2.1. Definitions and basic properties

Throughout this paper we use the following notation:

(2.1)\begin{equation} \mathbb{R}_+ := [0,+\infty) , \quad \mathbb{R}_{++} := (0,+\infty) .\end{equation}

We routinely work with the extended line $[-\infty,\,+\infty ]$, endowed with the order topology.

Definition 2.1 The Halász–Székely kernel (or HS kernel) is the following function of three variables $x \in \mathbb {R}_+$, $y \in \mathbb {R}_{+ + }$, and $c \in [0,\,1]$:

(2.2)\begin{equation} K(x,y,c) := \begin{cases} \log y + c^{{-}1} \log ( cy^{{-}1}x + 1 - c ) & \text{if } c>0, \\ \log y + y^{{-}1} x -1 & \text{if } c=0. \end{cases} \end{equation}

Proposition 2.2 The HS kernel has the following properties (see also Figure 1):

1. (a) The function $K \colon [0,\,+\infty ) \times (0,\,+\infty ) \times [0,\,1] \to [-\infty,\,+\infty )$ is continuous, attaining the value $-\infty$ only at the points $(0,\,y,\,1)$.

2. (b) $K(x,\,y,\,c)$ is increasing with respect to $x$.

3. (c) $K(x,\,y,\,c)$ is decreasing with respect to $c,$ and strictly decreasing when $x\neq y$.

4. (d) $K(x,\,y,\,1) = \log x$ is independent of $y$.

5. (e) $K(x,\,y,\,c) \ge \log x,$ with equality if and only if $x=y$ or $c=1$.

6. (f) For each $y>0,$ the function $K(\mathord {\cdot },\,y,\,0)$ is affine, and its graph is the tangent line to $\log x$ at $x=y$.

7. (g) $K(\lambda x,\, \lambda y,\, c) = K(x,\,y,\,c) + \log \lambda,$ for all $\lambda >0$.

Figure 1.

Graphs of the functions $K(\mathord {\cdot },\,y,\,c)$ for $y=2$ and $c \in \{0,\,1/3,\,2/3,\,1\}$.

Proof. Most properties are immediate from Definition 2.1. To check monotonicity with respect to $c$, we compute the partial derivative when $c>0$:

(2.3)\begin{equation} K_c(x,y,c) = \displaystyle\frac{1}{c^{2}} \left[\frac{c(y^{{-}1}x-1)}{cy^{{-}1}x + 1 - c} + \log \bigg(\frac{1}{cy^{{-}1}x + 1 - c}\bigg)\right];\end{equation}

since $\log t \le t-1$ (with equality only if $t=1$), we conclude that $K_c(x,\,y,\,c) \le 0$ (with equality only if $x=y$). Since $K$ is continuous, we obtain property (c). Property (e) is a consequence of properties (c) and (d).

Let $\mathcal {P}(\mathbb {R}_+ )$ denote the set of all Borel probability measures $\mu$ on $\mathbb {R}_+$. The following is the central concept of this paper:

Definition 2.3 Let $c \in [0,\,1]$ and $\mu \in \mathcal {P}(\mathbb {R}_+ )$. If $c=1$, then we require that the function $\log x$ is semi-integrableFootnote ¹ with respect to $\mu$. The Halász–Székely barycenter (or HS barycenter) with parameter $c$ of the probability measure $\mu$ is:

(2.4)\begin{equation} [\mu]_c := \exp \inf_{y>0} \displaystyle\int K(x,y,c) \, d\mu(x) , \end{equation}

where $K$ is the HS kernel (2.2).

First of all, let us see that the definition is meaningful:

• • If $c<1$, then for all $y>0$, the function $K( \mathord {\cdot },\, y,\, c)$ is bounded from below by $K(0,\,y,\,c)>-\infty$, and therefore, it has a well-defined integral (possibly $+\infty$); so $[\mu ]_c$ is a well-defined element of the extended half-line $[0,\,+\infty ]$.

• • If $c=1$, then by part (d) of Proposition 2.2, the defining formula (2.4) becomes:

  (2.5)\begin{equation} [\mu]_1 = \exp \displaystyle\int \log x \, d\mu(x) . \end{equation}
  The integral is a well-defined element of $[-\infty,\, +\infty ]$, so $[\mu ]_1$ is well defined in $[0,\,+\infty ]$.

Formula (2.5) means that the HS barycenter with parameter $c=1$ is the geometric barycenter; let us see that $c=0$ corresponds to the standard arithmetic barycenter:

Let ${{\mathcal {P}_\mathrm {HS}(\mathbb {R}_{+ })}}$ denote the subset formed by those $\mu \in \mathcal {P}(\mathbb {R}_+ )$ such that:

(2.6)\begin{equation} \displaystyle\int \log (1 + x) \, d\mu(x) < \infty \end{equation}

or, equivalently, $\int \log ^{+} x \, d\mu < \infty$ (we will sometimes write ‘$d\mu$’ instead of ‘$d\mu (x)$’).

Proof. The case $c=1$ being clear, assume that $c \in (0,\,1)$. Note that for all $y>0$, the expression

(2.7)\begin{equation} \left | K(x,y,c) - c^{{-}1} \log(x+1) \right| \end{equation}

is a bounded function of $x$, so the integrability of $K(x,\,y,\,c)$ and $\log (x+1)$ are equivalent.

Next, let us see that the standard properties one might expect for something called a ‘barycenter’ are satisfied. For any $x \ge 0$, we denote by $\delta _x$ the probability measure such that $\delta _x(\{x\})=1$.

As it will be clear later (see Example 2.13), the internality and the monotonicity properties (w.r.t. $\mu$ and w.r.t. $c$) are not strict.

2.2. Computation and critical phenomenon

In the remaining of this section, we discuss how to actually compute HS barycenters. In view of Proposition 2.5, we may focus on measures in ${{\mathcal {P}_\mathrm {HS}(\mathbb {R}_{+ })}}$. The mass of zero plays an important role. Given $c \in (0,\,1)$ and $\mu \in {{\mathcal {P}_\mathrm {HS}(\mathbb {R}_{+ })}}$, we use the following terminology, where $\mu (0) = \mu (\{0\})$:

(2.8)\begin{equation} \left\{\begin{array}{@{}cc} \textit{subcritical case:} & \mu(0) < 1-c \\ \textit{critical case:} & \mu(0) = 1-c \\ \textit{supercritical case:} & \mu(0) > 1-c \end{array}\right. \end{equation}

The next result establishes a way to compute $[\mu ]_c$ in the subcritical case; the remaining cases will be dealt with later in Proposition 2.11.

Proposition 2.7 If $\mu \in {{\mathcal {P}_\mathrm {HS}(\mathbb {R}_{+ })}},$ $c \in (0,\, 1)$ and $\mu (0) < 1-c$ (subcritical case), then the equation

(2.9)\begin{equation} \displaystyle\int K_y (x,\eta,c) \, d \mu(x) = 0 \end{equation}

has a unique positive and finite solution $\eta = \eta (\mu,\,c),$ and the $\inf$ in formula (2.4) is attained uniquely at $y=\eta ;$ in particular,

(2.10)\begin{equation} [\mu]_c = \exp \displaystyle\int K(x, \eta, c) \, d \mu(x). \end{equation}

Proof. Fix $\mu$ and $c$ as in the statement. We compute the partial derivative:

(2.11)\begin{equation} K_y (x,y,c) = \displaystyle\frac{\Delta(x,y)}{y} , \quad \text{where} \quad \Delta(x,y) := 1 - \frac{x}{cx+(1-c)y} . \end{equation}

Since $\Delta$ is bounded, we are allowed to differentiate under the integral sign:

(2.12)\begin{equation} \displaystyle\frac{d}{dy} \displaystyle\int K(x,y,c) \, d\mu = \int K_y (x,\eta,c) \, d \mu = \frac{\psi(y)}{y} , \end{equation}

where $\psi (y) := \int \Delta (x,\,y) \, d\mu$. The partial derivative

(2.13)\begin{equation} \Delta_y(x,y) = \displaystyle\frac{(1-c)x}{(cx+(1-c)y)^{2}} \end{equation}

is positive, except at $x=0$. Since $\mu \neq \delta _0$, the function $\psi$ is strictly increasing. Furthermore,

(2.14)\begin{equation} \lim_{y \to +\infty} \Delta(x,y) = 1 \quad \text{and} \quad \lim_{y \to 0^{+}} \Delta(x,y) = \begin{cases} 1-1/c & \text{if }x>0,\\ 1 & \text{if }x=0, \end{cases} \end{equation}

and so

(2.15)\begin{equation} \lim_{y \to +\infty} \psi(y) = 1 \quad \text{and} \quad \lim_{y \to 0^{+}} \psi(y) = 1 - \displaystyle\frac{1-\mu(0)}{c} < 0 ,\end{equation}

using the assumption $\mu (0)<1-c$. Therefore, there exists a unique $\eta >0$ that solves the equation $\psi (\eta )=0$, or equivalently equation (2.9). By (2.12), the function $y \mapsto \int K(x,\,y,\,c) \, d\mu$ decreases on $(0,\,\eta ]$ and increases on $[\eta,\,+\infty )$, and so attains its infimum at $\eta$. Formula (2.10) follows from the definition of $[\mu ]_c$.

