Proof. Since $\mathfrak {a}_\bullet$ is linearly bounded, there exists $c_0 \in \mathbb {R}_{>0}$ such that $\mathfrak {a}_{\lambda } \subset \mathfrak {m}^{ \lceil c_0\lambda \rceil }$ for all $\lambda >0$. Since $\mathfrak {b}_1$ is an $\mathfrak {m}$-primary ideal, there exists $d\in \mathbb {Z}_{>0}$ such that $\mathfrak {m}^{d} \subset \mathfrak {b}_1$. If we set $c:= 2d/c_0$, then

\[ \mathfrak{a}_{ c\lambda } \subset \mathfrak{m}^{\lceil 2d \lambda \rceil } \subset \mathfrak{m}^{d \lceil \lambda \rceil } \subset (\mathfrak{b}_1)^{\lceil \lambda \rceil} \subset \mathfrak{b}_{\lceil \lambda \rceil } \subset \mathfrak{b}_{\lambda} \]

for all $\lambda \geq 1$. (The second inclusion uses that $d\lceil \lambda \rceil \leq d(\lambda +1) \leq 2d \lambda \leq \lceil 2 d \lambda \rceil$).

Proof. For any $\lambda \in \mathbb {R}_{>0}$,

\[ v(\mathfrak{a}_\lambda) = \lim_{m \to \infty} \frac{ v(\mathfrak{a}_\lambda^m)}{m} \geq \lim_{m\to\infty} \biggl( \frac{ v\big(\mathfrak{a}_{\lfloor \lambda m\rfloor }\big)}{\lfloor \lambda m\rfloor } \cdot \frac{ \lfloor \lambda m \rfloor}{m} \biggr)= \lambda v(\mathfrak{a}_\bullet) , \]

where the inequality uses that $(\mathfrak {a}_{\lambda })^m \subset \mathfrak {a}_{\lambda m} \subset \mathfrak {a}_{ \lfloor \lambda m \rfloor }$. Therefore, $\mathfrak {a}_{\lambda } \subset \widetilde {\mathfrak {a}}_\lambda$, which is (1).

For (2), note that $v(\mathfrak {a}_\bullet ) \leq {v(\widetilde {\mathfrak {a}}_m ) }/{m} \leq {v(\mathfrak {a}_m)}/{m}$ for each $m \in \mathbb {Z}_{>0}$, where the first inequality follows from the definition of $\widetilde {\mathfrak {a}}_m$ and the second from (1). Sending $m\to \infty$ gives $v(\mathfrak {a}_\bullet ) \leq v(\widetilde {\mathfrak {a}}_\bullet )\leq v(\mathfrak {a}_\bullet )$, which implies (2). Statement (3) follows immediately from (2).

Proof. First, note that $v(\mathfrak {b}_{c \bullet }) =c v(\mathfrak {b}_\bullet )$ for all $v\in \mathrm {Val}_{R,\mathfrak {m}}$. Therefore, (1) implies (2) follows from Lemma 3.8(2), while (2) implies (1) follows from the definition of the saturation.

Proof. By [Reference Huneke and SwansonHS06, Theorem 9.3.5], there is a bijective map $\varphi _* \colon \mathrm {DivVal}_{\widehat {R},\widehat {\mathfrak {m}}} \to \mathrm {DivVal}_{R,\mathfrak {m}}$ that sends $\hat {v}$ to the valuation $v$ defined by composition

\[ {\rm Frac}(R)^\times \hookrightarrow {\rm Frac}(\hat{R})^\times \overset{\hat{v}}{\to} \mathbb{R} . \]

Note that $\hat {v}(\mathfrak {a} \hat {R}) = v(\mathfrak {a})$ for any ideal $\mathfrak {a}\subset R$.

Fix $\lambda \in \mathbb {R}_{>0}$. Using the previous observation twice and the definition of the saturation, we compute

\[ \hat{v} (\widetilde{\mathfrak{a}}_\lambda \hat{R} ) = v(\widetilde{\mathfrak{a}}_\lambda) \geq \lambda v (\mathfrak{a}_\bullet) = \lambda \hat{v}(\mathfrak{b}_\bullet) . \]

Thus, $\widetilde {\mathfrak {a}}_{\lambda } \hat {R} \subset \widetilde {\mathfrak {b}}_\lambda$. To prove the reverse inclusion, note that $\mathfrak {c} R = \widetilde {\mathfrak {b}}_\lambda$ for some ideal $\mathfrak {c}\subset R$, since $\widetilde {\mathfrak {b}}_\lambda$ is $\widehat {\mathfrak {m}}$-primary. As before, we compute

\[ v(\mathfrak{c})= \hat{v}(\widetilde{\mathfrak{b}}_\lambda) \geq \lambda \hat{v}(\mathfrak{b}_\bullet) =\lambda v (\mathfrak{a}_\bullet) . \]

Therefore, $\widetilde {\mathfrak {b}}_\lambda = \mathfrak {c} \hat {R} \subset \widetilde {\mathfrak {a}}_\lambda \hat {R}$.

