bigr)^{p/2}}\Bigr]\notag\\&= C_p {\mathbb E{}} \Bigl[{\Bigl({\int_x^y L^z_t\,\textrm{d} z}\Bigr)^{p/2}}\Bigr]\notag\\&= C_p (y-x)^{p/2}{\mathbb E{}} \Bigl[{\Bigl({\frac{1}{y-x}\int_x^y L^z_t\,\textrm{d} z}\Bigr)^{p/2}}\Bigr]\notag\\&\leqslant C_p (y-x)^{p/2}{\mathbb E{}} \Bigl[{{\frac{1}{y-x}\int_x^y \bigl({L^z_t}\bigr)^{p/2}\,\textrm{d} z}}\Bigr].\notag\\&= C_p (y-x)^{p/2}{\frac{1}{y-x}\int_x^y {\mathbb E{}}\bigl({L^z_t}\bigr)^{p/2}\,\textrm{d} z}.\end{align}

We have $L^z_t \leqslant L^z_1=L_\textbf{e}(z)$ , and thus (130) implies ${\mathbb E{}}(L^z_t)^{p/2}\leqslant C_p$ for all z; hence (146) yields

(147) \begin{align} {\mathbb E{}} |F_3(x)-F_3(y)|^p&\leqslant C_p (y-x)^{p/2}.\end{align}

Consequently, $F_3$ satisfies the Kolmogorov continuity criterion, and the version of it stated in [Reference Revuz and Yor49, I.(2.1)] shows that if p is chosen so large that $(p/2-1)/p > \alpha$ , then (147) implies

(148) \begin{align} {\mathbb E{}}\, \lVert{{F_3}}\rVert_{\mathcal H_\alpha[0,A]}\leqslant \bigl({ {\mathbb E{}}\, \lVert{{F_3}}\rVert_{\mathcal H_\alpha[0,A]}^p}\bigr)^{1/p}<\infty.\end{align}

Similarly, using the extension of the occupation times formula in [Reference Revuz and Yor49, VI.(1.15)], if again $0\leqslant x<y\leqslant A$ ,

(149) \begin{align} F_4(y)-F_4(x)&=2\int_0^t\boldsymbol1\{{{x<\textbf{e}(s)\leqslant y}}\}\textbf{v}(s)\,\textrm{d} s\notag\\&=2\int_0^t\boldsymbol1\{{{x<\textbf{e}(s)\leqslant y}}\}\Bigl({\frac{1}{\textbf{e}(s)} - \frac{\textbf{e}(s)}{1-s}}\Bigr)\,\textrm{d} s\notag\\&=2\int_x^y\,\textrm{d} z\int_0^t \Bigl({\frac{1}{z} - \frac{z}{1-s}}\Bigr)\,\textrm{d} L^z_s.\end{align}

We now simplify and assume $0\leqslant t\leqslant \frac12$ . Then (149) implies

(150) \begin{align}\bigl\lvert{ F_4(y)-F_4(x)}\bigr\rvert\leqslant2\int_x^y\Bigl({\frac{1}{z} + 2z}\Bigr)L^z_t \,\textrm{d} z\leqslant C\int_x^y \frac{1}{z} L^z_t \,\textrm{d} z\leqslant C\int_x^y \frac{L_\textbf{e}(z)}{z} \,\textrm{d} z.\end{align}

Let $p'\,:\!=\,1/\alpha$ and let $p\,:\!=\,(1-\alpha)^{-1}>1$ be the conjugate exponent. Then, by (150) and Hölder’s inequality,

(151) \begin{align}\bigl\lvert{ F_4(y)-F_4(x)}\bigr\rvert\leqslant C(y-x)^\alpha\Bigl({\int_x^y \frac{L_\textbf{e}(z)^p}{z^p} \,\textrm{d} z}\Bigr)^{1/p}.\end{align}

Consequently,

(152) \begin{align} \lVert{{F_4}}\rVert_{\mathcal H_\alpha[0,\infty)}\leqslant C\Bigl({\int_0^\infty \frac{L_\textbf{e}(z)^p}{z^p} \,\textrm{d} z}\Bigr)^{1/p}.\end{align}

Thus, using again (130),

(153) \begin{align}{\mathbb E{}}\, \lVert{{F_4}}\rVert_{\mathcal H_\alpha[0,\infty)}^p\leqslant C{\mathbb E{}}{\int_0^\infty \frac{L_\textbf{e}(z)^p}{z^p} \,\textrm{d} z}= C\int_0^\infty \frac{{\mathbb E{}}L_\textbf{e}(z)^p}{z^p} \,\textrm{d} z<\infty\end{align}

and thus

(154) \begin{align} {\mathbb E{}}\, \lVert{{F_4}}\rVert_{\mathcal H_\alpha[0,A]}\leqslant {\mathbb E{}}\, \lVert{{F_4}}\rVert_{\mathcal H_\alpha[0,\infty)}<\infty.\end{align}

Consequently, (144), (145), (148) and (154) yield

(155) \begin{align}{\mathbb E{}}\, \lVert{{L_t^{x}}}\rVert_{\mathcal H_\alpha[0,A]}<\infty\end{align}

for $0\leqslant t\leqslant\frac12$ .