Let us note that, as a consequence of (2.11), equation (2.9) is equivalent to:

(2.16)\begin{equation} \displaystyle\int \displaystyle\frac{x}{cx+(1-c)\eta} \, d \mu(x) = 1 . \end{equation}

We introduce the following auxiliary function, plotted in Figure 2:

(2.17)\begin{equation} B(c) := \begin{cases} 0 & \text{if } c=0, \\ c(1-c)^{\frac{1-c}{c}} & \text{if } 0< c<1, \\ 1 & \text{if } c=1. \end{cases} \end{equation}

Figure 2.

Graph of the function $B$ defined by (2.17).

The following alternative formula for the HS barycenter matches the original one from [Reference Halász and Székely15], and in some situations is more convenient:

Proposition 2.9 If $0 < c\le 1$ and $\mu \in {{\mathcal {P}_\mathrm {HS}(\mathbb {R}_{+ })}}$, then:

(2.18)\begin{equation} [\mu]_c = B(c) \, \inf_{r>0} \left\{ r^{-\frac{1-c}{c}} \exp [ c^{{-}1} \displaystyle\int \log (x+r) \, d\mu(x) ] \right\} . \end{equation}

Furthermore, if $\mu (0)<1-c$, then the $\inf$ is attained at the unique positive finite solution $\rho = \rho (\mu,\,c)$ of the equation

(2.19)\begin{equation} \displaystyle\int \displaystyle\frac{x}{x+\rho} \, d\mu(x) = c . \end{equation}

If $\mu$ is a Borel probability measure on $\mathbb {R}_{+ }$ not entirely concentrated at zero (i.e., $\mu \neq \delta _0$) then we denote by $\mu ^{+ }$ the probability measure obtained by conditioning on the event $\mathbb {R}_{+ + } =(0,\,\infty )$, that is,

(2.20)\begin{equation} \mu^{+}(U) := \displaystyle\frac{\mu(U \cap \mathbb{R}_{++})}{\mu(\mathbb{R}_{++})}, \quad \text{for every Borel set } U \subseteq \mathbb{R} . \end{equation}

Obviously, if $\mu \in {{\mathcal {P}_\mathrm {HS}(\mathbb {R}_{+ })}}$, then $\mu ^{+ } \in {{\mathcal {P}_\mathrm {HS}(\mathbb {R}_{+ })}}$ as well.

Proposition 2.10 Let $\mu \in {{\mathcal {P}_\mathrm {HS}(\mathbb {R}_{+ })}}$ and $p := 1-\mu (0)$. If $c \in (0,\,1]$ and $c \le p$ (critical or subcritical cases), then

(2.21)\begin{equation} [\mu]_c= \displaystyle\frac{B(c)}{B(c/p)} [\mu^{+}]_{c/p} . \end{equation}

Proof. Note that $\mu \neq \delta _0$, so the positive part $\mu ^{+ }$ is defined. Using formula (2.18), we have:

(2.22)\begin{align} [\mu]_c & = B(c) \inf_{r >0} r^{1-\frac{1}{c}} \exp \bigg(\frac{1}{c} \displaystyle\int \log (x+r\bigg) d \mu ) \end{align}

(2.23)\begin{align} [0] & = B(c) \inf_{r >0} r^{1-\frac{1}{c}} \exp \bigg(\frac{1}{c} ( \mu(0\bigg) \log r + \displaystyle\int_{\{x>0\} } \log (x+r) d \mu ) ) \end{align}

(2.24)\begin{align} [0] & = B(c) \inf_{r >0} r^{1-\frac{1}{c}+ \frac{1-p}{c}} \exp \bigg(\frac{1}{c} \int_{\{x>0\} } \log (x+r\bigg) d \mu ) \end{align}

(2.25)\begin{align} [0] & = B(c) \inf_{r >0} r^{1- \frac{p}{c}} \exp \bigg(\frac{1}{c/p} \int \log(x+r\bigg) d \mu^{+} ) . \end{align}

At this point, the assumption $c \le p$ guarantees that the barycenter $[\mu ^{+ }]_{c/p}$ is well defined, and using (2.18) again we obtain (2.21).

Finally, we compute the HS barycenter in the critical and supercritical cases:

Proof. In the critical case, we use (2.21) with $p=c$ and conclude.

In the supercritical case, we can assume that $\mu \neq \delta _0$. Note that $p< c$, thus, $\lim _{r \to 0^{+}} r^{1-\frac {p}{c}}=0$. Moreover, since $\log ^{+} x \in L^{1}(\mu )$, we have

(2.26)\begin{equation} \lim_{r \to 0^{+}} \displaystyle\int_{\{x>0\}} \log(x+r) \, d\mu = \int_{\{x>0\}} \log x \, d\mu <{+}\infty .\end{equation}

Therefore, using (2.25), we obtain $[\mu ]_c = 0$.

Propositions 2.7 and 2.11 allow us to compute HS barycenters in all cases. For emphasis, let us list explicitly the situations where the barycenter vanishes:

Example 2.13 Consider the family of probability measures:

(2.27)\begin{equation} \mu_p := (1-p) \delta_0 + p \delta_1 ,\quad 0\le p \le 1. \end{equation}

If $0< c<1$, then

(2.28)\begin{equation} [\mu_p]_c = \begin{cases} 0 & \quad \text{if } p< c \\ B(c) = c (1-c)^{\frac{1-c}{c}} & \quad \text{if } p=c \\ \frac{B(c)}{B(c/p)} = (1-c)^{\frac{1-c}{c}} p^{\frac{p}{c}} (p-c)^{\frac{c-p}{c}} & \quad \text{if } p>c . \end{cases} \end{equation}

These formulas were first obtained by Székely [Reference Székely26]. It follows that the function

(2.29)\begin{equation} (p,c) \in [0,1] \times [0,1] \mapsto [\mu_p]_c \end{equation}

(whose graph is shown on [Reference van Es28, p. 680]) is discontinuous at the points with $p=c>0$, and only at those points. We will return to the issue of continuity in § 4.

The practical computation of HS barycenters usually requires numerical methods. In any case, it is useful to notice that the function $\eta$ from Proposition 2.7 satisfies the internality property:

The proof is left to the reader.

3.1. HS means as repetitive symmetric means

The HS barycenter of a probability measure, introduced in the previous section, may now be specialized to the case of discrete equidistributed probabilities. So the HS mean of a tuple $\underline {x} = (x_1,\,\dots,\,x_n)$ of non-negative numbers with parameter $c \in [0,\,1]$ is defined as:

(3.1)\begin{equation} \mathsf{hsm}_c(\underline{x}) := \left[\displaystyle\frac{\delta_{x_1} + \dots + \delta_{x_n}}{n}\right]_c . \end{equation}

Using (2.18), we have more explicitly:

(3.2)\begin{equation} \mathsf{hsm}_c(\underline{x}) = B(c) \inf_{r>0} r^{-\frac{1-c}{c}} \prod_{j=1}^{n} (x_j + r)^\frac{1}{cn} \quad \text{(if $c>0$)}, \end{equation}

where $B$ is the function (2.17). On the other hand, recall that for $k \in \{1,\,\dots,\,n\}$, the $k$-th symmetric mean of the $n$-tuple $\underline {x}$ is:

(3.3)\begin{equation} \mathsf{sym}_k(\underline{x}) := \left(\displaystyle\frac{E^{(n)}_k(\underline{x})}{\binom{n}{k}}\right)^{\frac{1}{k}} , \end{equation}

where $E^{(n)}_k$ denotes the elementary symmetric polynomial of degree $k$ in $n$ variables.

