FALSE: pointwise-integrable generator constants justify ideal-critic gradient exchange

Set $a_n=4^{-n}$ for $n\ge 0$ and $p_n=2^{-n}$ for $n\ge 1$. Define a nonnegative piecewise-linear hat ${\alpha }_n$ on $[0,+\mathrm{\infty })$: it is zero outside $[a_{n+1},a_{n-1}]$, rises linearly from $0$ at $a_{n+1}$ to $1$ at $a_n$, then falls linearly to $0$ at $a_{n-1}$. On $[a_{n+1},a_n]$ only ${\alpha }_n,{\alpha }_{n+1}$ are nonzero and their sum is one. Thus ${\sum }_{n\ge 1}{\alpha }_n(r)=1$ for $0<r\le 1\mathrm{/}4$. At each fixed positive $r$ only finitely many hat supports meet it.

Put $B_n=[n-1\mathrm{/}8,n+1\mathrm{/}8]$ and $q(z)={\sum }_{n\ge 1}4p_n{\mathbf{1}}_{B_n}(z)$. Define $\rho$ on Borel subsets of $\mathbb{R}$ by $\rho (A)={\int }_{A}qd\lambda$, the restriction to Borel sets of the measure with density $q$. The density is Borel, and [F1]–[F4] give $\rho (\mathbb{R})={\sum }_{n}4p_n\lambda (B_n)={\sum }_n{p}_n=1$. Let $u_n(z)$ be one when $\mathrm{\mid }z-n\mathrm{\mid }\le 1\mathrm{/}8$, decrease linearly to zero when $1\mathrm{/}8<\mathrm{\mid }z-n\mathrm{\mid }<1\mathrm{/}4$, and be zero for $\mathrm{\mid }z-n\mathrm{\mid }\ge 1\mathrm{/}4$. It is $8$-Lipschitz; its supports are disjoint, and $u_n=1$ on $B_n$.

Set $b_n(\theta )=\theta {\alpha }_n(\mathrm{\mid }\theta \mathrm{\mid })\mathrm{/}{p}_n$ and $g_{\theta }(z)=1\mathrm{/}2+(1\mathrm{/}16){\sum }_n{u}_n(z)b_n(\theta )$. Since $\mathrm{\mid }{b}_n\mathrm{\mid }\le a_{n-1}\mathrm{/}{p}_n=2^{2-n}\le 2$ and at most one latent support is active, $3\mathrm{/}8\le g\le 5\mathrm{/}8$. Each $b_n$ is continuous piecewise quadratic with compact support and bounded one-sided slopes, hence globally Lipschitz. More explicitly $\mathrm{\mid }{\alpha }_n^{\mathrm{\prime }}\mathrm{\mid }\le 4\mathrm{/}(3a_n)$ and $\mathrm{\mid }\theta \mathrm{\mid }\le 4a_n$ on its support, so $\mathrm{\mid }{b}_n^{\mathrm{\prime }}\mathrm{\mid }\le (1+16\mathrm{/}3)\mathrm{/}{p}_n=19\mathrm{/}(3p_n)$ wherever the derivative exists; subdivision at finitely many corners gives this bound for all differences. Local finiteness in $z$ then proves joint local Lipschitzness of $g$ and Borel measurability of every section.

A valid measurable local joint constant is $L(\theta ,z)={\sum }_n{M}_n{\mathbf{1}}_{[a_{n+1},a_{n-1}]}(\mathrm{\mid }\theta \mathrm{\mid }){\mathbf{1}}_{(n-1\mathrm{/}2,n+1\mathrm{/}2)}(z)$, with $M_n=(16+19\mathrm{/}(3p_n))\mathrm{/}16$: the two product terms are bounded using $\mathrm{\mid }{b}_n\mathrm{\mid }\le 2$, the $8$-Lipschitz $u_n$, and the bound in step 2.1. At any fixed nonzero $\theta$ only finitely many parameter supports are active, so $\int L(\theta ,z)d\rho <\mathrm{\infty }$. At $\theta =0$ every $b_n$ vanishes on a neighborhood of zero, so for each fixed $z$ the locally finite sum is constant in a neighborhood of $(0,z)$; take $L(0,z)=0$. At latent or parameter support boundaries the displayed positive constants apply whenever needed, and outside the closed active supports a neighborhood has zero variation. Thus the original pointwise-local condition holds at every parameter.

