Proof of Gaussian Smoothing of a Bounded Borel Function on Euclidean Space: the Two Forms, Derivatives through the Kernel and through the Function, and Pointwise Convergence

Each result cited is universally quantified over the data in its own statement.

Throughout, $q$, $H$ and $M$ are as in the statement, and $g_{s}$, for $0<s\le1$, is the kernel of The Gaussian Kernels on Euclidean Space: Scaling, Derivatives up to Order Three, the Convolution Identity, Moments and Tails, that is, the Gaussian smoothing weight $\varphi_{\eta}$ of claim 1 of The Gaussian Smoothing Weight: Normalization, Derivatives, Exponential Tilting, Moments, and First-Order Remainder read in dimension $m=q$ with $\eta=s$. Elementary arithmetic and order in $\mathbb{R}$, including manipulations of finite sums, is that of The Real Numbers: Standing Notation and Background §background. The measure space is always $(\mathbb{R}^{q},\mathcal{B}(\mathbb{R}^{q}),\lambda_{q})$, and Borel for a map $\mathbb{R}^{q}\to\mathbb{R}$ means measurable with respect to $\mathcal{B}(\mathbb{R}^{q})$ and $\mathcal{B}(\mathbb{R})$, as in Probability Measures on Euclidean Space and Random Vectors: Standing Notation §borel-maps.

Step 0 (Conventions and standing facts). (a) Points. Points of $\mathbb{R}^{q}$ are read as $q$-tuples by Euclidean Points as Tuples of Real Numbers. Differences, sums and scalar multiples of points are formed coordinatewise, by clause 1 of Difference, Dot Product, and Orthogonality in $\mathbb{R}^n$, by Sum of Points of $\mathbb{R}^n$ and by Scalar Multiple of a Point of $\mathbb{R}^n$, and two points are equal exactly when all their coordinates agree, by claim 1 of Euclidean Points as Tuples of Real Numbers. Every identity between points used below, such as $y-(y-x)=x$, $(y_{0}-z)-(y_{0}-y)=y-z$ or $y-z=0_{\mathbb{R}^{q}}-(z-y)$, is verified in this way coordinate by coordinate, using only arithmetic in $\mathbb{R}$, and is not commented on further. For $k\in[q]$ let $e_{k}\in\mathbb{R}^{q}$ be the point whose $k$th coordinate is $1$ and whose other coordinates are $0$, which exists by claim 2 of Euclidean Points as Tuples of Real Numbers. Thus, for $p\in\mathbb{R}^{q}$, $i\in[q]$ and $t\in\mathbb{R}$, the point $p+te_{i}$ has $i$th coordinate $p_{i}+t$ and its other coordinates equal to those of $p$; in particular $(p+te_{i})+he_{i}=p+(t+h)e_{i}$, $(p+te_{i})-x=(p-x)+te_{i}$ and $p+0e_{i}=p$.

(b) Norms. By Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ §square, for $x,a\in\mathbb{R}^{q}$ we have $\sum_{k=1}^{q}(x_{k}-a_{k})^{2}=\lVert x-a\rVert^{2}$; for real $c$ and $k\in[q]$ we have $\lVert ce_{k}\rVert^{2}=c^{2}$, so $\lVert ce_{k}\rVert=|c|$ by the uniqueness in Existence and Uniqueness of the Nonnegative Square Root; in particular $\lVert(p+te_{i})-p\rVert=\lVert te_{i}\rVert=|t|$. By Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ §distance the Euclidean distance satisfies $d_{E}(x,a)=\lVert x-a\rVert$, by Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ §coordinate we have $|x_{k}|\le\lVert x\rVert$, by Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ §homogeneity we have $\lVert(-1)x\rVert=\lVert x\rVert$, and by Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ §triangle we have $\lVert x+a\rVert\le\lVert x\rVert+\lVert a\rVert$.