Since they originate from a barycenter, the HS means are repetition invariant Footnote ³ in the sense that, for any $m>0$,

(3.4)\begin{equation} \mathsf{hsm}_c ( \underline{x}^{(m)} ) = \mathsf{hsm}_c ( \underline{x} ) , \end{equation}

where $\underline {x}^{(m)}$ denotes the $nm$-tuple obtained by concatenation of $m$ copies of the $n$-tuple $\underline {x}$. No such property holds for the symmetric means, even allowing for adjustment of the parameter $k$. Nevertheless, if the number of repetitions tends to infinity, then the symmetric means tend to stabilize, and the limit is a HS mean; more precisely:

Theorem 3.1 If $\underline {x} = (x_1,\,\dots,\,x_n),$ $x_i \ge 0,$ and $1 \le k \le n,$ then:

(3.5)\begin{equation} \lim_{m \to \infty} \mathsf{sym}_{km} ( \underline{x}^{(m)} ) = \mathsf{hsm}_{k/n} (\underline{x}) . \end{equation}

Furthermore, the relative error goes to zero uniformly with respect to the $x_i$'s.

This theorem will be proved in the next subsection.

It is worthwhile to note that the Navas barycenter [Reference Navas22] is obtained as a ‘repetition limit’ similar to (3.5).

Example 3.2 Using Propositions 2.7 and 2.11, one computes:

(3.6)\begin{equation} \mathsf{hsm}_{1/2} (x,y) = \left(\displaystyle\frac{\sqrt{x}+\sqrt{y}}{2}\right)^{2} \, ; \end{equation}

Therefore,

(3.7)\begin{equation} \lim_{m \to \infty} \mathsf{sym}_{m} ( \underbrace{x, \dots, x}_{\text{$m$ times}}, \, \underbrace{y, \dots, y}_{\text{$m$ times}} ) = \left(\displaystyle\frac{\sqrt{x}+\sqrt{y}}{2}\right)^{2} .\end{equation}

The last equality was deduced in [Reference Cellarosi, Hensley, Miller and Wellens10, p. 31] from the asymptotics of Legendre polynomials.

Repeated symmetric means are particular instances of Whiteley means; in fact, in the notation of [Reference Bullen8, p. 344],

(3.8)\begin{equation} \mathsf{sym}_{km} ( \underline{x}^{(m)} ) = \mathfrak{M}_n^{[km,m]} (\underline{x}) . \end{equation}

Let us pose a problem:

There exists a partial result: when $k=1$ and $n=2$, [Reference Cellarosi, Hensley, Miller and Wellens10, Lemma 4.1] establishes eventual monotonicity.

3.2. Inequalities between symmetric means and HS means

The following is the first main result of this paper.

Theorem 3.4 If $\underline {x} = (x_1,\,\dots,\,x_n)$, $x_i \ge 0$, and $1 \le k \le n$, then

(3.9)\begin{equation} \mathsf{hsm}_{k/n}(\underline{x}) \le \mathsf{sym}_{k}(\underline{x}) \le \displaystyle\frac{\binom{n}{k}^{{-}1/k}}{B \bigg(\frac{k}{n} \bigg)} \, \mathsf{hsm}_{k/n}(\underline{x}) . \end{equation}

Let us postpone the proof to the next subsection. The factor at the RHS of (3.9) is asymptotically $1$ with respect to $k$; indeed:

Lemma 3.5 For all integers $n \ge k \ge 1$, we have

(3.10)\begin{equation} 1 \le \displaystyle\frac{\binom{n}{k}^{{-}1/k}}{B \bigg(\frac{k}{n} \bigg)} < (9k)^{\frac{1}{2k}} , \end{equation}

with equality if and only if $k=n$.

Proof. Let $c := k/n$ and

(3.11)\begin{equation} b := \binom{n}{k} [B(c)]^{k} = \binom{n}{k} c^{k} \, (1-c)^{n-k} .\end{equation}

By the Binomial Theorem, $b \le 1$, which yields the first part of (3.10). For the lower estimate, we use the following Stirling bounds (see [Reference Feller12, p. 54, (9.15)]), valid for all $n\ge 1$,

(3.12)\begin{equation} \sqrt{2\pi} \, n^{n+\frac{1}{2}} e^{{-}n} < n! < \sqrt{2\pi} \, n^{n+\frac{1}{2}} e^{{-}n+\frac{1}{12}} . \end{equation}

Then, a calculation (cf. [Reference Feller12, p. 184, (3.11)]) gives:

(3.13)\begin{equation} b > \displaystyle\frac{e^{-\frac{1}{6}}}{\sqrt{2 \pi c(1-c) n}} > \frac{1}{\sqrt{9k}} ,\end{equation}

from which the second part of (3.10) follows.

Theorem 3.4 and Lemma 3.5 imply that HS means (with rational values of the parameter) can be obtained as repetition limits of symmetric means:

Proof of Theorem 3.1. Applying Theorem 3.4 to the tuple $\underline {x}^{(m)}$, using observation (3.4) and Lemma 3.5, we have:

(3.14)\begin{equation} \mathsf{hsm}_{k/n}(\underline{x}) \le \mathsf{sym}_{km}(\underline{x}^{(m)}) \le (9km)^{\frac{1}{2km}} \,\mathsf{hsm}_{k/n}(\underline{x}) . \end{equation}

Making $m\to \infty$ we obtain (3.5).

Remark 3.6 If $k$ is fixed, then

(3.15)\begin{equation} \lim_{n \to \infty} \displaystyle\frac{\binom{n}{k}^{{-}1/k}}{B \bigg(\frac{k}{n}\bigg)} = \frac{e (k!)^{1/k}}{k} > 1,\end{equation}

and therefore, the bound from Theorem 3.7 may be less satisfactory. But in this case, we may use the alternative bound coming from Maclaurin inequality:

(3.16)\begin{equation} \mathsf{sym}_k(\underline{x}) \le \mathsf{sym}_1(\underline{x}) = \mathsf{hsm}_0(\underline{x}) . \end{equation}

3.3. Proof of Theorem 3.4

The two inequalities in (3.9) will be proved independently of each other. They are essentially contained in the papers [Reference Bochi, Iommi and Ponce5] (the first inequality) and [Reference Halász and Székely15] (the second one), though neither was stated explicitly. In the following Theorems 3.8 and 3.7, we also characterize the cases of equality, and in particular show that each inequality is sharp in the sense that the corresponding factors cannot be improved.

Let us begin with the second inequality, which is more elementary. By symmetry, there is no loss of generality in assuming that the numbers $x_i$ are ordered.

Theorem 3.7 If $\underline {x} = (x_1,\,\dots,\,x_n)$ with $x_1 \ge \cdots \ge x_n \ge 0,$ and $1 \le k \le n,$ then:

(3.17)\begin{equation} \mathsf{sym}_k(\underline{x}) \le \displaystyle\frac{\binom{n}{k}^{{-}1/k}}{B \bigg(\frac{k}{n} \bigg)} \mathsf{hsm}_{k/n}(\underline{x}) . \end{equation}

Furthermore, equality holds if and only if $x_{k+1} = 0$ or $k=n$.

