Proof of The Gaussian Kernels on Euclidean Space: Scaling, Derivatives up to Order Three, the Convolution Identity, Moments and Tails

Each result cited is universally quantified over the data in its own statement. Throughout, $s$ satisfies $0<s\le1$, and we write $\psi_{s}(z)=\exp(-\lVert z\rVert^{2}/(2s))$, so that $g_{s}=c_{s}\psi_{s}$, with $\int_{\mathbb{R}^{q}}\psi_{s}\,d\lambda_{q}=1/c_{s}$ by claim 1 of The Gaussian Smoothing Weight: Normalization, Derivatives, Exponential Tilting, Moments, and First-Order Remainder (read in dimension $m=q$ with $\eta=s$) and the homogeneity in claim 1 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion with the nonnegative factor $1/c_{s}$; that claim also gives $0<\psi_{s}\le1$, hence $0<g_{s}\le c_{s}$, the evenness $g_{s}(-z)=g_{s}(z)$, and, for every $a\in\mathbb{R}^{q}$, that $z\mapsto g_{s}(z-a)$ is Borel with $\int_{\mathbb{R}^{q}}g_{s}(z-a)\,dz=1$. By that lemma $\psi_{s}$ and $g_{s}$ are sequentially continuous, hence continuous on $\mathbb{R}^{q}$ by Continuity Between Metric Spaces is Equivalent to Sequential Continuity, hence Borel (Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets, preamble). Powers $(\sqrt{s})^{m}$ obey $(\sqrt{s})^{m}(\sqrt{s})^{n}=(\sqrt{s})^{m+n}$ (claim 1 of Real Powers Through the Exponential, and Elementary Asymptotic Tools: Monotonicity, Null Sequences of Negative Powers, Exponential Domination, Integer Rounding, and Square-Root and Exponential Inequalities, real and natural powers agreeing) and $(\sqrt{s})^{2}=\sqrt{s}\sqrt{s}=s$ (claim 1 of Properties of Natural Number Powers in a Field and Existence and Uniqueness of the Nonnegative Square Root), so that $s^{-m/2}s^{-n/2}=s^{-(m+n)/2}$ and $s^{-2/2}=s^{-1}$; and $\sqrt{a}\sqrt{b}=\sqrt{ab}$ for nonnegative $a,b$ by claim 5 of Real Powers Through the Exponential, and Elementary Asymptotic Tools: Monotonicity, Null Sequences of Negative Powers, Exponential Domination, Integer Rounding, and Square-Root and Exponential Inequalities. The exponential satisfies $\exp(u+v)=\exp(u)\exp(v)$, $\exp(u)>0$, and $\exp(u)\ge1+u$ for $u\ge0$ (claims 1, 2 and 4 of Basic Properties of the Exponential Function); in particular, for $u\ge0$, $\exp(u)\ge1$ and $\exp(-u)=1/\exp(u)\le1$ (the inverse of a number $\ge1$ is $\le1$, by claim 5 of Elementary Arithmetic in an Ordered Field applied to $1\le\exp(u)$ with the nonnegative multiplier $1/\exp(u)$, claim 4 there).

Claim 1 (scaling). Let $c=s^{-1/2}$. For $z\in\mathbb{R}^{q}$, $\lVert cz\rVert^{2}=c^{2}\lVert z\rVert^{2}=s^{-1}\lVert z\rVert^{2}$ 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, so $\psi_{1}(cz)=\exp(-\lVert z\rVert^{2}/(2s))=\psi_{s}(z)$. By claim 2 of Scaling Lebesgue Measure on ℝⁿ by a Diagonal Matrix, applied to the nonnegative Borel function $\psi_{1}$,

$$\frac{1}{c_{s}}=\int_{\mathbb{R}^{q}}\psi_{s}\,d\lambda_{q}=\int_{\mathbb{R}^{q}}\psi_{1}(cz)\,dz=|c|^{-q}\int_{\mathbb{R}^{q}}\psi_{1}\,d\lambda_{q}=\frac{(\sqrt{s})^{q}}{c_{1}},$$

since $|c|^{q}=(s^{-1/2})^{q}$ is the inverse of $(\sqrt{s})^{q}$. Hence $c_{s}=c_{1}\,s^{-q/2}$, and $g_{s}(z)=c_{s}\psi_{s}(z)=s^{-q/2}\,c_{1}\psi_{1}(s^{-1/2}z)=s^{-q/2}g_{1}(s^{-1/2}z)$.

