Proof of The Score Paired with the Identity, and the Laplacian of a Convex Potential Bounded by its Pairing with the Score

Each result cited is universally quantified over the data in its own statement. Write $\xi=\xi_{\mu}$ and fix a Borel representative of it. Since $\int\lVert\xi\rVert^{2}\,d\mu<\infty$ and $\mu$ is a probability measure, for every Borel $v:\mathbb{R}^{d}\to\mathbb{R}^{d}$ with $\int\lVert v\rVert^{2}\,d\mu<\infty$ the function $\lVert\xi\rVert\lVert v\rVert$ is $\mu$-integrable and $|\xi\cdot v|\le\lVert\xi\rVert\lVert v\rVert$ pointwise, by Hoelder's Inequality, for Two and for Finitely Many Factors §holder and Cauchy-Schwarz Inequality for the Euclidean Dot Product; in particular $\lVert\xi\rVert$ itself is $\mu$-integrable (take $v$ constant of norm $1$). Limits of integrals below are justified by the dominated convergence theorem with such dominating functions.

Claim 1. Let $\psi_{R}$ ($R\in\mathbb{N}$) and $M$ be as in Scaled Cutoffs and the Second-Moment Test Functions: Uniform Derivative Bounds and Agreement on a Ball with $q=d$. By Scaled Cutoffs and the Second-Moment Test Functions: Uniform Derivative Bounds and Agreement on a Ball §test, $\psi_{R}$ is smooth and compactly supported, hence a test function, so the definition of the score gives

The coordinates of $\nabla\psi_{R}(x)$ are $\partial_{i}\psi_{R}(x)$, bounded by $M\lVert x\rVert$, so $\lVert\nabla\psi_{R}(x)\rVert\le\sqrt{d}\,M\lVert x\rVert$ (as $\lVert v\rVert^{2}=\sum_{i}v_{i}^{2}$), and $|\Delta\psi_{R}|\le dM$. By Scaled Cutoffs and the Second-Moment Test Functions: Uniform Derivative Bounds and Agreement on a Ball §ball, $\nabla\psi_{R}(x)=x$ and $\Delta\psi_{R}(x)=d$ whenever $\lVert x\rVert<R$, hence for all $R>\lVert x\rVert$. So, as $R\to\infty$, $\xi\cdot\nabla\psi_{R}\to\xi\cdot\mathrm{id}$ pointwise with domination by $\sqrt{d}M\lVert\xi\rVert\lVert\mathrm{id}\rVert$, and $\Delta\psi_{R}\to d$ pointwise with domination by the constant $dM$. Dominated convergence gives $\langle\xi,\mathrm{id}\rangle_{\mu}=-\int d\,d\mu=-d$.

Claim 2. The set $G$ is nonempty ($\mu(G)=1$); fix $x_{0}\in G$ and $q_{0}\in\partial_{G}\varphi(x_{0})$ (The Subdifferential of a Convex Function on an Open Convex Set is Nonempty §nonempty) and let $L$ range over natural numbers with $\lVert q_{0}\rVert<L$. Let $\varphi^{L}$ be the Lipschitz truncation of $\varphi$ at level $L$, convex on $\mathbb{R}^{d}$ and Lipschitz with constant $L$ (The Lipschitz Truncation of a Convex Function: a Global Lipschitz Convex Minorant Agreeing with It Where the Slope is Small §minorant). By The Points of Twice Differentiability of a Convex Function: a Borel Set of Full Measure, and Borel Measurability of the Gradient and Hessian on It §full (with $d$ for $n$ and $U=\mathbb{R}^{d}$) there is $A^{L}\in\mathcal{B}(\mathbb{R}^{d})$ with $\lambda_{d}(\mathbb{R}^{d}\setminus A^{L})=0$, hence $\mu(A^{L})=1$ by absolute continuity, at every point of which $\varphi^{L}$ is twice differentiable; let $g^{L}$ and $\Delta^{L}$ be the Borel maps defined from $\varphi^{L}$ and $A^{L}$ exactly as $g$ and $\Delta$ are defined from $\varphi$ and $A$, and $0\le\Delta^{L}$ (The Points of Twice Differentiability of a Convex Function: a Borel Set of Full Measure, and Borel Measurability of the Gradient and Hessian on It §borel and The Points of Twice Differentiability of a Convex Function: a Borel Set of Full Measure, and Borel Measurability of the Gradient and Hessian on It §nonnegative). At $y\in A^{L}$, $g^{L}(y)$ is the only subgradient of $\varphi^{L}$ (Subgradients near a Point of Twice Differentiability of a Convex Function, and Invariance of the Second-Order Expansion under Lipschitz Truncation §singleton), so $\lVert g^{L}(y)\rVert\le L$ by Elementary Calculus of the Subdifferential of a Convex Function §bounded, $\varphi^{L}$ being Lipschitz with constant $L$ on every ball; off $A^{L}$, $g^{L}=0_{\mathbb{R}^{d}}$.