Proof. Our starting point is Vieta's formula:

(3.18)\begin{equation} z^{n} + \displaystyle\sum_{\ell = 1}^{n} E^{(n)}_\ell(x_1, \dots, x_n) z^{n-\ell} = \prod_{j=1}^{n} (x_j+z) .\end{equation}

Therefore, by Cauchy's formula, for any $r>0$:

(3.19)\begin{equation} E^{(n)}_k(x_1, \dots, x_n) = \frac{1}{2\pi \textbf{i}} {\unicode{x2232}}_{|z|=r} \frac{1}{z^{n-k+1}} \prod_{j=1}^{n} (x_j+z) \, dz . \end{equation}

That is,

(3.20)\begin{equation} E^{(n)}_k(x_1, \dots, x_n) = \displaystyle\frac{1}{2\pi} \displaystyle\int_{-\pi}^{\pi}(r e^{\textbf{i}\theta})^{{-}n+k} \prod_{j=1}^{n} (x_j + r e^{\textbf{i}\theta}) \, d\theta , \end{equation}

Taking absolute values,

(3.21)\begin{equation} E^{(n)}_k(x_1, \dots, x_n)\le\displaystyle\frac{1}{2\pi} \displaystyle\int_{-\pi}^{\pi} r^{{-}n+k} \prod_{j=1}^{n} |x_j + r e^{\textbf{i}\theta}| \, d\theta \le r^{{-}n+k} \prod_{j=1}^{n} (x_j + r) . \end{equation}

But these inequalities are valid for all $r>0$, and therefore:

(3.22)\begin{equation} E^{(n)}_k(x_1, \dots, x_n) \le \inf_{r>0} r^{{-}n+k} \prod_{j=1}^{n} (x_j + r) . \end{equation}

So formulas (3.2) and (3.3) imply inequality (3.17).

Now let us investigate the possibility of equality. We consider three mutually exclusive cases, which correspond to the classification (2.8):

(3.23)\begin{equation} \left. \begin{array}{ll@{}} \text{subcritical case:} & x_{k+1}>0 \\ \text{critical case:} & x_k > 0 = x_{k+1} \\ \text{supercritical case:} & x_k = 0 \end{array} \right\} \end{equation}

Using Proposition 2.11, in the critical case, we have:

(3.24)\begin{equation} \mathsf{sym}_k (\underline{x}) = \left(\displaystyle\frac{x_1\cdots x_k}{\binom{n}{k}}\right)^{\frac{1}{k}} \quad \text{and} \quad \mathsf{hsm}_{k/n}(\underline{x}) = \frac{(x_1\cdots x_k)^{\frac{1}{k}}}{B\bigg(\frac{k}{n}\bigg)} , \end{equation}

while in the supercritical case, the two means vanish together. So, in both cases, inequality (3.17) becomes an equality. Now suppose we are in the subcritical case; then the $\inf$ at the RHS of (3.22) is attained at some $r > 0$: see Proposition 2.9. On the other hand, for this (and actually any) value of $r$, the second inequality in (3.21) must be strict, because the integrand is non-constant. We conclude that, in the subcritical case, inequality (3.22) is strict, and therefore, (3.17) is strict.

The first inequality in (3.9) is a particular case of an inequality between two types of matrix means introduced in [Reference Bochi, Iommi and Ponce5], which we now explain. Let $A = (a_{i,j})_{i,j\in \{1,\dots,n\}}$ be a $n \times n$ matrix with non-negative entries. Recall that the permanent of $A$ is the ‘signless determinant’

(3.25)\begin{equation} \operatorname{per}(A) := \displaystyle\sum_{\sigma} \prod_{i=1}^{n} a_{i,\sigma(i)} , \end{equation}

where $\sigma$ runs on the permutations of $\{1,\, \dots,\, n\}$. Then the permanental mean of $A$ is defined as:

(3.26)\begin{equation} \mathsf{pm}(A) := \left(\displaystyle\frac{\operatorname{per}(A)}{n!}\right)^{\frac{1}{n}} . \end{equation}

On the other hand, the scaling mean of the matrix $A$ is defined as:

(3.27)\begin{equation} \mathsf{sm}(A) := \displaystyle\frac{1}{n^{2}} \inf_{u,v} \frac{u^{\top} A v}{\mathsf{gm}(u)\mathsf{gm}(v)} , \end{equation}

where $u$ and $v$ run on the set of strictly positive column vectors, and $\mathsf {gm}(\mathord {\cdot })$ denotes the geometric mean of the entries of the vector. Equivalently,

(3.28)\begin{equation} \mathsf{sm}(A) = \displaystyle\frac{1}{n} \inf_{v} \frac{\mathsf{gm}(Av)}{\mathsf{gm}(v)}; \end{equation}

see [Reference Bochi, Iommi and Ponce5, Remark 2.6].Footnote ⁴ By [Reference Bochi, Iommi and Ponce5, Theorem 2.17],

(3.29)\begin{equation} \mathsf{sm}(A) \le \mathsf{pm}(A) , \end{equation}

with equality if and only if $A$ has permanent $0$ or rank $1$. This inequality is far from trivial. Indeed, if the matrix $A$ is doubly stochastic (i.e. row and column sums are all $1$), then an easy calculation (see [Reference Bochi, Iommi and Ponce5, Prop. 2.4]) shows that $\mathsf {sm}(A) = \frac {1}{n}$, so (3.29) becomes $\mathsf {pm}(A) \ge \frac {1}{n}$, or equivalently,

(3.30)\begin{equation} \operatorname{per}(A) \ge \displaystyle\frac{n!}{n^{n}} \quad \text{(if $A$ is doubly stochastic).} \end{equation}

This lower bound on the permanent of doubly stochastic matrices was conjectured in 1926 by van der Waerden and, after a protracted series of partial results, proved around 1980 independently by Egorichev and Falikman: see [Reference Zhan30, Chapter 5] for the exact references and a self-contained proof, and [Reference Gurvits14] for more recent developments. Our inequality (3.29), despite being a generalization of Egorichev–Falikman's (3.30), is actually a relatively simple corollary of it: we refer the reader to [Reference Bochi, Iommi and Ponce5, § 2] for more information.Footnote ⁵

We are now in position to complete the proof of Theorem 3.4, i.e., to prove the second inequality in (3.9). The next result also characterizes the cases of equality.

Theorem 3.8 If $\underline {x} = (x_1,\,\dots,\,x_n)$ with $x_1 \ge \cdots \ge x_n \ge 0,$ and $1 \le k \le n,$ then:

(3.31)\begin{equation} \mathsf{hsm}_{k/n}(\underline{x}) \le \mathsf{sym}_k(\underline{x}) . \end{equation}

Furthermore, equality holds if and only if :

(3.32)\begin{equation} k = n \quad \text{or} \quad x_1 = \cdots = x_n \quad \text{or} \quad x_k = 0 . \end{equation}

Proof. Consider the non-negative $n \times n$ matrix:

(3.33)\begin{equation} A := \begin{pmatrix} x_1 & \cdots & x_1 & \, 1 \, & \, \cdots \, & \, 1 \, \\ \vdots & & \vdots & \vdots & & \vdots \\ x_n & \cdots & x_n & \, 1 \, & \, \cdots \, & \, 1 \, \end{pmatrix} \quad \text{with $n-k$ columns of $1$'s.} \end{equation}

Note that:

(3.34)\begin{equation} \operatorname{per}(A) = k! \, (n-k)! \, E^{(n)}_k(x_1,\dots,x_n)\end{equation}

and so

(3.35)\begin{align} \mathsf{pm}(A) & = \left(\displaystyle\frac{E^{(n)}_k(x_1,\dots,x_n)}{\binom{n}{k}}\right)^{\frac{1}{n}} \end{align}

(3.36)\begin{align} & = [\mathsf{sym}_k(\underline{x}) ]^{k/n} . \end{align}

Now let's compute the scaling mean of $A$ using formula (3.28). Assume that $k< n$. Given a column vector $v = \left (\begin {smallmatrix} v_1 \\ \vdots \\ v_n \end {smallmatrix}\right )$ with positive entries, we have:

(3.37)\begin{equation} \mathsf{gm}(Av) = \prod_{i=1}^{n} (sx_i+r)^{\frac{1}{n}} , \quad \text{where } s := \displaystyle\sum_{i=1}^{k} v_i \text{ and } r := \sum_{i=k+1}^{n} v_i .\end{equation}

