Proof of The Score of a Measure with a Positive Continuously Differentiable Density

Each result cited is universally quantified over the data in its own statement. The function $m$ is continuous on $\mathbb{R}^{d}$ (claim 3 of Euclidean Space Is Open in Itself), hence Borel by Probability Measures on Euclidean Space and Random Vectors: Standing Notation (ii), and takes values in $[0,\infty)$; so $\mu$ is the measure with density $m$ with respect to $\lambda_{d}$ in the sense of claim 3 of Image Measures and Measures with a Density: Measures, Integrals and Change of Variables, and that claim is in force: for a Borel $f:\mathbb{R}^{d}\to[0,\infty]$, $\int f\,d\mu=\int fm\,d\lambda_{d}$ in $[0,\infty]$, and a Borel $f:\mathbb{R}^{d}\to\mathbb{R}$ is $\mu$-integrable exactly when $fm$ is $\lambda_{d}$-integrable, with $\int f\,d\mu=\int fm\,d\lambda_{d}$. Points of $\mathbb{R}^{d}$ are tuples, the dot product is $u\cdot v=\sum_{i=1}^{d}u_{i}v_{i}$ and scalar multiples are coordinatewise (Difference, Dot Product, and Orthogonality in $\mathbb{R}^n$, Scalar Multiple of a Point of $\mathbb{R}^n$). Finite sums are those of Finite Sum Notation in a Field, with Properties of Finite Sums in force.

Step 1: $\eta$ is square-integrable. The map $\eta$ and the function $\lVert\eta\rVert^{2}$ are Borel, as recorded in the statement (the continuity of $\partial_{i}m$ there is clause 1 of Functions of Class Cᵏ on an Open Subset of ℝⁿ read through its clause 3, with claim 1 of Euclidean Continuity Agrees with Metric Continuity for Real-Valued Functions). The function $\lVert\eta\rVert^{2}$ is nonnegative with finite integral by hypothesis, so it is $\mu$-integrable by Integrable Functions and Their Lebesgue Integral; that is, $\eta$ is a square-integrable random vector on $(\mathbb{R}^{d},\mathcal{B}(\mathbb{R}^{d}),\mu)$ and its class lies in $L^{2}(\mu;\mathbb{R}^{d})$ (Square-Integrable Vector Fields Against a Probability Measure on Euclidean Space, and Test Functions: Standing Notation (ii)).

Step 2: the identity of claim 1. Fix $\psi\in C_{c}^{\infty}(\mathbb{R}^{d})$. By The Gradient of a Test Function is Bounded and Square-Integrable, and Its Laplacian Bounded and Integrable, Against Every Probability Measure (ii), $\Delta\psi$ is Borel and $\mu$-integrable, so $\int\Delta\psi\,d\mu=\int(\Delta\psi)\,m\,d\lambda_{d}$ with $(\Delta\psi)m$ integrable. For each $i\in[d]$, the function $\partial_{i}\psi$ is smooth (claim 3 of Basic Properties of Cᵏ Functions), hence of class $C^{1}$, and compactly supported by The Gradient of a Test Function is Bounded and Square-Integrable, and Its Laplacian Bounded and Integrable, Against Every Probability Measure (i); so Integration by Parts on Euclidean Space Against a Compactly Supported Function of Class $C^{1}$ (ii), applied with $f=m$ and $g=\partial_{i}\psi$, says that $m\,\partial_{i}\partial_{i}\psi$ and $(\partial_{i}m)\,\partial_{i}\psi$ are $\lambda_{d}$-integrable and

$$\int_{\mathbb{R}^{d}}m\,\partial_{i}\partial_{i}\psi\,d\lambda_{d}=-\int_{\mathbb{R}^{d}}(\partial_{i}m)\,\partial_{i}\psi\,d\lambda_{d}.$$

Pointwise, $(\Delta\psi)m=\sum_{i=1}^{d}m\,\partial_{i}\partial_{i}\psi$ by The Laplacian of a Twice Continuously Differentiable Function (i) and claim 3 of Properties of Finite Sums, and $\sum_{i=1}^{d}(\partial_{i}m)\,\partial_{i}\psi=\nabla m\cdot\nabla\psi$; moreover $\nabla m=m\,\eta$ coordinatewise (as $m\,m^{-1}\partial_{i}m=\partial_{i}m$), so $\nabla m\cdot\nabla\psi=\sum_{i=1}^{d}m\,\eta_{i}\,\partial_{i}\psi=m\,(\eta\cdot\nabla\psi)$, again by claim 3 of Properties of Finite Sums. By Linearity of the Lebesgue Integral over a Finite Sum of Integrable Functions (ii), applied with all coefficients equal to $1$ for the first equality below and to $-1$ for the second and third,

$$\int_{\mathbb{R}^{d}}\Delta\psi\,d\mu=\sum_{i=1}^{d}\int_{\mathbb{R}^{d}}m\,\partial_{i}\partial_{i}\psi\,d\lambda_{d}=-\sum_{i=1}^{d}\int_{\mathbb{R}^{d}}(\partial_{i}m)\,\partial_{i}\psi\,d\lambda_{d}=-\int_{\mathbb{R}^{d}}(\eta\cdot\nabla\psi)\,m\,d\lambda_{d}.$$