We have for simplicity assumed $t\leqslant\frac12$ . To complete the proof, we note that $\textbf{e}$ is invariant under reflection: $\textbf{e}(1-t)\overset{\textrm{d}}{=}\textbf{e}(t)$ (as processes), and thus $L^x_1-L^x_{1/2}\overset{\textrm{d}}{=} L^x_{1/2}$ (as processes in x). Consequently,

(156) \begin{align}{\mathbb E{}}\, \lVert{{L_\textbf{e}}}\rVert_{\mathcal H_\alpha[0,A]}= {\mathbb E{}}\, \lVert{{L_1^{x}}}\rVert_{\mathcal H_\alpha[0,A]}\leqslant 2 {\mathbb E{}}\, \lVert{{L_{1/2}^{x}}}\rVert_{\mathcal H_\alpha[0,A]} <\infty,\end{align}

showing (139). (The case $\frac12<t<1$ follows similarly.)

Finally, (139) shows that a.s., $L_\textbf{e}\in\mathcal H_\alpha[0,A]$ for every $A>0$ . Moreover, $L_\textbf{e}$ has finite support $[0, \sup\textbf{e}]$ , and thus $L_\textbf{e}\in\mathcal H_\alpha[0,\infty)$ .

We have for simplicity considered a finite interval [0, A] in Theorem 15.1, but the result can easily be extended to $\mathcal H_\alpha[0,\infty)$ .

Theorem 15.2. Let $0<\alpha<\frac12$ . Then

(157) \begin{align} {\mathbb E{}}\, \lVert{{L_\textbf{e}}}\rVert_{\mathcal H_\alpha[0,\infty)} <\infty.\end{align}

Proof. Note that the proof of Theorem 15.1 actually shows ${\mathbb E{}}\,\lVert{{L_\textbf{e}}}\rVert_{\mathcal H_\alpha[0,A]}^p<\infty$ for some $p>1$ , see (148) and (153). Moreover, the same proof applied to the interval $[m,m+1]$ shows that for any $m\geqslant1$ ,

(158) \begin{align} {\mathbb E{}}\,\lVert{{L_\textbf{e}}}\rVert_{\mathcal H_\alpha {[m,m+1]}}^p\leqslant Cm^p,\end{align}

with C independent of m, where the factor $m^p$ comes from (150). Furthermore, $L_\textbf{e}=0$ on $[m,m+1]$ unless $\sup\textbf{e}>m$ . The explicit formula for the distribution function of $\sup\textbf{e}$ , see e.g. [Reference Chung15, Reference Kennedy33] or [Reference Drmota17, p. 114], yields the well-known subgaussian decay

(159) \begin{align}{\mathbb P{}}\bigl({\sup\textbf{e}>x}\bigr)\leqslant e^{-cx^2},\qquad x\geqslant1.\end{align}

(See also [Reference Addario-Berry, Devroye and Janson1] for the corresponding result for heights of conditioned Galton–Watson trees.) Combining (158) and (159) with Hölder’s inequality, we obtain, with $1/q=1-1/p$ ,

(160) \begin{align} {\mathbb E{}}\,\lVert{{L_\textbf{e}}}\rVert_{\mathcal H_\alpha{[m,m+1]}}&\leqslant \Bigl({{\mathbb E{}}\,\lVert{{L_\textbf{e}}}\rVert_{\mathcal H_\alpha{[m,m+1]}}^p}\Bigr)^{1/p}{\mathbb P{}}\bigl({\lVert{{L_\textbf{e}}}\rVert_{\mathcal H_\alpha{[m,m+1]}}\neq0}\bigr)^{1/q}\notag\\&\leqslant Cm {\mathbb P{}}\bigl({\sup\textbf{e}>m}\bigr)^{1/q}\leqslant Cm e^{-cm^2},\end{align}

Finally, it is easy to see that

(161) \begin{align}\lVert{{L_\textbf{e}}}\rVert_{\mathcal H_\alpha[0,\infty)}\leqslant \sum_{m=0}^\infty\lVert{{L_\textbf{e}}}\rVert_{\mathcal H_\alpha{[m,m+1]}}\end{align}

and thus, by (139) and (160),

(162) \begin{align}{\mathbb E{}}\,\lVert{{L_\textbf{e}}}\rVert_{\mathcal H_\alpha[0,\infty)}\leqslant \sum_{m=0}^\infty{\mathbb E{}}\,\lVert{{L_\textbf{e}}}\rVert_{\mathcal H_\alpha {[m,m+1]}}\leqslant C + \sum_{m=1}^\infty Cme^{-cm^2}<\infty.\end{align}