Claim 2 (smoothness and derivatives). Let $\mathcal{C}$ be the set of functions $F:\mathbb{R}^{q}\to\mathbb{R}$ of the form $F=h\,g_{s}$ (pointwise product) with $h:\mathbb{R}^{q}\to\mathbb{R}$ smooth on $\mathbb{R}^{q}$. Every $F\in\mathcal{C}$ is continuous, as a product of continuous functions (claim 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space; a smooth $h$ is of class $C^{1}$, hence continuous, by Smooth Functions on an Open Subset of ℝⁿ and clause 1 of Functions of Class Cᵏ on an Open Subset of ℝⁿ). By claim 2 of The Gaussian Smoothing Weight: Normalization, Derivatives, Exponential Tilting, Moments, and First-Order Remainder the partial derivative $\partial_{i}g_{s}(z)=-\frac{z_{i}}{s}g_{s}(z)$ exists at every $z$, so by the product rule in claim 1 of Constants, Affine Functions, Sums, Products and Quotients of Cᵏ Functions, for $F=h\,g_{s}\in\mathcal{C}$,

$$\partial_{i}F(z)=\partial_{i}h(z)\,g_{s}(z)+h(z)\,\partial_{i}g_{s}(z)=\Bigl(\partial_{i}h(z)-\frac{z_{i}}{s}h(z)\Bigr)g_{s}(z),$$

and $\partial_{i}h-\frac{z_{i}}{s}h$ is smooth: $\partial_{i}h$ is of class $C^{k}$ for every $k$ by clause 2 of Functions of Class Cᵏ on an Open Subset of ℝⁿ applied to $h\in C^{k+1}$, hence smooth; the coordinate function $z\mapsto z_{i}$ and the constant $-\frac1s$ are smooth by claim 2 of Constants, Affine Functions, Sums, Products and Quotients of Cᵏ Functions; and products and sums of smooth functions are smooth by claim 3 there. Thus $\partial_{i}F\in\mathcal{C}$, in particular $\partial_{i}F$ is continuous, and so every $F\in\mathcal{C}$ is of class $C^{1}$ (clauses 1 and 3 of Functions of Class Cᵏ on an Open Subset of ℝⁿ) with all partial derivatives in $\mathcal{C}$. Let $T$ be the set of $k\in\mathbb{N}$ such that every member of $\mathcal{C}$ is of class $C^{k}$; then $1\in T$, and if $k\in T$ then every $F\in\mathcal{C}$ is of class $C^{1}$ with partial derivatives of class $C^{k}$, hence of class $C^{k+1}$ by clause 2 of Functions of Class Cᵏ on an Open Subset of ℝⁿ, so $k+1\in T$; by Principle of Induction for the Natural Numbers, $T=\mathbb{N}$, and every member of $\mathcal{C}$ is smooth. In particular $g_{s}=1\cdot g_{s}\in\mathcal{C}$ is smooth. The first formula is claim 2 of The Gaussian Smoothing Weight: Normalization, Derivatives, Exponential Tilting, Moments, and First-Order Remainder. Applying the displayed rule to $F=\partial_{i}g_{s}$, that is $h(z)=-\frac{z_{i}}{s}$, with $\partial_{j}h=-\frac{\delta_{ij}}{s}$ (the partial derivative of a coordinate function is $1$ or $0$ according as the indices agree or not, by Partial Derivative of a Real-Valued Function on an Open Subset of $\mathbb{R}^n$), gives

$$\partial_{j}\partial_{i}g_{s}(z)=\Bigl(-\frac{\delta_{ij}}{s}+\frac{z_{j}z_{i}}{s^{2}}\Bigr)g_{s}(z);$$

and applying it once more to $h(z)=\frac{z_{i}z_{j}}{s^{2}}-\frac{\delta_{ij}}{s}$, with $\partial_{k}h(z)=\frac{\delta_{ik}z_{j}+\delta_{jk}z_{i}}{s^{2}}$ (product rule, claim 1 of Constants, Affine Functions, Sums, Products and Quotients of Cᵏ Functions), gives