On the other hand, by the inequality of arithmetic and geometric means,

(3.38)\begin{equation} \mathsf{gm}(v) \le \left(\displaystyle\frac{s}{k}\right)^{\frac{k}{n}} \bigg(\frac{r}{n-k}\bigg)^{\frac{n-k}{n}} ,\end{equation}

with equality if $v_1 = \cdots = v_k = \frac {s}{k}$, $v_{k+1} = \cdots = v_n = \frac {r}{n-k}$. So, in order to minimize the quotient $\frac {\mathsf {gm}(Av)}{\mathsf {gm}(v)}$, it is sufficient to consider column vectors $v$ satisfying these conditions. We can also normalize $s$ to $1$, and (3.28) becomes:

(3.39)\begin{align} \mathsf{sm}(A) & = \inf_{r>0} \displaystyle \frac{\prod_{i=1}^{n} (x_i+r)^\frac{1}{n}}{ \bigg(\frac{1}{k}\bigg)^{\frac{k}{n}} \bigg(\frac{r}{n-k}\bigg)^{\frac{n-k}{n}}} \end{align}

(3.40)\begin{align} & = [\mathsf{hsm}_{k/n}(\underline{x})]^{k/n}\, \end{align}

by (3.2). This formula $\mathsf {sm}(A) = [\mathsf {hsm}_{k/n}(\underline {x})]^{k/n}$ also holds for $k=n$, taking the form $\mathsf {sm}(A) = (x_1 \cdots x_n)^{\frac {1}{n}}$; this can be checked either by adapting the proof above, or more simply by using the homogeneity and reflexivity properties of the scaling mean (see [Reference Bochi, Iommi and Ponce5]).

In conclusion, the matrix (3.33) has scaling and permanental means given by formulas (3.40) and (3.36), respectively, and the fundamental inequality (3.29) translates into $\mathsf {hsm}_{k/n}(\underline {x}) \le \mathsf {sym}_k(\underline {x})$, that is, (3.8).

Furthermore, equality holds if and only if the matrix $A$ defined by (3.33) satisfies $\mathsf {sm}(A) = \mathsf {pm}(A)$, by formulas (3.40) and (3.36). As mentioned before, $\mathsf {sm}(A) = \mathsf {pm}(A)$ if and only if $A$ has rank $1$ or permanent $0$ (see [Reference Bochi, Iommi and Ponce5, Theorem 2.17]). Note that $A$ has rank $1$ if and only if $k=1$ or $x_1=\dots =x_n$. On the other hand, by (3.36), $A$ has permanent $0$ if and only if $\mathsf {sym}_k(\underline {x})=0$, or equivalently $x_k=0$. So we have proved that equality $\mathsf {hsm}_{k/n}(\underline {x}) = \mathsf {sym}_k(\underline {x})$ is equivalent to condition (3.32).

We close this section with some comments on related results.

Remark 3.10 Given an arbitrary $n \times n$ non-negative matrix $A$, the permanental and scaling means satisfy the following inequalities (see [Reference Bochi, Iommi and Ponce5, Theorem 2.17]),

(3.41)\begin{equation} \mathsf{sm}(A) \leq \mathsf{pm}(A) \leq n(n!)^{{-}1/n} \mathsf{sm}(A). \end{equation}

The sequence $(n(n!)^{-1/n})$ is increasing and converges to $e$. In general, as $n$ tends to infinity the permanental mean does not necessarily converge to the scaling mean. However, there are some special classes of matrices for which this is indeed the case: for example, in the repetitive situation covered by the generalized Friedland limit [Reference Bochi, Iommi and Ponce5, Theorem 2.19]. Note that $\mathsf {hsm}_{k/n}(\underline {x})$ and $\mathsf {sym}_k(\underline {x})$ correspond to the $n/k-$th power of the scaling and permanental mean of the matrix $A$, respectively. Therefore, (3.9) can be regarded as an improvement of (3.41) for this particular class of matrices.

4.1. Convergence of measures

If $(X,\, \mathrm {d})$ is a separable complete metric space, let $C_\mathrm {b}(X)$ be the set of all continuous bounded real functions on $X$, and let $\mathcal {P}(X)$ denote the set of all Borel probability measures on $X$. Recall (see e.g. [Reference Parthasarathy23]) that the weak topology is a metrizable topology on $\mathcal {P}(X)$ according to with a sequence $(\mu _n)$ converges to some $\mu$ if and only if $\int \phi \, d \mu _n \to \int \phi \, d \mu$ for every test function $\phi \in C_\mathrm {b}(X)$; we say that $(\mu _n)$ converges weakly to $\mu$, and denote this by $\mu _n \rightharpoonup \mu$. The space $\mathcal {P}(X)$ is Polish, and it is compact if and only if $X$ is compact. Despite the space $C_\mathrm {b}(X)$ being huge (nonseparable w.r.t. its usual topology if $X$ is noncompact), by [Reference Parthasarathy23, Theorem II.6.6] we can nevertheless find a countable subset $\mathcal {C} \subseteq C_\mathrm {b}(X)$ such that, for all $(\mu _n)$ and $\mu$ in $\mathcal {P}(X)$,

(4.1)\begin{equation} \mu_n \rightharpoonup \mu \quad \Longleftrightarrow \quad \forall \phi \in \mathcal{C}, \ {\textstyle \displaystyle\int \phi \, d \mu_n \to \int \phi \, d \mu } . \end{equation}

The following result deals with sequences of integrals $\int \phi _n \, d\mu _n$ where not only the measures but also the integrands vary, and bears a resemblance to Fatou's Lemma:

Note that, as in Fatou's Lemma, the integrals in Proposition 4.1 can be infinite.

Proof. Without loss of generality, assume that $C=0$. Let $\lambda \in \mathbb {R}$ be such that $\lambda < \int \phi \, d\mu$. By the monotone convergence theorem, there exists $m \in \mathbb {N}$ such that $\int \min \{ \phi,\, m\} \, d\mu > \lambda$. For a function $\psi :X \to \mathbb {R}$, let $\hat {\psi }(x):=\min \{ \psi (x),\, m\}$. Note that

(4.2)\begin{equation} \displaystyle\int \phi_n \, d\mu_n \geq \int \hat{\phi}_n \, d\mu_n = \int (\hat{\phi}_n -\hat{\phi} )\, d\mu_n + \int \hat{\phi} \, d\mu_n . \end{equation}

By Prokhorov's theorem (see e.g. [Reference Parthasarathy23, Theorem 6.7]), the sequence $(\mu _n)$ forms a tight set, that is, for every $\epsilon > 0$, there exists a compact set $K \subseteq X$ such that $\mu _n(X \setminus K) \le \epsilon /(2\,m)$ for all $n$. Since $(\phi _n)$ converges uniformly on compact subsets to $\phi$, we obtain:

(4.3)\begin{equation} \lim_{n \to \infty} \left| \displaystyle\int_K (\hat{\phi}_n -\hat{\phi} )\, d\mu_n \right| \leq \lim_{n \to \infty} \sup_{x \in K} \left| \hat{\phi}_n(x) -\hat{\phi}(x) \right| =0 . \end{equation}

We also have:

(4.4)\begin{equation} \left| \displaystyle\int_{X \setminus K} (\hat{\phi}_n -\hat{\phi} )\, d\mu_n \right| \leq \mu_n(X \setminus K) \sup_ {x \in X \setminus K} \left| \hat{\phi}_n(x) -\hat{\phi}(x) \right| < \displaystyle\frac{\epsilon}{2m} 2m= \epsilon. \end{equation}

Since $(\mu _n)$ converges weakly to $\mu$, we have $\lim _{n \to \infty } \int \hat {\phi } \, d\mu _n =\int \hat {\phi } \, d\mu > \lambda$. Therefore, combining (4.2), (4.3) and (4.4), for sufficiently large values of $n$, we obtain $\int \phi _n \, d\mu _n > \lambda$. The result now follows.

The next direct consequence is useful.