Proof. Since $R/\mathfrak {a}_\lambda$ and $\widehat {R}/\mathfrak {b}_\lambda$ are isomorphic as Artinian rings, $\ell (R/\mathfrak {a}_\lambda )=\ell (\widehat {R}/\mathfrak {b}_\lambda )$ for each $\lambda >0$. Therefore, the equality of multiplicities holds.

Proof. Suppose the statement is false. Then there exists $v\in \mathrm {Val}_{R,\mathfrak {m}}^+$ such that $v(\mathfrak {b}_\bullet ) < v(\mathfrak {a}_\bullet )$. After scaling $v$, we may assume $v(\mathfrak {a}_\bullet )=1$. Therefore, there exists $l \in \mathbb {Z}_{>0}$ so that $v(\mathfrak {b}_{l}) /l < v(\mathfrak {a}_\bullet )=1$. Hence, we may choose $f\in \mathfrak {b}_l\setminus \mathfrak {a}_l$ such that $k:= v(f)< l$.

Now, consider the map

\[ \phi: R \to \mathfrak{b}_{l m}/ \mathfrak{a}_{lm}\quad \text{sending}\quad g \mapsto f^m g. \]

We claim $\ker (\phi ) \subset \mathfrak {a}_{ (l-k)m } (v)$. Indeed, if $g\in \ker \phi$, then $f^m g \in \mathfrak {a}_{lm}$. Hence,

\[ v(g) =v(f^m g) - v(f^m) \geq v(\mathfrak{a}_{l m}) - km \geq l m - km = (l-k) m \]

as desired. Using the claim, we deduce

(3.1)\begin{equation} \ell( \mathfrak{b}_{l m}/ \mathfrak{a}_{lm}) \geq \ell (R/ \ker \phi) \geq \ell ( R/ \mathfrak{a}_{ (l-k)m } (v)) . \end{equation}

Finally, we compute

\[ \mathrm{e}(\mathfrak{a}_\bullet) - \mathrm{e}(\mathfrak{b}_\bullet) = \lim_{ m \to \infty} \frac{ \ell (\mathfrak{b}_{lm } / \mathfrak{a}_{lm})}{(lm)^n /n!} \geq \lim_{m\to\infty} \frac{\ell ( R/ \mathfrak{a}_{ (l-k) m }(v) )}{ (lm)^n /n!} = \frac{(l-k)^n \,\mathrm{vol}(v)}{l^n} >0 , \]

where the first inequality is by (3.1) and the second uses that $v$ has positive volume. This contradicts our assumption that $\mathrm {e}(\mathfrak {a}_\bullet )=\mathrm {e}(\mathfrak {b}_\bullet )$.

Proof. We claim that there exists a sequence of proper birational morphisms

\[ \cdots\to X_3 \to X_2\to X_1\to X_0 = X \]

such that the each $X_i$ is normal and the sheafs $\mathfrak {a}_{l} \cdot \mathcal {O}_{X_m}$ and $\mathfrak {b}_{l}\cdot \mathcal {O}_{X_m}$ are line bundles when $l\leq m$. Such a sequence can be constructed inductively as follows. Let $X_{i} \to X_{i-1}$ be defined by the composition

\[ X_{i} \to X_{i,2}\to X_{i,1} \to X_{i-1}, \]

where $X_{i,1}$ is the blowup of $\mathfrak {a}_i \cdot \mathcal {O}_{X_{i-1}}$, $X_{i,2}$ is the blowup of $\mathfrak {b}_i \cdot \mathcal {O}_{X_{i,1}}$, and $X_i$ the normalization of $X_{i,2}$. Note that $X_{i,2} \to X$ is proper, since $X_{i,2}\to X_{i-1}$ is a blowup and $X_{i-1}\to X$ is proper by our inductive assumption. Since $X$ is Nagata and $X_{i,2}\to X$ is finite type, $X_{i,2}$ is Nagata by [Sta, Lemma 035A]. Therefore, $X_i\to X_{i,2}$ is finite by the definition of Nagata, and we conclude the composition $X_i \to X_{i-1}$ is proper, which completes the proof of the claim.