$$\partial_{k}\partial_{j}\partial_{i}g_{s}(z)=\Bigl(\frac{\delta_{ik}z_{j}+\delta_{jk}z_{i}}{s^{2}}-\frac{z_{k}}{s}\Bigl(\frac{z_{i}z_{j}}{s^{2}}-\frac{\delta_{ij}}{s}\Bigr)\Bigr)g_{s}(z)=\Bigl(-\frac{z_{i}z_{j}z_{k}}{s^{3}}+\frac{\delta_{ij}z_{k}+\delta_{ik}z_{j}+\delta_{jk}z_{i}}{s^{2}}\Bigr)g_{s}(z),$$

the stated formulas.

Claim 3 (bounds). Step (a): an elementary inequality. For every real $u\ge0$,

$$u\,\exp(-u^{2}/4)\le1,\qquad u^{2}\exp(-u^{2}/4)\le4,\qquad u^{3}\exp(-u^{2}/4)\le512 .$$

Indeed, $\exp(u^{2}/4)\ge1+u^{2}/4\ge u$, the last by $0\le(u/2-1)^{2}=u^{2}/4-u+1$ (claim 5 of Zero Products and Elementary Identities in a Field and claim 2 of Nonnegativity of Squares in an Ordered Field), and $\exp(u^{2}/4)\ge1+u^{2}/4\ge u^{2}/4$; multiplying by $\exp(-u^{2}/4)>0$ (claim 5 of Elementary Arithmetic in an Ordered Field) gives the first two bounds. For the third, if $u\le8$ then $u^{3}\le8^{3}=512$ (claim 5 of Properties of Natural Number Powers in a Field, monotonicity of powers of nonnegative numbers) and $\exp(-u^{2}/4)\le1$; if $8\le u$ then $\exp(u^{2}/4)=\exp(u^{2}/8)^{2}\ge(1+u^{2}/8)^{2}\ge(u^{2}/8)^{2}=u^{4}/64$, so $u^{3}\exp(-u^{2}/4)\le64u^{3}/u^{4}=64/u\le8\le512$.

Step (b): pointwise bounds with a Gaussian factor. Fix $z$, put $u=\lVert z\rVert/\sqrt{s}\ge0$, so that $\lVert z\rVert=\sqrt{s}\,u$ and $\lVert z\rVert^{2}/(2s)=u^{2}/2=u^{2}/4+u^{2}/4$; hence $\psi_{s}(z)=\exp(-u^{2}/4)\exp(-u^{2}/4)$ and, by step (a), for $m\in\{1,2,3\}$,

$$\lVert z\rVert^{m}\,\psi_{s}(z)=(\sqrt{s})^{m}\,u^{m}\exp(-u^{2}/4)\,\exp(-u^{2}/4)\le D_{m}(\sqrt{s})^{m}\exp\bigl(-\lVert z\rVert^{2}/(4s)\bigr),$$

with $D_{1}=1$, $D_{2}=4$, $D_{3}=512$. Since $|z_{i}|\le\lVert z\rVert$ for every $i$ (claim 4 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square) and $|\delta_{ij}|\le1$, the formulas of claim 2 and the triangle inequality (claims 4 and 5 of Properties of the Absolute Value in an Ordered Field) give, with $E_{s}(z)=\exp(-\lVert z\rVert^{2}/(4s))\le1$,

$$g_{s}(z)\le c_{s}E_{s}(z),\qquad|\partial_{i}g_{s}(z)|\le\frac{\lVert z\rVert}{s}c_{s}\psi_{s}(z)\le c_{s}\,D_{1}\frac{\sqrt{s}}{s}E_{s}(z)=c_{1}D_{1}\,s^{-(q+1)/2}E_{s}(z),$$ $$|\partial_{j}\partial_{i}g_{s}(z)|\le\Bigl(\frac{\lVert z\rVert^{2}}{s^{2}}+\frac1s\Bigr)c_{s}\psi_{s}(z)\le c_{1}(D_{2}+1)\,s^{-(q+2)/2}E_{s}(z),\qquad |\partial_{k}\partial_{j}\partial_{i}g_{s}(z)|\le\Bigl(\frac{\lVert z\rVert^{3}}{s^{3}}+\frac{3\lVert z\rVert}{s^{2}}\Bigr)c_{s}\psi_{s}(z)\le c_{1}(D_{3}+3D_{1})\,s^{-(q+3)/2}E_{s}(z),$$