Remark 15.3. The proofs above apply also to the local time of the Brownian bridge. The main differences are that we consider functions on $(-\infty,\infty)$ and that the term $1/\textbf{e}(t)$ in (142) disappears. This shows that, writing ${L}_{\textbf{b}}(x)$ for the local time of the Brownian bridge at time $t=1$ ,

(163) \begin{align} {\mathbb E{}}\,\lVert{{{L}_{\textbf{b}}}}\rVert_{\mathcal H_\alpha{(-\infty,\infty)}} <\infty.\end{align}

By Vervaat [Reference Vervaat51], the Brownian excursion can be constructed from a Brownian bridge by shifts in both time and space, and thus $L_\textbf{e}$ is a (random) shift of ${L}_{\textbf{b}}$ . Consequently, with this coupling,

(164) \begin{align}\lVert{{L_\textbf{e}}}\rVert_{\mathcal H_\alpha[0,\infty)} =\lVert{{{L}_{\textbf{b}}}}\rVert_{\mathcal H_\alpha{(-\infty,\infty)}},\end{align}

and thus (157) and (163) are equivalent. This yields an alternative proof of Theorem 15.2.

We have so far considered $\alpha<1/2$ . For $\alpha\geqslant1/2$ , it is well known that Brownian motion is a.s. not locally Hölder( $\alpha$ ); see e.g. [Reference Revuz and Yor49, I.(2.7)] or [Reference Mörters and Peres43, Section 1.2] for even more precise results. Moreover, the same holds for the local time $\bar{L}^x_t$ of Brownian motion, regarded as a function of the space variable x (for any fixed $t>0$ ); we do not know an explicit reference but this follows easily, for example, from the first Ray–Knight theorem [Reference Revuz and Yor49, XI.(2.2)], which says that if we stop at $\tau_1$ , the hitting time of 1, then the process $(\bar{L}^{1-x}_{\tau_1})$ , $0\leqslant x\leqslant 1$ , is a 2-dimensional squared Bessel process, which (at least away from 0) has the same smoothness as a Brownian motion (since it can be written as the sum of squares of two independent Brownian motions); we omit the details.

Similarly, $L_\textbf{e}=L^x_1$ , which is the local time of a standard Brownian excursion is a.s. not Hölder( $\frac12$ ). One way to see this is that if we stop a Brownian motion when its local time at 0 reaches 1, i.e., at $\tau\,:\!=\,\inf{\{{t\,:\,\bar{L}^0_t=1}\}}$ , then the part of the Brownian motion before time $\tau$ and above any fixed $\delta>0$ is a.s. included in a finite number of excursions, and these are independent, conditioned on the number of them and their lengths. Hence, if the local time of a Brownian excursion were Hölder( $\frac12$ ) with positive probability, then so would $\bar{L}_\tau^x$ , restricted to $x\geqslant\delta$ , be, and then $\bar{L}_t^x$ , $x\geqslant\delta$ , would be Hölder( $\frac12$ ) with positive probability for some rational t, and thus for all $t>0$ by scaling, which contradicts the argument above.

15.1. Distance profile

We next consider the asymptotic distance profile $\Lambda_\textbf{e}$ . We first note that the nice re-rooting invariance property of the standard Brownian excursion $\textbf{e}$ presented in Section 2.3 yields the following corollary of Theorem 15.2.

Corollary 15.4. Let $0<\alpha<\frac12$ . Then

(165) \begin{align} {\mathbb E{}}\, \lVert{{\Lambda_\textbf{e}}}\rVert_{\mathcal H_\alpha[0,\infty)} <\infty.\end{align}

In particular, the random function $\Lambda_\textbf{e}\in\mathcal H_\alpha[0,\infty)$ a.s.

Proof. It follows from the identity (24) that

(166) \begin{align} \lVert{{\Lambda_\textbf{e}}}\rVert_{\mathcal H_\alpha[0,\infty)} \leqslant \int_{0}^{1} \lVert{{ L_{\textbf{e}^{[s]}}} }\rVert_{\mathcal H_\alpha[0,\infty)} \,\textrm{d} s,\end{align}

where for every $s \in{[0,1]}$ , $L_{\textbf{e}^{[s]}}$ denotes the local time of the process $\textbf{e}^{[s]}$ defined in (22), which is distributed as a standard Brownian excursion (23). Hence, Theorem 15.2 yields

(167) \begin{align} {\mathbb E{}}\, \lVert{{\Lambda_\textbf{e}}}\rVert_{\mathcal H_\alpha[0,\infty)} \leqslant \int_{0}^{1} {\mathbb E{}}\, \lVert{{ L_{\textbf{e}^{[s]}}}}\rVert_{\mathcal H_\alpha[0,\infty)} \,\textrm{d} s={\mathbb E{}}\, \lVert{{ L_\textbf{e}}}\rVert_{\mathcal H_\alpha[0,\infty)} <\infty.\end{align}

However, it turns out that the averaging in (24) actually makes $\Lambda_\textbf{e}$ smoother than $L_\textbf{e}$ ; we have the following stronger result, which improves Corollary 15.4.