(c) The kernel. Let $0<s\le1$. By claim 1 of The Gaussian Smoothing Weight: Normalization, Derivatives, Exponential Tilting, Moments, and First-Order Remainder, $g_{s}$ is Borel and sequentially continuous, $0<g_{s}(z)$ and $g_{s}(0_{\mathbb{R}^{q}}-z)=g_{s}(z)$ for every $z$, and for every $a\in\mathbb{R}^{q}$ the map $z\mapsto g_{s}(z-a)$ is Borel with $\int_{\mathbb{R}^{q}}g_{s}(z-a)\,dz=1$; for $a=0_{\mathbb{R}^{q}}$ this gives $\int_{\mathbb{R}^{q}}g_{s}\,d\lambda_{q}=1$. A nonnegative function has vanishing negative part, so by Integrable Function and the Lebesgue Integral these functions are integrable, with integral $1$. Consequently, for $y,z\in\mathbb{R}^{q}$, since $y-z=0_{\mathbb{R}^{q}}-(z-y)$,

$$g_{s}(y-z)=g_{s}(z-y),$$

and $z\mapsto g_{s}(y-z)$ is Borel and integrable with integral $1$. By The Gaussian Kernels on Euclidean Space: Scaling, Derivatives up to Order Three, the Convolution Identity, Moments and Tails §derivatives the partial derivative of $g_{s}$ with respect to the $i$th variable exists at every $z$, with $\partial_{i}g_{s}(z)=-\frac{z_{i}}{s}\,g_{s}(z)$, so by (b)

$$|\partial_{i}g_{s}(z)|\le s^{-1}\,\lVert z\rVert\,g_{s}(z)\qquad(z\in\mathbb{R}^{q},\ i\in[q]);\tag{C1}$$

and $\partial_{i}g_{s}$ is sequentially continuous by claim 2 of The Gaussian Smoothing Weight: Normalization, Derivatives, Exponential Tilting, Moments, and First-Order Remainder. By The Gaussian Kernels on Euclidean Space: Scaling, Derivatives up to Order Three, the Convolution Identity, Moments and Tails §moments the function $z\mapsto\lVert z\rVert\,g_{s}(z)$ is integrable with $\int_{\mathbb{R}^{q}}\lVert z\rVert\,g_{s}(z)\,dz\le\sqrt{q\,s}$.

(d) Integrals. An integrable function is measurable by Integrable Function and the Lebesgue Integral, and a measurable $f:\mathbb{R}^{q}\to\mathbb{R}$ is integrable exactly when $\int_{\mathbb{R}^{q}}|f|\,d\lambda_{q}<\infty$, by Measure Spaces and the Lebesgue Integral: Standing Notation §integral; for measurable $f$ the function $|f|$ is measurable by claim 4 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, so $|f|$ is integrable whenever $f$ is. Linearity, monotonicity and the estimate $\bigl|\int f\bigr|\le\int|f|$ are those of Linearity and Monotonicity of the Lebesgue Integral §nonnegative and Linearity and Monotonicity of the Lebesgue Integral §integrable. In particular, if $f$ is Borel, $G$ is integrable and $|f(z)|\le G(z)$ for every $z$, then $\int|f|\,d\lambda_{q}\le\int G\,d\lambda_{q}<\infty$ by monotonicity of the nonnegative integral, so $f$ is integrable, and $\bigl|\int f\,d\lambda_{q}\bigr|\le\int G\,d\lambda_{q}$. We refer to this as domination. A finite sum of integrable functions with real coefficients is integrable, with the corresponding integral, by Linearity and Monotonicity of the Lebesgue Integral §integrable and induction on the number of summands.

Step 1 (Claim 1: the second form). Let $0<s\le1$ and $y\in\mathbb{R}^{q}$, and put $u(z)=g_{s}(y-z)\,H(z)$. By Step 0(c) the map $z\mapsto g_{s}(y-z)$ is Borel, and $H$ is Borel, so $u$ is Borel by claim 3 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions. Since $|H(z)|\le M$ and $g_{s}(y-z)>0$, we have $|u(z)|\le M\,g_{s}(y-z)$, and the right-hand side is integrable with integral $M$ by Step 0(c) and Linearity and Monotonicity of the Lebesgue Integral §integrable. By domination, $u$ is integrable and

$$\Bigl|\int_{\mathbb{R}^{q}}g_{s}(y-z)\,H(z)\,dz\Bigr|\le\int_{\mathbb{R}^{q}}|u|\,d\lambda_{q}\le M .$$