Next, consider a proper birational morphism $Y\to X$ with $Y$ normal and factoring as

\[ Y \overset{\pi_m}{\longrightarrow} X_m \longrightarrow X, \]

For $l\leq m$, there exist Cartier divisors $G_l$ and $G'_l$ on $Y$ such that

\[ \mathfrak{a}_l \cdot \mathcal{O}_{Y} = \mathcal{O}_{Y} (G_l) \quad \text{ and } \quad \mathfrak{b}_l\cdot\mathcal{O}_{Y}=\mathcal{O}_{Y}(G_l') . \]

Set $F_l:= {G_l}/{l}$ and $F'_l:= {G'_l}/{l}$, which are relatively nef over $X$ and satisfy

\[ F_l=- \sum_{E \subset Y} \frac{\mathrm{ord}_E(\mathfrak{a}_l)}{l} E \quad \text{ and }\quad F'_l =- \sum_{E \subset Y} \frac{\mathrm{ord}_E(\mathfrak{b}_l)}{l} E, \]

where the sums run through prime divisors $E\subset Y$. Throughout the proof, we will without mention replace $Y$ with higher birational models so it factors through certain $X_m\to X$.

Given $\epsilon >0$, by Proposition 2.6 and (2.1), there exists $m_0>0$ such that

(3.2)\begin{equation} - F_{m_0}^{n-1} \cdot F_{m_0} =\mathrm{e}(\mathfrak{a}_{m_0})/{m_0}^n< \mathrm{e}(\mathfrak{a}_\bullet) + \epsilon. \end{equation}

For a multiple $m_1$ of $m_0$,

\[ -(F_{m_0})^{n-1} \cdot F_{m_1}< \mathrm{e}(\mathfrak{a}_\bullet) + \epsilon , \]

since $F_{m_0} \leq F_{m_1}$ and $F_{m_0}$ is nef over $X$. Now, we compute

\begin{align*} - (F_{m_0})^{n-1} \cdot F_{m_1} + (F_{m_0})^{n-1} \cdot F'_{m_1} &= \sum_{E\subset Y } \big(\!- \mathrm{ord}_{E} (F_{m_1}) +\mathrm{ord}_{E} (F'_{m_1}) \big) (F_{m_0})^{n-1} \cdot E\\ &= \sum_{E\subset X_{m_0} } \big(\!-\mathrm{ord}_{E} (F_{m_1}) +\mathrm{ord}_{E} (F'_{m_1}) \big) ({\pi_{m_0}}_* F_{m_0})^{n-1} \cdot E, \end{align*}

where the second equality uses that $F_{m_0}$ is the pullback of ${\pi _{m_0}}_* F_{m_0}$ and the projection formula [Reference KleimanKle05, Proposition B.16]. Since ${\pi _{m_0}}_* F_{m_0}$ is nef over $X$ and

\[ \lim_{m\to \infty} \big(\!-\mathrm{ord}_{E} (F_m) + \mathrm{ord}_{E}(F'_m) \big) = \mathrm{ord}_E(\mathfrak{a}_\bullet) -\mathrm{ord}_E(\mathfrak{b}_\bullet) \geq 0 , \]

we may choose $m_1$ sufficiently large and divisible by $m_0$ such that

\[ -(F_{m_0})^{n-1} \cdot F'_{m_1} \leq - F_{m_0}^{n-1} \cdot F_{m_1} + \epsilon. \]

Therefore,

\[ - (F_{m_0})^{n-2} \cdot F'_{m_1} \cdot F_{m_0} = - (F_{m_0})^{n-1} \cdot F'_{m_1} \leq \mathrm{e}(\mathfrak{a}_\bullet) + 2\epsilon , \]

where the first equality is Proposition 2.5.

For any multiple $m_2$ of $m_1$,

\[ - (F_{m_0})^{n-2} \cdot F'_{m_1} \cdot F_{m_2} \leq \mathrm{e}(\mathfrak{a}_\bullet) + 2\epsilon, \]

since $F_{m_0} \leq F_{m_2}$ and the terms $F'_{m_1}$ and $F_{m_0}$ are nef over $X$. Similarly to the previous paragraph, we compute

\begin{align*} &- (F_{m_0})^{n-2} \cdot F'_{m_1} \cdot F_{m_2} - (F_{m_0})^{n-2} \cdot F'_{m_1} \cdot F'_{m_2}\\ &\qquad =\sum_{E \subset X_{m_1} } \big(\!- \mathrm{ord}_{E} (F_{m_2})+ \mathrm{ord}_{E} (F'_{m_2}) \big) ({ ({\pi_{m_1}}_* F_{m_0})^{n-1} \cdot \pi_{m_1}}_*F'_{m_1}) \cdot E. \end{align*}

and, hence, we may choose $m_2$ sufficiently large and divisible by $m_1$ so that

\[ -(F_{m_0})^{n-2} \cdot F'_{m_1} \cdot F'_{m_2} <-(F_{m_0})^{n-2} \cdot F'_{m_1} \cdot F_{m_2}+ \epsilon. \]

Thus,

\[ -(F_{m_0})^{n-2} \cdot F'_{m_1} \cdot F'_{m_2} < \mathrm{e}(\mathfrak{a}_\bullet) +3\epsilon. \]