Theorem 15.5. The asymptotic distance profile $\Lambda_{{\textbf{e}}}\in \mathcal H_\alpha$ for every $\alpha<1$ , a.s. Furthermore, $\Lambda_\textbf{e}$ is a.s. absolutely continuous and with a derivative $\Lambda_{{\textbf{e}}}'$ (defined a.e.) that belongs to $L^p(\mathbb R)$ for every $p<\infty$ .

In order to prove Theorem 15.5, we first prove an estimate of the Fourier transform of $\Lambda_\textbf{e}$ . We choose to define Fourier transforms as, for $f\in L^1(\mathbb R)$ ,

(168) \begin{align}\widehat f(\xi)\,:\!=\,\int_{-\infty}^\infty f(x) e^{\textrm{i} \xi x}\,\textrm{d} x,\qquad -\infty<\xi<\infty.\end{align}

Note that by (21), the Fourier transform $\widehat{\Lambda_\textbf{e}}$ can be written as

(169) \begin{align} \widehat{\Lambda_\textbf{e}}(\xi)\,:\!=\,\int_0^\infty e^{\textrm{i}\xi x}\Lambda_\textbf{e}(x)\,\textrm{d} x=\iint_{s,t\in{[0,1]}} e^{\textrm{i}\xi\textsf{d}(s,t;\,\textbf{e})}\,\textrm{d} s\,\textrm{d} t.\end{align}

Lemma 15.6. There exists a constant C such that

(170) \begin{align} {\mathbb E{}}\,|\widehat{\Lambda_\textbf{e}}(\xi)|^2 \leqslant C\bigl({\xi^{-4}\land 1}\bigr),\qquad -\infty<\xi<\infty. \end{align}

Proof. Note first that the profile $\Lambda_\textbf{e}$ is a (random) non-negative function with integral 1, e.g., by (4). Hence, for every $\xi$ ,

(171) \begin{align} \bigl\lvert{\widehat{\Lambda_\textbf{e}}(\xi)}\bigr\rvert\leqslant\int_0^\infty \bigl\lvert{\Lambda_\textbf{e}(x)}\bigr\rvert\,\textrm{d} x=1.\end{align}

Thus, (170) is trivial for $|\xi|\leqslant1$ . Assume in the remainder of the proof $|\xi|\geqslant1$ .

By (169),

(172) \begin{align} {\mathbb E{}}\,\bigl\lvert{\widehat{\Lambda_\textbf{e}}(\xi)}\bigr\rvert^2&={\mathbb E{}} \idotsint_{t_1,\dots t_4\in{[0,1]}} e^{\textrm{i} \xi (\textsf{d}(t_1,t_2;\,\textbf{e})-\textsf{d}(t_3,t_4;\,\textbf{e}))}\,\textrm{d} t_1\dotsm\,\textrm{d} t_4\notag\\&={\mathbb E{}} e^{\textrm{i} \xi (\textsf{d}(U_1,U_2;\,\textbf{e})-\textsf{d}(U_3,U_4;\,\textbf{e}))},\end{align}

where $U_1,\dots,U_4$ are i.i.d. uniform U(0,1) random variables, independent of $\textbf{e}$ .

Recall from Section 2.3 that the Brownian excursion $\textbf{e}$ defines the continuum random tree $T_{\textbf{e}}$ , with a quotient map $\rho_{\textbf{e}}\,:\,{[0,1]}\to T_{\textbf{e}}$ . Let $\overline{U}_i\,:\!=\,\rho_\textbf{e}(U_i)$ , $i=1,\dots,4$ , be the points in $T_{\textbf{e}}$ corresponding to $U_i$ ; these are i.i.d. uniformly random points in $T_{\textbf{e}}$ , and $\textsf{d}(U_i,U_j;\,\textbf{e})$ equals the distance $\textsf{d}(\overline{U}_i,\overline{U}_j)$ between $\overline{U}_i$ and $\overline{U}_j$ in the real tree $T_{\textbf{e}}$ . Then (172) becomes

(173) \begin{align} {\mathbb E{}}\,\bigl\lvert{\widehat{\Lambda_\textbf{e}}(\xi)}\bigr\rvert^2&={\mathbb E{}} e^{\textrm{i} \xi (\textsf{d}(\overline{U}_1,\overline{U}_2)-\textsf{d}(\overline{U}_3,\overline{U}_4))}.\end{align}

Furthermore, we can simplify the calculations by rerooting $T_{\textbf{e}}$ at $\overline{U}_4$ ; this preserves the distribution, see (23), and $\overline{U}_1,\dots,\overline{U}_3$ are still independent uniformly random points in the tree; hence, we also have, with $o=\rho_\textbf{e}(0)$ the root of $T_{\textbf{e}}$ .