using $c_{s}=c_{1}s^{-q/2}$ (claim 1), $\psi_{s}=E_{s}^{2}\le E_{s}$ (since $0<E_{s}\le1$) in the first bound and in the $\delta_{ij}/s$ term of the third, and the power rules of the preamble (for instance $(\sqrt{s})^{3}/s^{3}=s^{-3/2}$ and $\sqrt{s}/s^{2}=s^{-3/2}$, so both terms of the last bound carry the same power of $s$). Dropping the factor $E_{s}\le1$ gives the uniform bounds with $A_{0}=c_{1}$, $A_{1}=c_{1}D_{1}$, $A_{2}=c_{1}(D_{2}+1)$, $A_{3}=c_{1}(D_{3}+3D_{1})$, which depend only on $q$ (through $c_{1}$).

Step (c): square-integral bounds. Each of $g_{s}$, $\partial_{i}g_{s}$, $\partial_{j}\partial_{i}g_{s}$, $\partial_{k}\partial_{j}\partial_{i}g_{s}$ is continuous (claim 2), hence Borel, and so is its square (claim 3 of Arithmetic of Measurable Real- and [0,∞]-Valued Maps). Since $E_{s}(z)^{2}=\exp(-\lVert z\rVert^{2}/(2s))=\psi_{s}(z)$, writing $\partial^{\beta}g_{s}$ for any one of these four functions and $|\beta|\in\{0,1,2,3\}$ for its order, squaring the bounds of step (b) gives $(\partial^{\beta}g_{s})^{2}\le A_{|\beta|}^{2}\,s^{-(q+|\beta|)}\,\psi_{s}$ and $s^{-(q+|\beta|)}=(s^{-(q+|\beta|)/2})^{2}$; by claim 1 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion and $\int\psi_{s}=1/c_{s}=s^{q/2}/c_{1}$,

$$\int_{\mathbb{R}^{q}}(\partial^{\beta}g_{s})^{2}\,d\lambda_{q}\le A_{|\beta|}^{2}\,s^{-(q+|\beta|)}\,\frac{(\sqrt{s})^{q}}{c_{1}}=\frac{A_{|\beta|}^{2}}{c_{1}}\,s^{-(q+2|\beta|)/2},$$

a finite number, so the square is integrable (Integrable Functions and Their Lebesgue Integral), and $B_{|\beta|}=A_{|\beta|}^{2}/c_{1}$ serves.

Claim 4 (convolution identity). Let $0<s\le\tfrac12$, so that $s/2\le1$ and $2s\le1$, and let $x,x'\in\mathbb{R}^{q}$, $m=\tfrac12(x+x')$. For $y\in\mathbb{R}^{q}$ put $a=y-x$ and $b=y-x'$, so $a+b=2(y-m)$ and $a-b=x'-x$. Expanding $\lVert a\pm b\rVert^{2}=(a\pm b)\cdot(a\pm b)$ by the bilinearity and symmetry of the dot product (The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite, with $\lVert v\rVert^{2}=v\cdot v$ by The Euclidean Norm on $\mathbb{R}^n$) gives $\lVert a\rVert^{2}+\lVert b\rVert^{2}=\tfrac12\lVert a+b\rVert^{2}+\tfrac12\lVert a-b\rVert^{2}=2\lVert y-m\rVert^{2}+\tfrac12\lVert x-x'\rVert^{2}$, using 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 and $\lVert x'-x\rVert=\lVert x-x'\rVert$ (claim 2 there with $\mathbb{R}^n$ with the Euclidean Distance Is a Complete Metric Space with Coordinatewise Convergence). Hence

$$g_{s}(y-x)\,g_{s}(y-x')=c_{s}^{2}\exp\Bigl(-\frac{\lVert y-m\rVert^{2}}{2(s/2)}\Bigr)\exp\Bigl(-\frac{\lVert x-x'\rVert^{2}}{2(2s)}\Bigr)=\frac{c_{s}^{2}}{c_{s/2}\,c_{2s}}\,g_{s/2}(y-m)\,g_{2s}(x-x'),$$

and $c_{s/2}c_{2s}=c_{1}^{2}\bigl(\sqrt{s/2}\,\sqrt{2s}\bigr)^{-q}=c_{1}^{2}(\sqrt{s^{2}})^{-q}=c_{1}^{2}s^{-q}=c_{s}^{2}$ by claim 1 and $\sqrt{a}\sqrt{b}=\sqrt{ab}$. So $y\mapsto g_{s}(y-x)g_{s}(y-x')$ equals $g_{2s}(x-x')\cdot g_{s/2}(y-m)$, a nonnegative multiple of a Borel function with integral $1$; it is integrable with integral $g_{2s}(x-x')$ by claim 1 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion.