We will also need a slightly stronger notion of convergence. Let $\mathcal {P}_1(X) \subseteq \mathcal {P}(X)$ denote the set of measures $\mu$ with finite first moment, that is,

(4.5)\begin{equation} \displaystyle\int \mathrm{d}(x,x_0) \, d\mu(x) < \infty , \end{equation}

Here and in what follows, $x_0 \in X$ is a basepoint which we consider as fixed, the particular choice being entirely irrelevant. We metrize $\mathcal {P}_1(X)$ with the Kantorovich metric (see e.g. [Reference Villani29, p. 207]):

(4.6)\begin{equation} W_1 (\mu, \nu) := \sup \left\{ \displaystyle\int \psi \, d(\mu-\nu) \;\mathord{;}\; \psi \colon X \to [{-}1,1] \text{ is $1$-Lipschitz} \right\} . \end{equation}

Of course, the Kantorovich metric depends on the original metric $\mathrm {d}$ on $X$; in fact, it ‘remembers’ it, since $W_1(\delta _x ,\, \delta _y) = \mathrm {d}(x,\,y)$. The metric space $(\mathcal {P}_1(X),\, W_1)$ is called $1$-Wasserstein space; it is separable and complete. Unless $X$ is compact, the topology on $\mathcal {P}_1(X)$ is stronger than the weak topology. In fact, we have the following characterizations of convergence:

The next lemma should be compared to Corollary 4.2:

Proof. Fix $\epsilon >0$. By part (d) of Theorem 4.3, there exists $R>0$ such that, for all sufficiently large $n$,

(4.8)\begin{equation} \displaystyle\int_{X \setminus B_R(x_0)} [1+\mathrm{d}(x,x_0)]\, d\mu_n (x) \le \epsilon . \end{equation}

Then, we write:

(4.9)\begin{align} & \displaystyle\int \phi_n \, d\mu_n - \int \phi \, d\mu = \int (\phi_n-\phi) \, d\mu_n + \int \phi \, d(\mu_n-\mu)\nonumber\\ & \quad=\underbrace{\int_{X \setminus B_R(x_0)} (\phi_n-\phi) \, d\mu_n}_{\unicode{x2460}}^{\vphantom{1}}} + \underbrace{\int_{B_R(x_0)} (\phi_n-\phi) \, d\mu_n}_{\unicode{x2461}}^{\vphantom{1}}} + \underbrace{\int \phi \, d(\mu_n-\mu)}_{\unicode{x2462}}^{\vphantom{1}}}. \end{align}

By part (b) of Theorem 4.3, the term ${\unicode{x2462}}$ tends to $0$ as $n \to \infty$. By the assumption of uniform convergence on bounded sets, $\left|{\unicode{x2461}}\right| \le \sup _{B_{2R}(x_0)} |\phi _n - \phi |$ tends to $0$ as well. Finally,

(4.10)\begin{equation} \left| {\unicode{x2460}} \right| \le \displaystyle\int_{X \setminus B_{R}(x_0)} (|\phi_n| + |\phi| ) \, d\mu_n \le 2 C \int_{X \setminus B_R(x_0)} [1+\mathrm{d}(x,x_0)]\, d\mu_n (x) \le 2C \epsilon ,\end{equation}

for all sufficiently large $n$. Since $\epsilon >0$ is arbitrary, we conclude that $\int \phi _n \, d\mu _n \to \int \phi \, d\mu$, as claimed.

Let us now consider the specific case of the metric space $(X,\,\mathrm {d}) = (\mathbb {R}_+,\, \mathrm {d}_\mathrm {HS})$, where:

(4.11)\begin{equation} \mathrm{d}_\mathrm{HS}(x,y) := \left| \log(1+x) - \log(1+y) \right| . \end{equation}

Then the finite-first-moment condition (4.5) becomes our usual integrability condition (2.6), so the $1$-Wasserstein space $\mathcal {P}_1(X)$ becomes ${{\mathcal {P}_\mathrm {HS}(\mathbb {R}_{+ })}}$.

4.2. Lower and upper semicontinuity

The HS barycenter is definitely not continuous with respect to the weak topology, since the complement of ${{\mathcal {P}_\mathrm {HS}(\mathbb {R}_{+ })}}$ is dense in $\mathcal {P}(\mathbb {R}_+ )$, and the barycenter is $\infty$ there (by Proposition 2.5). Nevertheless, lower semicontinuity holds, except in the critical configuration:

Theorem 4.5 For every $(\mu,\,c) \in \mathcal {P}(\mathbb {R}_+ ) \times [0,\,1),$ we have:

(4.12)\begin{equation} \liminf_{(\tilde \mu, \tilde c) \to (\mu,c)} [\tilde \mu]_{\tilde c} \ge [\mu]_c \quad \text{unless} \quad \mu(0)=1-c \text{ and } [\mu]_c > 0 .\end{equation}

To be explicit, the inequality above means that for every $\lambda < [\mu ]_c$, there exists a neighborhood $\mathcal {U} \subseteq \mathcal {P}(\mathbb {R}_+ ) \times [0,\,1)$ of $(\mu,\,c)$ with respect to the product topology (weak $\times$ standard) such that $[\tilde \mu ]_{\tilde c} > \lambda$ for all $(\tilde \mu,\, \tilde c) \in \mathcal {U}$.

Proof. Let us first note that:

(4.13)\begin{equation} \mu(0)=1-c \text{ and } [\mu]_c > 0 \quad \Rightarrow \quad \liminf_{(\tilde \mu, \tilde c) \to (\mu,c)} [\tilde \mu]_{\tilde c} < [\mu]_c . \end{equation}

Indeed, if $c_n := c +1/n$, then $(\mu,\, c_n) \to (\mu,\, c)$. By Proposition 2.11, we have that $[\mu ]_{c_n}=0$. Thus,

(4.14)\begin{equation} 0= \lim_{n \to \infty} [\mu]_{c_n} =\liminf_{(\tilde \mu, \tilde c) \to (\mu,c)} [\tilde \mu]_{\tilde c} < [\mu]_c .\end{equation}

We now consider the converse implication: given $(\mu,\,c) \in \mathcal {P}(\mathbb {R}_+ ) \times [0,\,1)$ such that $\mu (0) \neq 1-c$ or $[\mu ]_c = 0$, we want to show that $\liminf _{(\tilde \mu, \tilde c) \to (\mu,c)} [\tilde \mu ]_{\tilde c} \ge [\mu ]_c$. There are some trivial cases:

• • If $[\mu ]_c=0$, then the conclusion is obvious.

• • If $\mu (0)> 1-c$, then $[\mu ]_c=0$ by Proposition 2.11, and again the conclusion is clear.

In what follows, we assume that $\mu (0)<1-c$ and $[\mu ]_c>0$. Fix a sequence $(\mu _n,\,c_n)$ converging to $(\mu,\,c)$. We need to prove that:

(4.15)\begin{equation} \liminf_{n \to \infty} [\mu_n]_{c_n} \ge [\mu]_c . \end{equation}

We may also assume without loss of generality that $[\mu _n]_{c_n}<\infty$ for each $n$. We divide the proof in two cases:

Case $c>0$: We can also assume that $c_n >0$ for every $n$. By Proposition 2.5, the hypothesis $[\mu _n]_{c_n}<\infty$ means that $\mu _n \in {{\mathcal {P}_\mathrm {HS}(\mathbb {R}_{+ })}}$. By Portmanteau's Theorem [Reference Parthasarathy23, Theorem 6.1(c)], $\limsup _{n \to \infty } \mu _n(0) \leq \mu (0) < 1-c$. Thus, for sufficiently large values of $n$ we have $\mu _n(0) < 1-c_n$. In this setting, the HS barycenter $[\mu _n]_{c_n}$ can be computed by Proposition 2.7. Recall from (2.16) that $\eta (\tilde {\mu },\,\tilde {c})$ denotes the unique positive solution of the equation $\int \frac {x}{\tilde {c}x + (1-\tilde {c}) \eta } \, d\tilde {\mu } = 1$. Note that $\eta (\mu,\,c)$ is well defined even in the case $[\mu ]_{c}= \infty$ (see Proposition 2.8). We claim that:

(4.16)\begin{equation} \lim_{n \to \infty} \eta(\mu_n, c_n) = \eta(\mu,c) . \end{equation}

Proof Proof of the claim

Fix numbers $\alpha _0$, $\alpha _1$ with $0< \alpha _0 < \eta (\mu,\,c) < \alpha _1$. Then a monotonicity property shown in the proof of Proposition 2.7 gives:

(4.17)\begin{equation} \displaystyle\int \displaystyle\frac{x}{cx + (1-c) \alpha_0} \, d\mu > 1 > \int \frac{x}{cx + (1-c) \alpha_1} \, d\mu . \end{equation}

Note the uniform bounds:

(4.18)\begin{equation} 0 \le \displaystyle\frac{x}{c_n x + (1-c_n) \alpha_i} \le (\inf_n c_n)^{{-}1} . \end{equation}

So, using Corollary 4.2, we see that $\int \frac {x}{c_n x + (1-c_n) \alpha _i} \, d\mu _n \to \int \frac {x}{cx + (1-c) \alpha _i} \, d\mu$. In particular, for all sufficiently large $n$,

(4.19)\begin{equation} \displaystyle\int \displaystyle\frac{x}{c_n x + (1-c_n) \alpha_0} \, d\mu_n > 1 > \int \frac{x}{c_n x + (1-c_n) \alpha_1} \, d\mu_n , \end{equation}

and thus, $\alpha _0 < \eta (\mu _n,\,c_n) < \alpha _1$, proving the claim (4.16).

For simplicity, write $y_n := \eta (\mu _n,\, c_n)$. By Proposition 2.2.(b),

(4.20)\begin{equation} K(x, y_n, c_n) \ge K(0, y_n, c_n) = \log y_n + c_n^{{-}1} \log(1-c_n) \ge - C \end{equation}

for some finite $C$, since $\sup _n c_n < 1$ and $\inf _n y_n > 0$. This allows us to apply Proposition 4.1 and obtain:

(4.21)\begin{equation} \liminf_{n \to \infty} \displaystyle\int K(x, y_n, c_n) \, d\mu_n \ge \int K(x, y_\infty, c) \, d\mu , \end{equation}

where $y_\infty := \lim _{n\to \infty } y_n = \eta (\mu,\,c)$. Using formula (2.10), we obtain (4.15). This completes the proof of Theorem 4.5 in the case $c>0$.

Case $c=0$: By Proposition 4.1 we obtain, $\liminf \int x \, d\mu _n \ge \liminf \int x \, d\mu$, that is, $\liminf [\mu _n]_0 \ge [\mu ]_c$. So we can assume that $c_n >0$ for every $n$, like in the previous case.

In order to prove (4.15) in the case $c=0$, let us fix an arbitrary positive $\lambda < [\mu ]_0 = \int x \, d \mu$, and let us show that $[\mu _n]_{c_n} > \lambda$ for every sufficiently large $n$. By the monotone convergence theorem, there exists $m \in \mathbb {N}$ such that $\int \min (x,\,m) \, d\mu (x) > \lambda$. Let $\hat {\mu }$ (respectively $\hat {\mu }_n$) be the pushforward of the measure $\hat {\mu }$ (respectively $\mu _n$) by the map $x \mapsto \min (x,\,m)$. Then $[\hat {\mu }]_0 > \lambda$ and, by Proposition 2.6.(c), $[\hat {\mu }_n]_{c_n} \le [\mu _n]_{c_n}$. Furthermore, we have $\hat \mu _n \rightharpoonup \hat \mu$, since for every $f \in C_\mathrm {b}(\mathbb {R}_+ )$,

(4.22)\begin{equation} \displaystyle\int f \, d\hat{\mu}_n =\int f(\min(x,m)) \, d\mu_n(x) \to\int f(\min(x,m)) \, d\mu(x) =\int f \, d\hat{\mu} .\end{equation}

So, to simplify the notation, we remove the hats and assume that the measures $\mu _n$, $\mu$ are all supported in the interval $[0,\,m]$.

The numbers $\eta (\mu _n,\,c_n)$ and $\eta (\mu,\,c)$ are well defined, as in the previous case, and furthermore they belong to the interval $[0,\,m]$, by Lemma 2.14. On the other hand, by Proposition 4.1,

(4.23)\begin{equation} \liminf_{n \to \infty} \displaystyle\int \displaystyle\frac{x}{c_n x + (1-c_n) \lambda} \, d\mu_n \ge \int \frac{x}{\lambda} \, d\mu > 1 . \end{equation}

It follows that $\eta (\mu _n,\,c_n) > \lambda$ for all sufficiently large $n$. We claim that $\eta (\mu _n,\,c_n) \to \eta (\mu,\,0)$, as before in (4.16). The proof is the same, except that the upper bound in (4.18) becomes infinite and must be replaced by the following estimate:

(4.24)\begin{equation} \left. \begin{array}{l@{}} 0 \le x \le m \\ \alpha_i \ge \lambda \end{array} \right\} \quad \Rightarrow \quad 0 \le \displaystyle\frac{x}{c_n x + (1-c_n) \alpha_i} \le \frac{m}{c_n m + (1-c_n) \alpha_i} \le \frac{m}{\lambda} . \end{equation}

So a repetition of the previous arguments yields (4.16), then (4.20) and (4.21), and finally (4.15). Therefore, Theorem 4.5 has been proved in both cases $c>0$ and $c=0$.

Next, let us investigate the behaviour of the HS barycenter on the product space ${{\mathcal {P}_\mathrm {HS}(\mathbb {R}_{+ })}} \times [0,\,1]$, where ${{\mathcal {P}_\mathrm {HS}(\mathbb {R}_{+ })}}$ is endowed with the topology defined in the end of § 4.1.

Theorem 4.6 For every $(\mu,\,c) \in {{\mathcal {P}_\mathrm {HS}(\mathbb {R}_{+ })}} \times [0,\,1]$, we have:

(4.25)\begin{align} \limsup_{(\tilde \mu, \tilde c) \to (\mu,c)} [\tilde \mu]_{\tilde c} & \le [\mu]_c \text{ unless } c=0 \text{ and } [\mu]_0 < \infty , \end{align}

(4.26)\begin{align} \liminf_{(\tilde \mu, \tilde c) \to (\mu,c)} [\tilde \mu]_{\tilde c} & \ge [\mu]_c \text{ unless } \mu(0)=1-c \text{ and } [\mu]_c > 0 . \end{align}

Proof of part (4.25) of Proposition 4.6. Let us start by proving the following implication:

(4.27)\begin{equation} \mu \in {{\mathcal{P}_\mathrm{HS}(\mathbb{R}_{+})}}, \ c=0, \text{ and } [\mu]_0 < \infty \quad \Rightarrow \quad \limsup_{(\tilde \mu, \tilde c) \to (\mu,c)} [\tilde \mu]_{\tilde c} > [\mu]_c . \end{equation}

Consider measures $\mu _n := \frac {n-1}{n} \, \mu + \frac {1}{n} \, \delta _n$. Clearly, $\mu _n \rightharpoonup \mu$; moreover,

(4.28)\begin{equation} \displaystyle\int \log(1+x) \, d(\mu-\mu_n)(x) = \displaystyle\frac{\log(1+n)}{n} \to 0 .\end{equation}

So using characterization (c) of Theorem 4.3, we conclude that $\mu _n \to \mu$ in the topology of ${{\mathcal {P}_\mathrm {HS}(\mathbb {R}_{+ })}}$. On the other hand, $[\mu _n]_0 = \frac {n-1}{n} \, [\mu ]_0 + 1 \to [\mu ]_0 + 1$. This proves (4.27).