(174) \begin{align} {\mathbb E{}}\,\bigl\lvert{\widehat{\Lambda_\textbf{e}}(\xi)}\bigr\rvert^2&={\mathbb E{}} e^{\textrm{i} \xi (\textsf{d}(\overline{U}_1,\overline{U}_2)-\textsf{d}(\overline{U}_3,o))}.\end{align}

To calculate (174), it suffices to consider the (real) subtree of $T_{\textbf{e}}$ spanned by the root o and $\overline{U}_{1},\dots,\overline{U}_3$ . [Reference Aldous4, Corollary 22] showed that the distribution of the random real tree $\widetilde{{\mathcal T}}_k$ spanned in this way by $k\geqslant1$ i.i.d. uniform random points $\overline{U}_1,\dots,\overline{U}_k\in T_{\textbf{e}}$ can be described as follows. Let the shape $\tau$ of $\widetilde{{\mathcal T}}_k$ be the tree regarded as a combinatorial tree, i.e., ignoring the edge lengths. Then $\tau$ is a.s. a rooted binary tree with k leaves, labelled by $1,\dots,k$ (corresponding to $\overline{U}_1,\dots,\overline{U}_k$ ). Let $\bar{\mathfrak{T}}^{k}$ be the set of possible shapes, i.e., the set of binary trees with k labelled leaves. Then $|\bar{\mathfrak{T}}^{k}|=(2k-3)!!$ . Each shape $\tau$ has $k-1$ internal vertices, and thus 2k vertices and $2k-1$ edges. For each shape $\tau$ , label the edges $1,\dots,2k-1$ in some order, and for each tree $\widetilde{{\mathcal T}}_k$ with shape $\tau$ , let $\ell_1,\dots,\ell_{2k-1}$ be the corresponding edge lengths. Then $\widetilde{{\mathcal T}}_k$ is described by $(\tau,\ell_1,\dots,\ell_k)$ , and the distribution of $(\tau,\ell_1,\dots,\ell_k)$ is given by the density, for $\tau\in\bar{\mathfrak{T}}^{k}$ and $\ell_i>0$ ,

(175) \begin{align}2^{2k} s e^{-2s^2}\,\textrm{d}\ell_1\dotsm\,\textrm{d}\ell_{2k-1},\qquad \text{with } s\,:\!=\,\ell_1+\dotsm\ell_{2k-1}.\end{align}

Recall that our normalisation differs from [Reference Aldous4], where $T_{2\textbf{e}}$ is used instead of $T_{\textbf{e}}$ , and thus all edges are twice as long; hence (175) is obtained from [4, (33)] by a trivial change of variables.

For $k=3$ we have $|\bar{\mathfrak{T}}^{3}|=3!!=3$ different shapes. It is easy to verify that in each of them, we may label the five edges such that $\textsf{d}(\overline{U}_1,\overline{U}_2)-\textsf{d}(\overline{U}_3,o)=\ell_1+\ell_2-\ell_3-\ell_4$ . Consequently, (174) and (175) yield, with s as in (175),

(176) \begin{align} {\mathbb E{}}\,\bigl\lvert{\widehat{\Lambda_\textbf{e}}(\xi)}\bigr\rvert^2&= 3\int_{\mathbb R_+^5} e^{\textrm{i}\xi(\ell_1+\ell_2-\ell_3-\ell_4)} 2^6 s e^{-2s^2}\,\textrm{d}\ell_1\dotsm\,\textrm{d}\ell_5\notag\\&=192 \int_0^\infty \psi(s) s e^{-2s^2}\,\textrm{d} s,\end{align}

where, with $\Sigma_5(s)\,:\!=\,{\{{(\ell_1,\dots,\ell_5)\,:\,\ell_i>0 \text{ and } \ell_1+\dots+\ell_5=s}\}}$ ,

(177) \begin{align} \psi(s)\,:\!=\, \int_{\Sigma_5(s)} e^{\textrm{i}\xi(\ell_1+\ell_2-\ell_3-\ell_4)}\,\textrm{d}\ell_1\dotsm\,\textrm{d}\ell_4.\end{align}

We can easily find the Laplace transform of $\psi(s)$ : for $\lambda>0$ ,

(178) \begin{align} \int_0^\infty e^{-\lambda s}\psi(s)\,\textrm{d} s &= \int_{\mathbb R_+^5} e^{-\lambda(\ell_1+\dots+\ell_5)}e^{\textrm{i}\xi(\ell_1+\ell_2-\ell_3-\ell_4)}\,\textrm{d}\ell_1\dotsm\,\textrm{d}\ell_5\notag\\&=\frac{1}{(\lambda-\textrm{i}\xi)^2(\lambda+\textrm{i}\xi)^2\lambda}.\end{align}