Repeating in this way gives

\[ -F'_{m_1} \cdot F'_{m_2} \cdot \cdots \cdot F'_{m_n} < \mathrm{e}(\mathfrak{a}_\bullet)+ (n+1)\epsilon , \]

where each $m_i$ divides $m_{i+1}$. Since each $F'_{m_i}$ is nef over $X$ and $F'_{m_{i}} \leq F'_{m_{i+1}}$, we see that

\[ -(F'_{m_n})^n < \mathrm{e}(\mathfrak{a}_\bullet)+ n\epsilon. \]

Using that $\mathrm {e}(\mathfrak {b}_\bullet ) = \inf _{m} \mathrm {e}(\mathfrak {b}_m)/m^n =\inf _{m} -(F'_{m})^n$ by Lemma 2.6 and (2.1), we see that $\mathrm {e}(\mathfrak {b}_\bullet ) < \mathrm {e}(\mathfrak {a}_\bullet ) + (n+1)\epsilon$. Therefore, $\mathrm {e}(\mathfrak {b}_\bullet )\leq \mathrm {e}(\mathfrak {a}_\bullet )$.

Proof of Theorem 1.4 By Propositions 3.10 and 3.11, we may assume $(R,\mathfrak {m})$ is complete. This condition will be needed below to apply Proposition 3.13.

By Propositions 3.12 and 3.13, $\mathrm {e}(\mathfrak {a}_\bullet ) = \mathrm {e}(\mathfrak {b}_\bullet )$ if and only if $v(\mathfrak {a}_\bullet ) = v(\mathfrak {b}_\bullet )$ for all $v\in \mathrm {DivVal}_{R,\mathfrak {m}}$. By Proposition 3.9, the latter condition holds if and only if $\widetilde {\mathfrak {a}}_\bullet = \widetilde {\mathfrak {b}}_\bullet$.

Proof. By Lemma 3.8, $\mathfrak {a}_\bullet \subset \widetilde {\mathfrak {a}}_\bullet$ and $\widetilde {\mathfrak {a}}$ is saturated. Thus, Theorem 1.4 implies $\mathrm {e}(\mathfrak {a}_\bullet ) = \mathrm {e}(\widetilde {\mathfrak {a}}_\bullet )$.

Proof. If $\mathfrak {a}_\bullet$ is linearly bounded, then there exists $c>0$ such that $\mathfrak {a}_{\lambda } \subset \mathfrak {m}^{\lceil c \lambda \rceil }$ for all $\lambda >0$. Thus, $\mathrm {e}(\mathfrak {a}_\bullet ) \geq c^{n} \mathrm {e}(\mathfrak {m}) >0$ as desired.

Next, assume $\mathrm {e}(\mathfrak {a}_\bullet )>0$. We claim that there exists $v\in \mathrm {DivVal}_{R,\mathfrak {m}}$ such that $v(\mathfrak {a}_\bullet )>0$. If not, then $\widetilde {\mathfrak {a}}_\lambda = \mathfrak {m}$ for all $\lambda >0$. Using Corollary 3.16, we then see $\mathrm {e}(\mathfrak {a}_\bullet ) = \mathrm {e}(\widetilde {\mathfrak {a}}_\bullet ) =0$, which is a contradiction. Now, fix $v\in \mathrm {DivVal}_{R,\mathfrak {m}}$ with $v(\mathfrak {a}_\bullet ) >0$. Using that $\mathfrak {a}_\bullet \subset \mathfrak {a}_{\bullet }(v)$ and Proposition 2.7, we conclude $\mathrm {e}(\mathfrak {a}_\bullet )\geq \mathrm {e}(\mathfrak {a}_\bullet (v)) >0$.

Proof. Suppose the statement is false. Then there exists some $\lambda \in \mathbb {R}_{>0}$ such that $\mathfrak {a}_\lambda \subsetneq \widetilde {\mathfrak {a}}_{\lambda }$. Thus, there exist $f\in \widetilde {\mathfrak {a}}_{\lambda }$ and $i \in I$ such that $v_i(f) < \lambda c_i$. We claim $v_i(\widetilde {\mathfrak {a}}_\bullet )< v_i(\mathfrak {a}_\bullet )$. Indeed,

\[ v_i(\widetilde{\mathfrak{a}}_\bullet)= \lim_{m \to \infty} \frac{ v_i(\widetilde{\mathfrak{a}}_{ \lfloor \lambda m\rfloor })}{ \lfloor \lambda m \rfloor } \leq \lim_{m\to \infty} \frac{ v_i(f^m)}{ \lfloor \lambda m \rfloor } = \frac{ v_i(f) }{\lambda}< c_i, \]

where the second equality uses that $f^m \in (\widetilde {\mathfrak {a}}_{\lambda })^m \subset \widetilde {\mathfrak {a}}_{\lambda m} \subset \widetilde {\mathfrak {a}}_{\lfloor \lambda m \rfloor }$. Since $v_i(\mathfrak {a}_\bullet ) \geq c_i$, the claim holds.