Claim 5 (moments and tails). For $i\in[q]$ apply claim 4 of The Gaussian Smoothing Weight: Normalization, Derivatives, Exponential Tilting, Moments, and First-Order Remainder with $a=e_{i}$, the $i$th standard basis vector: $Z_{a}(z)=z_{i}/s$ and $\kappa_{a}=\lVert e_{i}\rVert^{2}/s=1/s$, so $\int_{\mathbb{R}^{q}}\frac{z_{i}^{2}}{s^{2}}g_{s}(z)\,dz=\frac1s$, whence $\int_{\mathbb{R}^{q}}z_{i}^{2}g_{s}(z)\,dz=s$ by claim 1 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (multiplication by the nonnegative constant $s^{2}$). Since $\lVert z\rVert^{2}=\sum_{i=1}^{q}z_{i}^{2}$ (The Euclidean Norm on $\mathbb{R}^n$), the additivity in claim 1 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion, applied $q-1$ times by induction on the number of summands (the inductive set being the set of $n\le q$ for which the integral of $\sum_{i=1}^{n}z_{i}^{2}g_{s}$ is $ns$, with Principle of Induction for the Natural Numbers and claim 1 of Properties of Finite Sums), gives $\int\lVert z\rVert^{2}g_{s}=qs$, finite, so $z\mapsto\lVert z\rVert^{2}g_{s}(z)$ (Borel by claim 3 of Arithmetic of Measurable Real- and [0,∞]-Valued Maps and Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets (iv)) is integrable. Next, let $t=\sqrt{qs}$, which is positive since $qs>0$ (claim 5 of Elementary Order Arithmetic in an Ordered Field with claim 3 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field) and $t^{2}=qs$. For every $z$, $0\le(\lVert z\rVert-t)^{2}=\lVert z\rVert^{2}-2t\lVert z\rVert+t^{2}$ (claim 5 of Zero Products and Elementary Identities in a Field and claim 2 of Nonnegativity of Squares in an Ordered Field), so $2t\lVert z\rVert\le\lVert z\rVert^{2}+t^{2}$ (claim 3 of Elementary Arithmetic in an Ordered Field). Multiplying by $g_{s}(z)\ge0$ (claim 5 of Elementary Arithmetic in an Ordered Field) and integrating, by claim 1 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion with $\int g_{s}=1$ and the first moment identity,

$$2t\int_{\mathbb{R}^{q}}\lVert z\rVert g_{s}(z)\,dz\le qs+t^{2}=2t^{2},$$

so the integral of the nonnegative Borel function $z\mapsto\lVert z\rVert g_{s}(z)$ is finite, the function is integrable, and $\int\lVert z\rVert g_{s}(z)\,dz\le t=\sqrt{qs}$ (multiplying by the positive $(2t)^{-1}$, claim 5 of Elementary Arithmetic in an Ordered Field). Finally let $R>0$. For $z\in A_{R}$ one has $R<\lVert z\rVert$, hence $R^{2}<\lVert z\rVert^{2}$ (claim 1 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field) and $1<\lVert z\rVert^{2}/R^{2}$ (claim 10 of Elementary Order Arithmetic in an Ordered Field with the positive factor $R^{-2}$); for $z\notin A_{R}$ the indicator vanishes. Thus $\mathbf{1}_{A_{R}}(z)\,g_{s}(z)\le R^{-2}\lVert z\rVert^{2}g_{s}(z)$ for every $z$, and the monotonicity and homogeneity in claim 1 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion give $\int\mathbf{1}_{A_{R}}g_{s}\,d\lambda_{q}\le R^{-2}\,qs$, as claimed.