A partial fraction expansion of (178) yields,

(179) \begin{align} \int_0^\infty e^{-\lambda s}\psi(s)\,\textrm{d} s &=\frac{a(\xi)}{\lambda}+\frac{b_+(\xi)}{\lambda-\textrm{i}\xi}+\frac{b_-(\xi)}{\lambda+\textrm{i}\xi}+\frac{c_+(\xi)}{(\lambda-\textrm{i}\xi)^2}+\frac{c_-(\xi)}{(\lambda+\textrm{i}\xi)^2},\end{align}

for some coefficients $a(\xi),b_\pm(\xi),c_\pm(\xi)$ . It is easy to calculate these, but it suffices to note that (178) is homogeneous of degree $-5$ in $(\lambda,\xi)$ , and thus $a(\xi)=a\xi^{-4}$ , $b_\pm(\xi)=b_\pm\xi^{-4}$ and $c_\pm(\xi)=c_\pm\xi^{-3}$ for some complex $a,b_\pm,c_\pm$ . By inverting the Laplace transform (179), we find

(180) \begin{align} \psi(s)=a\xi^{-4} + b_+\xi^{-4}e^{\textrm{i}\xi s}+ b_-\xi^{-4}e^{-\textrm{i}\xi s}+ c_+\xi^{-3} se^{\textrm{i}\xi s} + c_-\xi^{-3} se^{-\textrm{i}\xi s}.\end{align}

Finally, we substitute (180) in (176). We define

(181) \begin{align}h_m(s)\,:\!=\,s^m e^{-2s^2}\boldsymbol1\{{{s>0}}\},\end{align}

and obtain, recalling (168),

(182) \begin{align} {\mathbb E{}}\,\bigl\lvert{\widehat{\Lambda_\textbf{e}}(s)}\bigr\rvert^2=192\xi^{-4}\bigl({a\hat h_1(0)+b_+\hat h_1(\xi)+b_-\hat h_1(-\xi)+c_+\xi\hat h_2(\xi)+c_-\xi\hat h_2(-\xi)}\bigr).\end{align}

We have $\hat h_1(\xi)=O(1)$ since $h_1$ is integrable, and $\xi\hat h_2(\xi)=\textrm{i}\widehat{h_2'}(\xi)=O(1)$ since $h_2$ is differentiable with integrable derivative. Consequently, the result (170) follows.

Remark 15.7. It is easy to see that in the proof above, $a=1$ and $\hat h_1(0)=\int_0^\infty s e^{-2s^2}\,\textrm{d} s=1/4$ ; moreover, $\hat h_1(\xi)=O(\xi^{-2})$ and $\hat h_2(\xi)=O(\xi^{-3})$ . Hence, the proof yields

(183) \begin{align} {\mathbb E{}}\,\bigl\lvert{\widehat{\Lambda_\textbf{e}}(\xi)}\bigr\rvert^2=48\xi^{-4} + O\bigl({\xi^{-6}}\bigr).\end{align}

Thus, the estimate in Lemma 15.6 is sharp.

Remark 15.8. The result (170) (or (183)) can also be obtained in the same way from (172). However, for $k=4$ we have $|\bar{\mathfrak{T}}^{4}|=5!!=15$ different shapes that we have to consider, and they yield several different terms in (176). The leading term in the partial fraction expansion (179) will now be $a\xi^{-4}/\lambda^3$ .

Proof of Theorem 15.5. Let $0<\alpha<3/2$ . Then, by Lemma 15.6,

(184) \begin{align} {\mathbb E{}} \int_{-\infty}^\infty \bigl\lvert{|\xi|^{\alpha}\widehat{\Lambda_\textbf{e}}(\xi)}\bigr\rvert^2\,\textrm{d}\xi&=2 \int_0^\infty |\xi|^{2\alpha}{\mathbb E{}}\,\bigl\lvert{\widehat{\Lambda_\textbf{e}}(\xi)}\bigr\rvert^2\,\textrm{d}\xi\notag\\&\leqslant C \int_0^1 \xi^{2\alpha}\,\textrm{d}\xi+ C\int_1^\infty \xi^{2\alpha-4}\,\textrm{d}\xi <\infty.\end{align}

Hence, a.s., the function $\Lambda_\textbf{e}$ belongs to the (generalised) Sobolev space ${\mathcal L}^2_{{\alpha}}$ (also called potential space; many different notations exist), which is defined as the space of all functions $f\in L^2(\mathbb R)$ such that