By Proposition 3.12 and the claim, $\mathrm {e}(\widetilde {\mathfrak {a}}_\bullet )< \mathrm {e}(\mathfrak {a}_\bullet )$. The latter contradicts Theorem 1.4.

Proof. For $\mathfrak {m}$-primary ideals $\mathfrak {c}\subset R$ and $\mathfrak {d}\subset R$,

\[ (\mathfrak{c} + \mathfrak{d} )\widehat{R} = \mathfrak{c} \widehat{R} + \mathfrak{d} \widehat{R} \quad \text{ and } \quad (\mathfrak{c} \cap \mathfrak{d} )\widehat{R} = \mathfrak{c} \widehat{R} \cap \mathfrak{d} \widehat{R} . \]

Therefore,

\[ (\mathfrak{a}_{a, 0} \cap \mathfrak{a}_{b,1} ) \widehat{R} = \mathfrak{b}_{a, 0} \cap \mathfrak{b}_{b,1} \quad \text{ and } \quad \mathfrak{a}_{a,t} \widehat{R} = \mathfrak{b}_{a, t} . \]

Statements (1) and (2) follow from these equalities.

Proof. By Lemma 4.4, it suffices to prove the result when $(R,\mathfrak {m})$ is complete. In this case we continue with the notation from the previous proof, but additionally fix $N>\max \{{a}/({1-t}), {b}/{t}\}$. Now, consider the filtration $\mathcal {F}^\bullet _t$ of $V$ defined by $\mathcal {F}_t^\lambda V:= {\rm im} ( \mathfrak {a}_{\lambda,t} \to V)$ for each $\lambda \in \mathbb {R}$. Observe that

(4.2)\begin{equation} \mathcal{F}_t^{ma} V = \sum_{ ma = (1-t) \mu + t \nu } \mathcal{F}_0^\mu V \cap \mathcal{F}_1^\nu V = {\rm span} \langle s_i \mid (1-t)\lambda_{i,0} +t \lambda_{i,1} \geq ma \rangle, \end{equation}

where the sum runs through all $\mu, \nu \in \mathbb {R}$ satisfying $ma=(1-t)\mu +t \nu$. Now, we compute

(4.3)\begin{align} \ell \big( R/\mathfrak{a}_{ma,t} \big) &= \dim_\kappa V / \mathcal{F}_t^{ma} V = \dim_\kappa V- \dim_\kappa \mathcal{F}_t^{ma} V \nonumber\\ &= \# \{ i \mid (1-t) \lambda_{i,0} + t\lambda_{i,1} < ma \} = \frac{m^n}{n!}\mu_m\big( \{ (1-t)x + ty < a\} \big) \end{align}

where the first equality uses that $\mathfrak {m}^{NCm} \subseteq \mathfrak {a}_{ma/(1-t),0} + \mathfrak {a}_{mb/t,1} \subset \mathfrak {a}_{ma,t}$, the third (4.2), and the fourth (4.1). Therefore, the first formula in the proposition holds.

To verify the second formula, we fix $N> \max \{a,b\}$ and similarly compute

\begin{align*} \dim_\kappa R/( \mathfrak{a}_{ma,0} \cap \mathfrak{a}_{mb,1} ) &= \dim_\kappa V / (\mathcal{F}_0^{ma} V \cap \mathcal{F}_1^{mb} V) = \dim_\kappa V- \dim_\kappa \mathcal{F}_0^{ma} V \cap \mathcal{F}_1^{mb} V\\ &= \# \{ j \mid \lambda_{i,0} < ma \text{ or } \lambda_{i,1} < mb \} = \frac{m^n}{n!}\mu_m\big( \{ x< a \} \cup \{y < b \} \big), \end{align*}

where the first equality uses that $\mathfrak {m}^{NCm} \subset \mathfrak {a}_{ma,0} \cap \mathfrak {a}_{mb,1}$.

Proof. We claim that $H_m$ converges to $H$ in $L^1_{ {\rm loc}}( \mathbb {R}^2)$. Assuming the claim, then $H_m$ converges to $H$ as distributions, and, hence, $\mu _m$ converges to $\mu := - ({ \partial ^2 H}/{\partial x \partial y})$ as distributions as well. Since each $\mu _m$ is a measure, [Reference HörmanderHör03, Theorem 2.1.9] implies $\mu$ is a measure and $\mu _m$ converges to $\mu$ weakly as measures.