Next, let us prove the converse implication. So, let us fix $(\mu,\,c)$ such that $c\neq 0$ or $[\mu ]_0 = \infty$, and let us show that if $(\mu _n,\, c_n)$ is any sequence in ${{\mathcal {P}_\mathrm {HS}(\mathbb {R}_{+ })}} \times [0,\,1]$ converging to $(\mu,\,c)$, then $\limsup _{n \to \infty } [\mu _n]_{c_n} \le [\mu ]_c$. This is obviously true if $[\mu ]_c = \infty$, so let us assume that $[\mu ]_c < \infty$. Then our assumption becomes $c>0$, so by removing finitely many terms from the sequence $(\mu _n,\, c_n)$, we may assume that $\inf _n c_n > 0$. Fix some finite number $\lambda > [\mu ]_c$. By Definition 2.3, there is some $y_0>0$ such that $\int K(x,\,y_0,\,c) \, d\mu (x) < \log \lambda$. The sequence of continuous functions $\frac {|K(x,\,y_0,\,c_n)|}{1+\log (1+x)}$ is uniformly bounded, as a direct calculation shows. Furthermore, $K(\mathord {\cdot },\,y_0,\, c_n) \to K(\mathord {\cdot },\,y_0,\,c)$ uniformly on compact subsets of $\mathbb {R}_+$. So Lemma 4.4 ensures that $\int K(x,\,y_0,\,c_n) \, d\mu _n(x) \to \int K(x,\,y_0,\,c) \, d\mu (x)$. Now it follows from Definition 2.3 that $[\mu _n]_{c_n} < \lambda$ for all sufficiently large $n$. Since $\lambda >[\mu ]_c$ is arbitrary, we conclude that $\limsup _{n \to \infty } [\mu _n]_{c_n} \le [\mu ]_c$, as we wanted to show.

Proof of part (4.26) of Proposition 4.6. First, we prove that, for all $(\mu,\,c) \in {{\mathcal {P}_\mathrm {HS}(\mathbb {R}_{+ })}} \times [0,\,1]$,

(4.29)\begin{equation} \mu(0)=1-c, \text{ and } [\mu]_c > 0 \quad \Rightarrow \quad \liminf_{(\tilde \mu, \tilde c) \to (\mu,c)} [\tilde \mu]_{\tilde c} < [\mu]_c . \end{equation}

Consider measures $\mu _n := \frac {n-1}{n} \, \mu + \frac {1}{n} \, \delta _0$. Clearly, $\mu _n \to \mu$ in the topology of ${{\mathcal {P}_\mathrm {HS}(\mathbb {R}_{+ })}}$. By Proposition 2.11.(b), $[\mu _n]_c=0$. This proves (4.29). For $c \in [0,\,1)$ the converse is a direct consequence of Theorem 4.5, since the topology on ${{\mathcal {P}_\mathrm {HS}(\mathbb {R}_{+ })}}$ is stronger. If $c=1$ and $[\mu ]_1=0$ then the result is obvious. If $[\mu ]_1>0$ then, as the example above shows, the result does not hold.

4.3. Continuity for extremal values of the parameter

Theorem 4.6 shows that the HS barycenter map on ${{\mathcal {P}_\mathrm {HS}(\mathbb {R}_{+ })}} \times [0,\,1]$ is not continuous at the pairs $(\mu,\,0)$ (except if $[\mu ]_0 = \infty$), nor at the pairs $(\mu,\,1)$ (except if $[\mu ]_1 = 0$). Let us see that continuity can be ‘recovered’ if we impose extra integrability conditions and work with stronger topologies.

If we use the standard distance $\mathrm {d}(x,\,y) := |x-y|$ on the half-line $X = \mathbb {R}_+ = [0,\,+\infty )$, then the resulting $1$-Wasserstein space is denoted $\mathcal {P}_1(\mathbb {R}_+ )$. On the other hand, using the distance

(4.30)\begin{equation} \mathrm{d}_\mathrm{g}(x,y) := \left| \log x - \log y \right| \quad \text{on the open half-line } X = \mathbb{R}_{++} , \end{equation}

the corresponding $1$-Wasserstein space will be denoted $\mathcal {P}_\mathrm {g}(\mathbb {R}_{+ + })$. We consider the latter space as a subset of $\mathcal {P}(\mathbb {R}_+ )$, since any measure $\mu$ on $\mathbb {R}_{+ + }$ can be extended to $\mathbb {R}_+$ by setting $\mu (0) = 0$. The topologies we have just defined on the spaces $\mathcal {P}_1(\mathbb {R}_+ )$ and $\mathcal {P}_\mathrm {g}(\mathbb {R}_{+ + })$ are stronger than the topologies induced by ${{\mathcal {P}_\mathrm {HS}(\mathbb {R}_{+ })}}$; in other words, the inclusion maps below are continuous:

(4.31)

Note that the ‘arithmetic barycenter’ $[\mathord {\cdot }]_0$ is finite on $\mathcal {P}_1(\mathbb {R}_+ )$, while the ‘geometric barycenter’ $[\mathord {\cdot }]_1$ is finite and non-zero on $\mathcal {P}_\mathrm {g}(\mathbb {R}_{+ + })$.

Finally, let us establish continuity of the HS barycenter for the extremal values of the parameter with respect to these new topologies:

Proof of part (a) of Proposition 4.7. Note that if $c_n=0$ for every $n \in \mathbb {N}$, the result is direct from the definition of the topology in $\mathcal {P}_1(\mathbb {R}_+ )$ (use characterization (c) of Theorem 4.3). We assume now that $c_n>0$, for every $n \in \mathbb {N}$. It is a consequence of Theorem 4.5 that:

(4.32)\begin{equation} \liminf_{(\mu_n, c_n) \to (\mu,0)} [\mu_n]_{c_n} \ge [\mu]_0.\end{equation}

The same proof as that of part (4.25) of Proposition 4.6 can be used to prove,

(4.33)\begin{equation} \limsup_{(\mu_n, c_n) \to (\mu,0)} [\mu_n]_{c_n} \le [\mu]_0.\end{equation}

Indeed, in the topology of $\mathcal {P}_1(\mathbb {R}_+ )$ the HS kernels $K(x,\,y,\,c_n)$ satisfy the assumptions of Lemma 4.4. For this, it suffices to notice that, for any fixed value $y_0>0$, the sequence of continuous functions $\frac {|K(x,\,y_0,\,c_n)|}{1+x}$ is uniformly bounded, and that $K(x,\,y_0,\,c_n) \to K(x,\,y_0,\,0)$ uniformly on compact subsets of $\mathbb {R}_+$.

Proof of part (b) of Proposition 4.7 In the case that $c_n=1$ for every $n \in \mathbb {N}$, the result is direct from the topology in $\mathcal {P}_\mathrm {g}(\mathbb {R}_{+ + })$ (use characterization (c) of Theorem 4.3). We assume that $c_n <1$, for every $n$. It is a consequence of (4.25) of Proposition 4.6 that:

(4.34)\begin{equation} \limsup_{(\mu_n, c_n) \to (\mu,1)} [\mu_n]_{c_n} \le [\mu]_1. \end{equation}

Recall that the HS barycenter is decreasing in the variable $c$; see Proposition 2.6.(e). In particular, $[\mu _n]_{c_n} \geq [\mu _n]_1$, for every $n \in \mathbb {N}$. Noticing that $\log x$ is a test function for the convergence in the topology of $\mathcal {P}_\mathrm {g}(\mathbb {R}_{+ + })$, we obtain:

(4.35)\begin{align} \liminf_{n \to \infty} [\mu_n]_{c_n} & \geq \liminf_{n \to \infty} [\mu_n]_1\\ & =\exp ( \liminf_{n \to \infty} \displaystyle\int \log x \, d \mu_n ) = \exp \int \log x \, d \mu = [\mu]_1. \nonumber \end{align}

The result follows combining (4.34) and (4.35).

The following observation complements part (b) of Proposition 4.7, since it provides a sort of lower semicontinuity property at $c=1$ under a weaker integrability condition:

Lemma 4.8 Let $c \in [0,\,1)$ and let $\mu \in \mathcal {P}(\mathbb {R}_+ )$ be such that $\log ^{-}(x) \in L^{1}(\mu )$. Then:

(4.36)\begin{equation} [\mu]_c \ge \exp \displaystyle\int ( c^{{-}1} \log^{+}(x) - \log^{-}(x) ) \, d\mu(x) .\end{equation}