(185) \begin{align}\lVert{f}\rVert_{{\mathcal L}^2_{\alpha}}^2\,:\!=\,\lVert{f}\rVert_{L^2}^2+\int_{-\infty}^\infty \bigl\lvert{|\xi|^{\alpha}\widehat{\Lambda_\textbf{e}}(\xi)}\bigr\rvert^2\,\textrm{d}\xi<\infty,\end{align}

see, e.g. [Reference Stein50, Chapter V] or [Reference Bergh and Löfström8, Chapter 6]. Furthermore, this Sobolev space equals the Besov space $B^{2,2}_{{\alpha}}$ , see again [Reference Stein50, Chapter V (there denoted $\Lambda_\alpha^{2,2}$ )] or [Reference Bergh and Löfström8, Chapter 6.] $B^{2,2}_{\alpha}$ is an $L^2$ version of $\mathcal H_\alpha$ , and for $0<\alpha<1$ , $B^{2,2}_{{\alpha}}$ may be defined as the set of functions in $L^2$ such that

(186) \begin{align}\lVert{f}\rVert_{B^{2,2}_{\alpha}}^2\,:\!=\,\lVert{f}\rVert_{L^2}^2+ \int_0^\infty \left({\frac{\lVert{f(\cdot+u)-f(\cdot)}\rVert_{L^2(\mathbb R)}}{u^\alpha}}\right)^2 \frac{\,\textrm{d} u}{u} <\infty;\end{align}

the equivalence (within constant factors) of (186) with (185) follows easily from the Plancherel formula. (For a much more general result, see [Reference Bergh and Löfström8, Theorem 6.2.5] or [Reference Stein50, (V.60)].) For $1<\alpha<2$ , (177)) fails, but $B^{2,2}_{\alpha}={\{{f\in L^2\,:\,f'\in B^{2,2}_{{\alpha-1}}}\}}$ , which again equals ${\mathcal L}^2_{\alpha}$ .

If $1\leqslant\alpha<3/2$ , it follows that the derivative (in distribution sense) $\Lambda_\textbf{e}'\in{\mathcal L}^2_{{\alpha-1}}$ . By the Sobolev (or Besov) embedding theorem [Reference Bergh and Löfström8, Theorem 6.5.1], see also [Reference Stein50, Theorem V.2] and, for a simplified version, [Reference Bousquet-Mélou and Janson14, Lemma 6.2], this implies $\Lambda_\textbf{e}'\in L^p(\mathbb R)$ with $1/p = 1/2-(\alpha-1)$ . Since $\alpha$ may be chosen arbitrarily close to $3/2$ , it follows that $\Lambda_\textbf{e}'\in L^p$ for every $p<\infty$ . (Recall that $\Lambda_\textbf{e}$ , and thus $\Lambda_\textbf{e}'$ , has compact support, so small p is not a problem.) The continuous function $\Lambda_\textbf{e}$ thus has a distributional derivative $\Lambda_\textbf{e}'$ that is integrable, which implies that $\Lambda_\textbf{e}$ is the integral of this derivative, so $\Lambda_\textbf{e}$ is absolutely continuous, with a derivative in the usual sense a.e, which equals the distributional derivative $\Lambda_\textbf{e}'$ .

Finally, $\Lambda_\textbf{e}\in\mathcal H_\alpha$ for $\alpha<1$ by $\Lambda_\textbf{e}'\in L^p$ (with $p=1/(1-\alpha)$ ) and Hölder’s inequality, or directly by $\Lambda_\textbf{e}\in{\mathcal L}^2_{{\alpha+1/2}}=B^{2,2}_{{\alpha+1/2}}$ and the Besov embedding theorem [Reference Bergh and Löfström8, Theorem 6.5.1], [Reference Stein50, V.6.7], noting $\mathcal H_\alpha=B^{\infty,\infty}_{{\alpha}}$ .

Remark 15.9. For the Fourier transform $\widehat{L_\textbf{e}}$ , we have instead of (170) and (183) the analogous estimate

(187) \begin{align} {\mathbb E{}}\,|\widehat{L_\textbf{e}}(\xi)|^2 \leqslant C\bigl({\xi^{-2}\land 1}\bigr),\qquad -\infty<\xi<\infty, \end{align}

and, more precisely,

(188) \begin{align} {\mathbb E{}}\,\bigl\lvert{\widehat{L_\textbf{e}}(\xi)}\bigr\rvert^2=4\xi^{-2} + O\bigl({\xi^{-4}}\bigr).\end{align}

These are proved by the same argument as above, now with $\bigl({d(\overline{U}_1,o)-d(\overline{U}_2,o)}\bigr)$ in the exponent in (173) and using (175) with $k=2$ ; we omit the details.

Note that the exponent of $\xi$ is $-2$ in (188), but $-4$ in (174). Hence, $\widehat{\Lambda_\textbf{e}}(\xi)$ decays faster than $\widehat{L_\textbf{e}}(\xi)$ (at least in an average sense), which intuitively means that $\Lambda_\textbf{e}$ is smoother than $L_\textbf{e}$ , as seen in the results above.