It remains to prove the above claim. First, observe that $H_m$ converges to $H$ pointwise by the definition of the multiplicity as a limit. Next, fix an integer $N>0$. For $-N\leq x,y< N$, observe that

\[ 0 \leq H_m(x,y) \leq \frac{ \ell \big( R/ \mathfrak{m}^{N Cm} \big) }{ m^n/n! } \leq \sup_{m \geq 1} \frac{\ell \big( R/ \mathfrak{m}^{NCm}\big) }{ m^n/n! } <+\infty . \]

Therefore, the dominated convergence theorem implies that

\[ \lim_{m \to \infty} \int_{ (-N,N)\times (-N,N) } |H_m-H| =0. \]

Since $N>0$ was arbitrary, $H_m$ converges to $H$ in $L^1_{ {\rm loc}}( \mathbb {R}^2)$ as desired.

Proof. Fix $\epsilon >0$. Since $(\mu _{m})_{m\geq 1}$ converges to $\mu$ weakly as measures by Proposition 4.6,

\begin{align*} \limsup \mu_m\big( \{ x(1-t) +y t< a-\epsilon \} \big) &\leq \mu\big( \{ x(1-t) +y t< a\}\big)\\ &\leq \liminf \mu_m \big( \{x(1-t)+yt < a\} \big) \end{align*}

Using Proposition 4.5(1) to compute the lim sup and lim inf, we deduce

\[ (a-\epsilon)^n\mathrm{e}(\mathfrak{a}_{\bullet,t}) = \mathrm{e}(\mathfrak{a}_{(a-\epsilon)\bullet,t}) \leq \mu\big( \{ x(1-t)+yt< a\}\big) \leq \mathrm{e}(\mathfrak{a}_{a\bullet,t})= a^n \mathrm{e}(\mathfrak{a}_{\bullet,t}). \]

Sending $\epsilon \to 0$ completes the proof of (1). The proof of the second is similar, but uses Proposition 4.5(2).

Proof. If (1) holds, then

\[ \mu\big(\{ x< 1\} \big) = \mu \big(\{ x <1\} \cup \{ y < c \} \big) = \mu\big( \{y< c \}\big). \]

Proposition 4.7 then implies (2) holds.

For the reverse implication, assume (2) holds. We claim

\[ H(x,y)= \begin{cases} \mathrm{e}(\mathfrak{a}_{x\bullet,0} ), & \text{if } cx\geq y,\\ \mathrm{e}(\mathfrak{a}_{ y\bullet,1} ), & \text{if } y\geq cx. \end{cases} \]

Indeed, if $cx\geq y$, then

\[ \mathrm{e}(\mathfrak{a}_{x\bullet ,0} ) \leq H(x,y) \leq \mathrm{e}(\mathfrak{a}_{x \bullet,0 } \cap \mathfrak{a}_{cx\bullet ,1}) = \mathrm{e}(\mathfrak{a}_{ x\bullet ,0}) . \]

Thus, the claim holds when $cx\geq y$ and similar reasoning treats the case when $cx< y$. Using the claim, we conclude (1) holds.

Proof. By Propositions 4.3 and 4.6, $\mathrm {supp}(\mu ) \subset \{({1}/{D}) x \leq y \leq D x\} \cup \{ x=0\} \cup \{y =0 \}$. Note that

\[ \mu( \{x =0 \}) = \lim_{ a \to 0^+} \mu (\{ 0\leq x< a \} ) = \lim_{a \to 0^+} \mathrm{e}( \mathfrak{a}_{a \bullet,0}) = \lim_{a \to 0^+} a^n \mathrm{e}( \mathfrak{a}_{ \bullet,0}) = 0, \]

where the second equality is by Proposition 4.7. Similar reasoning shows $\mu (\{ y=0 \})$. Thus, (1) holds. Statement (2) follows from the fact that $H$ is homogeneous of degree $n$, meaning $H(cx,cy) = c^n H(x,y)$ for each $c>0$ and $(x,y) \in \mathbb {R}^2$.

Proof. Using that $\mathbb {R}\cdot j ({1}/({c+1}))= \{cx=y \}$ and the definition of $\widetilde {\mu }$, we see that (1) holds if and only if ${\rm supp}(\mu \vert _{\{0\leq x+y<1\}}) \subset \{cx=y \}$. Since $\mu$ is homogeneous by Proposition 4.9(2), ${\rm supp} (\mu \vert _{\{0\leq x+y<1\}}) \subset \{xc=y\}$ if and only if ${\rm supp}(\mu ) \subset \{cx=y\}$. By Lemma 4.8, the latter condition holds if and only if (2) holds.