Now $\eta$ and $\nabla\psi$ are square-integrable random vectors on $(\mathbb{R}^{d},\mathcal{B}(\mathbb{R}^{d}),\mu)$, so by Square-Integrable Random Vectors: Coordinates, Operations, Almost Sure Equality and the Mean-Square Form (ii) the function $\eta\cdot\nabla\psi$ is Borel and $\mu$-integrable, with $\int\eta\cdot\nabla\psi\,d\mu=\langle\eta,\nabla\psi\rangle_{\mu}$ by The Space of Square-Integrable Random Vectors (iii); by claim 3 of Image Measures and Measures with a Density: Measures, Integrals and Change of Variables, $\int(\eta\cdot\nabla\psi)\,m\,d\lambda_{d}=\int\eta\cdot\nabla\psi\,d\mu$. Hence $\int\Delta\psi\,d\mu=-\langle\eta,\nabla\psi\rangle_{\mu}$, which is claim 1.

Step 3: claim 2. By Step 2 and The Cauchy–Schwarz Inequality in a Real Inner Product Space in the inner product space $L^{2}(\mu;\mathbb{R}^{d})$,

$$\Bigl|\int_{\mathbb{R}^{d}}\Delta\psi\,d\mu\Bigr|=|\langle\eta,\nabla\psi\rangle_{\mu}|\le\lVert\eta\rVert_{\mu}\,\lVert\nabla\psi\rVert_{\mu}\qquad(\psi\in C_{c}^{\infty}(\mathbb{R}^{d})),$$

using $|-s|=|s|$ (claim 2 of Properties of the Absolute Value in an Ordered Field); so $\mu$ has finite Fisher information with $C=\lVert\eta\rVert_{\mu}$. The functional $\ell(\psi)=-\int\Delta\psi\,d\mu$ of Finite Fisher Information, the Score and the Fisher Information of a Probability Measure (ii) is linear in $\psi$ (The Gradient of a Test Function is Bounded and Square-Integrable, and Its Laplacian Bounded and Integrable, Against Every Probability Measure (iii) and claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion), satisfies $|\ell(\psi)|\le\lVert\eta\rVert_{\mu}\lVert\nabla\psi\rVert_{\mu}$ by the display above, and satisfies $\langle\eta,\nabla\psi\rangle_{\mu}=\ell(\psi)$ for every $\psi$ by Step 2; so Basic Properties of the Tangent Space: Closed Subspace, the Identity Map Belongs to It, Second-Moment Limits, and Representation of Bounded Functionals on Gradients (iv) and Basic Properties of the Tangent Space: Closed Subspace, the Identity Map Belongs to It, Second-Moment Limits, and Representation of Bounded Functionals on Gradients (v) apply with $C=\lVert\eta\rVert_{\mu}$, the element $\xi$ of the former being $\xi_{\mu}$ by Finite Fisher Information, the Score and the Fisher Information of a Probability Measure (ii), and give $\xi_{\mu}=P_{T_{\mu}}\eta$ and $\lVert\xi_{\mu}\rVert_{\mu}\le\lVert\eta\rVert_{\mu}$, whence $\mathcal{I}(\mu)=\lVert\xi_{\mu}\rVert_{\mu}^{2}\le\lVert\eta\rVert_{\mu}^{2}$ by claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field. Finally, for every $x$, $\lVert\eta(x)\rVert=|m(x)^{-1}|\,\lVert\nabla m(x)\rVert=m(x)^{-1}\lVert\nabla m(x)\rVert$ by claim 5 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square ($m(x)^{-1}>0$ by claim 7 of Elementary Order Arithmetic in an Ordered Field), so $\lVert\eta(x)\rVert^{2}m(x)=m(x)^{-1}\lVert\nabla m(x)\rVert^{2}$, and claim 3 of Image Measures and Measures with a Density: Measures, Integrals and Change of Variables for the nonnegative Borel function $\lVert\eta\rVert^{2}$ gives $\lVert\eta\rVert_{\mu}^{2}=\int\lVert\eta\rVert^{2}\,d\mu=\int\lVert\eta\rVert^{2}m\,d\lambda_{d}=\int m^{-1}\lVert\nabla m\rVert^{2}\,d\lambda_{d}$, which is the displayed integral; this proves claim 2.

Step 4: claim 3. If $\eta\in T_{\mu}$, the last assertion of Basic Properties of the Tangent Space: Closed Subspace, the Identity Map Belongs to It, Second-Moment Limits, and Representation of Bounded Functionals on Gradients (v) gives $\xi_{\mu}=\eta$, and then $\mathcal{I}(\mu)=\lVert\eta\rVert_{\mu}^{2}$ by Finite Fisher Information, the Score and the Fisher Information of a Probability Measure (iii).