Note also that the result of Vervaat [Reference Vervaat51] mentioned in Remark 15.3 implies that $|\widehat{L_\textbf{e}}(\xi)|=|\widehat{{L}_{\textbf{b}}}(\xi)|$ . The expectation ${\mathbb E{}}|\widehat{{L}_{\textbf{b}}}(\xi)|^2$ can easily be calculated and estimated directly, since $\textbf{b}$ is a Gaussian process, which leads to another proof of (187) and (188); we leave the details to the reader.

The proof above shows that

(189) \begin{align}\Lambda_\textbf{e}\in B^{2,2}_{\alpha},\qquad \alpha<3/2,\end{align}

and thus

(190) \begin{align}\Lambda_\textbf{e}'\in B^{2,2}_{\alpha},\qquad \alpha<1/2.\end{align}

More precisely, it follows from the proof that

(191) \begin{align}{\mathbb E{}}\,\lVert{\Lambda_\textbf{e}'}\rVert_{B^{2,2}_{{\alpha-1}}}^2\leqslant C{\mathbb E{}}\,\lVert{\Lambda_\textbf{e}}\rVert_{B^{2,2}_{\alpha}}^2<\infty\qquad\text{for $\alpha<3/2$},\end{align}

but not for any $\alpha\geqslant3/2$ .

For comparison, (188) implies that

(192) \begin{align}{\mathbb E{}}\,\lVert{L_\textbf{e}}\rVert_{B^{2,2}_{\alpha}}^2<\infty\qquad\text{for $\alpha<1/2$},\end{align}

but not for $\alpha\geqslant\frac12$ . In this case, the range of $\alpha$ in (192) is thus the same as the range of Hölder continuity, note that since $\mathcal H_\alpha=B^{\infty,\infty}_{\alpha}$ for $\alpha\in(0,1)$ , this range equals the set of $\alpha\in(0,1)$ such that $L_\textbf{e}\in B^{\infty,\infty}_{\alpha}$ . Another, similar, example is the Brownian bridge $\textbf{b}$ ; a simple calculation shows that ${\mathbb E{}}\, \lVert{\textbf{b}}\rVert_{B^{2,2}_{{\alpha}}}^2<\infty$ for $\alpha<\frac12$ , but not for larger $\alpha$ .

This suggests (but does not prove) that $\Lambda_\textbf{e}$ is even smoother than shown by Theorem 15.5. However, the derivative $\Lambda_\textbf{e}'$ is not Hölder continuous on $(-\infty,\infty)$ , as might be guessed from the analogy between (191) and (192). In fact, the (two-sided) derivative $\Lambda_\textbf{e}'$ does not exist at 0, and thus $\Lambda_\textbf{e}'$ is not even continuous, at least not at 0.

Theorem 15.10. $\Lambda_{{\textbf{e}}}$ does not a.s. have a (two-sided) derivative at 0.

To see this, we note first the following.

Lemma 15.11. For every $x\geqslant0$ ,

(193) \begin{align} {\mathbb E{}}\Lambda_{{\textbf{e}}}(x) = {\mathbb E{}} L_{{\textbf{e}}}(x) = 4x {e}^{-2x^2}. \end{align}

Proof. The result for $L_\textbf{e}$ is well known; it was shown by [Reference Chung15, (6.2)], and it is equivalent to the case $k=1$ of (175).

The result for $\Lambda_\textbf{e}$ follows by (24) and (23).

Proof of Theorem 15.10. Suppose that $\Lambda_\textbf{e}$ is differentiable at 0. Since $\Lambda_\textbf{e}(x)=0$ for $x<0$ , the derivative has to be 0, and thus $\Lambda_\textbf{e}(x)/x\to0$ as $x\searrow0$ . Furthermore, Theorem 14.6 shows that ${\mathbb E{}}\bigl({\Lambda_\textbf{e}(x)/x}\bigr)^2 \leqslant C$ , and thus $\Lambda_\textbf{e}(x)/x$ is uniformly integrable for $x>0$ . Consequently, if $\Lambda_\textbf{e}$ were a.s. differentiable at 0, then ${\mathbb E{}}\bigl({\Lambda_\textbf{e}(x)/x}\bigr)\to0$ as $x\searrow0$ , which contradicts Lemma 15.11.

Nevertheless, it is quite possible that $\Lambda_\textbf{e}$ is continuously differentiable on $[0,\infty)$ , with a one-sided derivative at 0. We end with some open problems suggested by the results above.

Problem 15.12.

1. (i) Is $\Lambda_\textbf{e}$ a.s. Lipschitz, i.e., is the derivative $\Lambda_\textbf{e}'$ a.s. bounded?

2. (ii) Does $\Lambda_\textbf{e}'(0)$ exist a.s. as a right derivative?

3. (iii) Is the derivative $\Lambda_\textbf{e}'$ a.s. continuous on $[0,\infty)$ ?

4. (iv) Is the derivative $\Lambda_\textbf{e}'$ a.s. in $\mathcal H_\alpha[0,\infty)$ for some $\alpha>0$ ? For every $\alpha<\frac12$ ?