Let $c_n(\theta )=1\mathrm{/}2+b_n(\theta )\mathrm{/}16$. The density $q$ vanishes off the disjoint union of the intervals $B_n$, and on each $B_n$ the function $g_{\theta }$ is constant with value $c_n(\theta )$. For $N\ge 1$, $q_N={\sum }_{n\le N}4p_n{\mathbf{1}}_{B_n}$ increases pointwise to $q$, and $g_{\theta }{q}_N$ is simple with integral ${\sum }_{n\le N}4p_n{c}_n(\theta )\lambda (B_n)={\sum }_{n\le N}{p}_n{c}_n(\theta )$ by [F4, F10]. By [F2], $\int g_{\theta }qd\lambda ={\sum }_{n\ge 1}{p}_n{c}_n(\theta )$; by [F9], this equals $\int g_{\theta }d\rho$. For each fixed $\theta$ only finitely many $b_n(\theta )$ are nonzero, while [F1] gives ${\sum }_n{p}_n=1$, so $\int g_{\theta }d\rho =1\mathrm{/}2+(1\mathrm{/}16){\sum }_n{p}_n{b}_n(\theta )=1\mathrm{/}2+(\theta \mathrm{/}16){\sum }_n{\alpha }_n(\mathrm{\mid }\theta \mathrm{\mid })$. Therefore this integral is $1\mathrm{/}2+\theta \mathrm{/}16$ for $0<\mathrm{\mid }\theta \mathrm{\mid }\le 1\mathrm{/}4$ and is $1\mathrm{/}2$ at $\theta =0$.

Put $\iota (y)=(0,y)$ and let ${\gamma }_{\theta }={\iota }_{\mathrm{\#}}{P}_{\theta }$. The map $\iota$ is continuous in the product topology, so [F8] makes ${\gamma }_{\theta }$ a Borel probability; for every Borel $A\subseteq X$ its marginals satisfy ${\gamma }_{\theta }(A\times X)={\mathbf{1}}_{\{0\in A\}}={\delta }_0(A)$ and ${\gamma }_{\theta }(X\times A)=P_{\theta }(A)$, so it is a coupling. Any coupling $\gamma$ is concentrated on $\{0\}\times X$ by its first marginal. For Borel $C\subseteq X\times X$, let $B_C={\iota }^{-1}(C)$, which is Borel by continuity; concentration gives $\gamma (C)=\gamma (C\cap (\{0\}\times X))=\gamma (\{0\}\times B_C)$. The first marginal also gives $\gamma ((X\setminus \{0\})\times B_C)=0$, so the second marginal gives $\gamma (\{0\}\times B_C)=\gamma (X\times B_C)=P_{\theta }(B_C)$. Hence $\gamma (C)={\gamma }_{\theta }(C)$ and this coupling is unique. Applying [F8] to $\iota$ and to $g_{\theta }$ gives its cost ${\int }_{X}yd{P}_{\theta }(y)=\int g_{\theta }d\rho$, so $V(\theta )=W_1^X({\delta }_0,P_{\theta })=\int g_{\theta }d\rho$.

For any normalized $1$-Lipschitz $h:[0,1]\to \mathbb{R}$, [F8] applied to the constant-zero map and to $g_{\theta }$ gives $\int hd{\delta }_0=h(0)=0$ and $\int hd{P}_{\theta }=\int h(g_{\theta }(z))d\rho (z)$. The function $f(x)=-x$ is normalized and $1$-Lipschitz, and its signed dual objective is $\int fd{\delta }_0-\int fd{P}_{\theta }=\int g_{\theta }d\rho =V(\theta )$ by step 4.1. The compact KR duality [F7] therefore makes $f$ an attaining critic, in particular at $\theta =0$. By step 3.1, each fixed latent section is constant near zero, so ${\mathrm{\partial }}_{\theta }f(g_{\theta }(z)){\mathrm{\mid }}_0=0$ for every $z$; this derivative is measurable and integrable with expectation zero. Step 3.2 gives $V^{\mathrm{\prime }}(0)=1\mathrm{/}16$, so the two defined terms in the proposed identity are unequal.

Suppose there were a common-neighborhood generator envelope at zero: an open neighborhood $U$ of zero, a nonnegative measurable $K$ with $\int Kd\rho <\mathrm{\infty }$, and a measurable set $E$ with $\rho (E)=1$ such that $\mathrm{\mid }{g}_{\theta }(z)-g_{{\theta }^{\mathrm{\prime }}}(z)\mathrm{\mid }\le K(z)\mathrm{\mid }\theta -{\theta }^{\mathrm{\prime }}\mathrm{\mid }$ for every $z\in E$ and every $\theta ,{\theta }^{\mathrm{\prime }}\in U$. Since $a_n\to 0$, there is $n_0$ such that $a_n\in U$ for every $n\ge n_0$. For every such $n$ and $z\in B_n\cap E$, the envelope then gives $K(z)\ge \mathrm{\mid }{g}_{a_n}(z)-g_0(z)\mathrm{\mid }\mathrm{/}{a}_n=1\mathrm{/}(16p_n)$, because $u_n=1$ on $B_n$ and ${\alpha }_n(a_n)=1$. By [F6], $\rho (B_n\setminus E)\le \rho (E^c)=0$, so $\rho (B_n\cap E)=\rho (B_n)=p_n$ by step 1.2. The sets $B_n\cap E$ are disjoint and each contributes at least $1\mathrm{/}16$ to $\int Kd\rho$. Summing these nonnegative contributions, using [F2] to pass from finite partial sums to their limit, forces $\int Kd\rho =\mathrm{\infty }$, a contradiction. The counterexample therefore violates precisely the stronger corrected envelope premise while satisfying the original one.