Proof. For simplicity, we assume $t \leq \tfrac {1}{2}$, which ensures $g_t(z): =(1-t)z +t(1-z)$ is non-decreasing. (The case when $t>\tfrac {1}{2}$ can be proved similarly, but using left-hand approximations below.) For each $m\geq 1$, consider the simple function

\[ g_{t,m}(z) = \sum_{k=1}^{2^m} g_{t}(k/2^m) \cdot 1_{I_{k,m} } \]

where $I_{1,m} := [0,{1}/{2^m}]$ and $I_{k,m}:= ( ({k-1})/{2^m}, {k}/{2^m}]$ when $1< k\leq 2^m$. Since $g_{t,m} \geq g_{t,m+1}$ and $g_{t,m}\to g$ pointwise as $m\to \infty$, the monotone convergence theorem implies

\[ \int g_t(z)^{-n} \, {d} \widetilde{\mu} = \lim_{m \to \infty} \int g_{t,m}(z)^{-n} \, {d} \widetilde{\mu}. \]

(Above we are using that $g_t(z)^{-n}$ is defined and continuous on $(z,t) \in \mathrm {supp} (\widetilde {\mu }) \times [0,1]$).) Now, set $A_{k,m} := g_{t}(k/2^m)^{-1}\cdot ( {\rm Cone}(I_{k,m}) \cap \{x+y < 1\} ) \subset \mathbb {R}^2$ and $A_m = \cup _{k=1}^{2^m} A_{k,m}$. We compute

\[ \int g_{t,m}(z)^{-n} \, {d} \widetilde{\mu} = \sum_{k=1}^{2^m} g_{t}(k/2^m)^{-n} \widetilde{\mu}(I_{k,m}) = \sum_{k=1}^{2^m} \mu\big( A_{k,m} \big) = \mu(A_m) , \]

where the second equality uses Proposition 4.9(2). Next, notice that $A_{m} \subset A_{m+1}$ and

\begin{align*} \bigcup_{m \geq 1} A_m &= \bigcup_{m \geq 1} \bigcup_{z\in [0,1] }g_{t,m}(z)^{-1} \cdot ({\rm Cone}(\{z\}) \cap \{ x+y <1 \})\\ &= \bigcup_{z\in [0,1]} g_{t}(z)^{-1} \cdot ({\rm Cone}(\{z\}) \cap \{x+y <1 \}) = \mathbb{R}^2_{>0} \cap \{(1-t) x+ty <1 \}. \end{align*}

Combining the previous two computations, we conclude

\[ \int g_t(z)^{-n} \, {d} \widetilde{\mu}= \lim_{m\to \infty} \mu(A_m) = \mu \big( \mathbb{R}^2_{>0} \cap \{ (1-t)x+ ty <1 \} \big) = \mathrm{e}(\mathfrak{a}_{\bullet,t}) , \]

where the final equality is Proposition 4.7(1).

Proof of Theorem 1.1 Set $g(z,t) := (1-t)z + t(1-z)$. By Lemma 4.11,

\[ E(t) = \int g(z,t)^{-n} \, {d}\widetilde{\mu} . \]

Since $g(z,t) = 0$ if and only if $z= {t}/({2t-1})$, there exists $\epsilon >0$ such that

\[ g(z,t) \neq 0\quad \text{for all } (z,t) \in \biggl[ \frac{1}{D+1}-\epsilon,\frac{D}{D+1} +\epsilon\biggr] \times [-\epsilon, 1+\epsilon] . \]

Since $g(z,t)$ is also smooth on $\mathbb {R}^2$, $g(z,t)^{-n}$ is smooth and has bounded derivatives on $({1}/({D+1})-\epsilon,{D}/({D+1})+\epsilon ) \times (-\epsilon, 1+\epsilon )$. Using that $\mathrm {{\rm supp}}\,{\tilde {\mu } } \subset [{1}/({D+1}),{D}/({D+1})]$, the Leibniz integral formula implies $\int g(z,t)^{-n}\, {d} \widetilde {\mu }$ is smooth on $(-\epsilon,1+\epsilon )$, which proves (1). The formula additionally implies

\[ E'(t)= n \int (2z-1)g(z,t)^{-n-1}\, {d} \tilde{\mu} \quad \text{and} \quad E''(t) = n(n+1) \int (2z-1)^2g(z,t)^{-n-2} \, {d} \tilde{\mu}. \]

To prove (2), it suffices to show $({d^2}/{dt^2}) (E(t)^{-1/n} )\leq 0$ for all $t\in (0,1)$. Note that

\[ \frac{d^2}{dt^2} \big(E(t)^{-1/n} \big) = \frac{(n+1) E'(t)^2- n E(t)E''(t) }{ n^2E(t)^{1/n+2} } . \]

To show the latter quantity is non-positive, we compute

(4.4)\begin{align} nE(t) E''(t) &= n^2(n+1) \biggl( \int \big| g(z,t)^{-n} \big| \, {d} \tilde{\mu} \biggr) \biggl( \int \big| (2z-1)^2{g(z,t)^{-n-2}} \big| \, {d} \tilde{\mu} \biggr) \nonumber\\ &\geq n^2 (n+1) \biggl( \int \big| (2z-1) g(z,t)^{-n-1} \big| \, {d} \tilde{\mu}\biggr)^2 \geq (n+1) E'(t)^2 , \end{align}

where the first inequality is by the Cauchy–Schwarz inequality. Therefore, $({d^2}/{dt^2}) (E(t)^{-1/n}) \leq 0$ for all $t\in (0,1)$, which proves (2).