Step A: the inequality for $\varphi^{L}$. Let $\rho$ be a mollifier kernel of radius $1$ (Existence of Mollifier Kernels of Every Radius) and $F_{m}=\varphi^{L}*\rho_{1/m}$ for $m\in\mathbb{N}$. By Mollification of a Lipschitz Convex Function: Smooth Convex Approximations with Bounded Gradients Converging Where the Subgradient is Unique §regularity, $F_{m}$ is smooth and convex with $\lVert DF_{m}\rVert\le L$, so $|\partial_{i}F_{m}|\le L$ (claim 4 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$); by Mollification at a Point of Twice Differentiability: Convergence of the Mollified Gradient and Hessian, and a Hessian Bound for Lipschitz Functions §bound, with its constant $K$ for the dimension $d$, the radius $1$ and the kernel $\rho$, $|\partial_{j}\partial_{i}F_{m}|\le KLm$. Hence Gradients of Functions with Bounded Derivatives Belong to the Tangent Space; Constants Are Tangent; the Score Identity; the Score Has Mean Zero; Translation of Tangent Fields §score-identity applies with the single bound $\max\{L,KLm\}$ for all first and second partial derivatives:

the diagonal entries of the positive semidefinite Hessian $D^{2}F_{m}(x)$ being nonnegative. For $y\in A^{L}$, Mollification at a Point of Twice Differentiability: Convergence of the Mollified Gradient and Hessian, and a Hessian Bound for Lipschitz Functions §convergence with $\varepsilon_{m}=m^{-1}$ gives $DF_{m}(y)\to g^{L}(y)$ and $D^{2}F_{m}(y)\to D^{2}\varphi^{L}(y)$ in $\mathcal{S}(d)$, hence $\Delta F_{m}(y)\to\Delta^{L}(y)$, the diagonal entries converging by Limits and Bounded Sequences of Symmetric Real Matrices §quadratic-form with $z=e_{i}$. Since $\mu(A^{L})=1$ and $|\xi\cdot DF_{m}|\le L\lVert\xi\rVert$, dominated convergence gives $\int\xi\cdot DF_{m}\,d\mu\to\langle\xi,g^{L}\rangle_{\mu}$. For $y\notin A^{L}$, $\Delta^{L}(y)=0\le\Delta F_{m}(y)$. Hence $\liminf_{m}\Delta F_{m}\ge\mathbf{1}_{A^{L}}\Delta^{L}$ pointwise, and Fatou's Lemma gives

Step B: $L\to\infty$. Let $A^{\ast}=A\cap\bigcap_{L}A^{L}$, a Borel set with $\mu(A^{\ast})=1$ (countably many $\mu$-full sets; claim 4 of Basic Properties of a Measure). Fix $y\in A^{\ast}$. If $L>\lVert g(y)\rVert=\lVert D\varphi(y)\rVert$, then Subgradients near a Point of Twice Differentiability of a Convex Function, and Invariance of the Second-Order Expansion under Lipschitz Truncation §truncation shows that $\varphi^{L}$ is twice differentiable at $y$ with first-order coefficient $D\varphi(y)$ and Hessian $D^{2}\varphi(y)$; by the uniqueness recorded in Differential Calculus and Convexity on Euclidean Open Sets: Standing Notation §twice-differentiable, $g^{L}(y)=g(y)$ and $\Delta^{L}(y)=\Delta(y)$. If $L\le\lVert g(y)\rVert$, then $\lVert g^{L}(y)\rVert\le L\le\lVert g(y)\rVert$. So $|\xi\cdot g^{L}|\le\lVert\xi\rVert\lVert g\rVert$ on $A^{\ast}$, and $\xi\cdot g^{L}\to\xi\cdot g$ and $\Delta^{L}\to\Delta$ pointwise on $A^{\ast}$ (the sequences being eventually constant). Dominated convergence gives $\langle\xi,g^{L}\rangle_{\mu}\to\langle\xi,g\rangle_{\mu}$, and Fatou's lemma applied to $\mathbf{1}_{A^{\ast}}\Delta^{L}\ge0$, together with (1), gives

where $\int\Delta\,d\mu=\int\mathbf{1}_{A^{\ast}}\Delta\,d\mu$ and $\int\mathbf{1}_{A^{L}}\Delta^{L}\,d\mu=\int\Delta^{L}\,d\mu$ hold because the integrands agree $\mu$-almost everywhere (The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §comparison). In particular $\int\Delta\,d\mu<\infty$, so the nonnegative Borel function $\Delta$ is integrable with respect to $\mu$.