To prove (3), note that $E(t)$ is linear if and only if $({d^2}/{dt^2}) (E(t)^{-1/n} )=0$ for all $t\in (0,1)$. The latter holds if and only if the inequalities in (4.4) are all equalities. The inequalities are equalities if and only if $(2z-1)g(z,t)^{-(n+2)/2}$ and $g(z,t)^{-n/2}$ are linearly dependent in $L^{1}(\widetilde {\mu })$. Equivalently, $2z-1$ and $g(z,t)$ are linearly dependent in $L^{1} (\widetilde {\mu })$. The linear dependence holds exactly when $\widetilde {\mu }$ is supported at a single point. By Lemma 4.10 and Corollary 3.17, the latter condition is equivalent to the existence of $c\in \mathbb {R}$ such that $\widetilde {\mathfrak {a}}_{\bullet, 0}=\widetilde {\mathfrak {a}}_{c\bullet,1}$.

Proof. We claim $v(\mathfrak {c}_\bullet )= v(\mathfrak {a}_{ \bullet,t} )$ for all $v\in \mathrm {Val}_{R,\mathfrak {m}}$, where $\mathfrak {c}_\bullet$ is the $\mathfrak {m}$-filtration defined by $\mathfrak {c}_\lambda := \sum _{i=0}^{\lceil \lambda \rceil } (\mathfrak {a}_{(\lceil \lambda \rceil -i)/(1-t),0 } \cap \mathfrak {a}_{i/t,0})$. Assuming the claim holds,

\begin{align*} \mathrm{lct}(X,D;\mathfrak{a}_{\bullet, t} ) &=\mathrm{lct}(X,D;\mathfrak{c}_\bullet)\\ &\leq \mathrm{lct}(X,D;\mathfrak{a}_{\bullet/(1-t),0}) + \mathrm{lct}(X,D;\mathfrak{a}_{\bullet/t,1}) = (1-t) \mathrm{lct}(X,D;\mathfrak{a}_{\bullet,0}) + t \mathrm{lct}(X,D;\mathfrak{a}_{\bullet,1}), \end{align*}

where the inequality is [Reference Xu and ZhuangXZ21, Theorem 3.11]. It remains to verify the claim. Since $\mathfrak {c}_{m} \subset \mathfrak {a}_{t,m}$ for each integer $m>0$, $v(\mathfrak {c}_{\bullet }) \geq v(\mathfrak {a}_{t,\bullet })$ for all $v\in \mathrm {Val}_{R,\mathfrak {m}}$. Next fix $\mu, \nu \in \mathbb {R}$ such that ${\mu (1-t)+\nu t = m}$. Set $r:= \mu (1-t) = m - \nu t$. For each $\ell \in \mathbb {Z}_{>0}$,

\begin{align*} (\mathfrak{a}_{\mu,0} \cap \mathfrak{a}_{\nu,1})^\ell \subset ( \mathfrak{a}_{\ell \mu,0} \cap \mathfrak{a}_{\ell \nu,1}) &= (\mathfrak{a}_{\ell r/(1-t),1} \cap \mathfrak{a}_{\ell (m-r)/t,1}) \subset (\mathfrak{a}_{\lfloor \ell r \rfloor /(1-t),0}\\ &\quad \cap \mathfrak{a}_{\big(\ell m -1 - \lfloor \ell r \rfloor \big)/t,1} ) \subset \mathfrak{c}_{\ell m -1 }. \end{align*}

Using the previous inclusions, we see that

\[ v(\mathfrak{a}_{\mu,0} \cap \mathfrak{a}_{\nu,1}) = \lim_{\ell \to \infty} \frac{v( (\mathfrak{a}_{\mu,0} \cap \mathfrak{a}_{\nu,1})^\ell)}{ \ell} \geq \lim_{\ell\to \infty} \frac{v(\mathfrak{c}_{\ell m-1}) }{ \ell} = v(\mathfrak{c}_\bullet) . \]

We can now compute

\[ v(\mathfrak{a}_{\bullet,t}) = \lim_{m\to \infty} \frac{v(\mathfrak{a}_{m,t})}{m} = \lim_{m\to \infty} \min_{ \mu(1-t)+\nu t= m} \biggl\{ \frac{v(\mathfrak{a}_{\mu,0} \cap \mathfrak{a}_{\nu,1})}{m} \biggr\} \geq v(\mathfrak{c}_\bullet) . \]

Therefore, $v(\mathfrak {a}_{\bullet,t})= v(\mathfrak {c}_\bullet )$ as desired.