Step 2 (Claim 1: the first form and the equality). With $s$, $y$ and $u$ as in Step 1, for every $x\in\mathbb{R}^{q}$ we have $u(y-x)=g_{s}(y-(y-x))\,H(y-x)=H(y-x)\,g_{s}(x)$. Claim 3 of Translation and Reflection Invariance of Lebesgue Measure on $\mathbb{R}^n$, applied with $n=q$, $a=y$ and $f=u$ (Borel and integrable by Step 1), shows that $x\mapsto u(y-x)$ is integrable and that $\int_{\mathbb{R}^{q}}u(y-x)\,dx=\int_{\mathbb{R}^{q}}u\,d\lambda_{q}$. Hence $x\mapsto H(y-x)\,g_{s}(x)$ is integrable, so Borel by Step 0(d), and

$$\int_{\mathbb{R}^{q}}H(y-x)\,g_{s}(x)\,dx=\int_{\mathbb{R}^{q}}g_{s}(y-z)\,H(z)\,dz .$$

This common value is $H_{s}(y)$, and $|H_{s}(y)|\le M$ by Step 1. This proves claim 1.

Step 3 (An envelope for kernels shifted by at most one). Fix $0<s\le1$ and $y_{0}\in\mathbb{R}^{q}$. Put $C_{s}=\exp\bigl(q^{2}/(2s)\bigr)$, with $\exp$ the exponential function, define $G:\mathbb{R}^{q}\to\mathbb{R}$ by $G(x)=(\lVert x\rVert+q+1)\,g_{s}(x)$, and define $\Gamma:\mathbb{R}^{q}\to\mathbb{R}$ by

$$\Gamma(z)=C_{s}\sum_{k=1}^{q}\Bigl(G\bigl((y_{0}-qe_{k})-z\bigr)+G\bigl((y_{0}+qe_{k})-z\bigr)\Bigr).$$

We show that $\Gamma$ is nonnegative and integrable, and that for every $y\in\mathbb{R}^{q}$ with $\lVert y-y_{0}\rVert\le1$, every $z\in\mathbb{R}^{q}$ and every $i\in[q]$,

$$g_{s}(y-z)\le\Gamma(z),\qquad\lVert y-z\rVert\,g_{s}(y-z)\le\Gamma(z),\qquad|g_{s}(y-z)\,H(z)|\le M\,\Gamma(z),\qquad|\partial_{i}g_{s}(y-z)\,H(z)|\le M\,s^{-1}\,\Gamma(z).\tag{E}$$

Integrability. By Step 0(c) and Linearity and Monotonicity of the Lebesgue Integral §integrable, $G=\lVert\cdot\rVert\,g_{s}+(q+1)\,g_{s}$ is integrable, hence Borel by Step 0(d), and $G\ge0$. For each $b\in\mathbb{R}^{q}$, claim 3 of Translation and Reflection Invariance of Lebesgue Measure on $\mathbb{R}^n$ with $a=b$ and $f=G$ shows that $z\mapsto G(b-z)$ is integrable. Hence $\Gamma$, a finite sum of such functions multiplied by $C_{s}$, is integrable by Step 0(d), and $\Gamma\ge0$ because $C_{s}>0$ by claim 2 of Basic Properties of the Exponential Function.

Pointwise bounds. Let $y,z\in\mathbb{R}^{q}$ with $\lVert y-y_{0}\rVert\le1$, and put $v=y_{0}-z$ and $a=y_{0}-y$, so that $v-a=y-z$ and, by Step 0(b), $\lVert a\rVert=\lVert(-1)(y-y_{0})\rVert=\lVert y-y_{0}\rVert\le1$. Recall that $a\cdot v=\sum_{k=1}^{q}a_{k}v_{k}$ by clause 2 of Difference, Dot Product, and Orthogonality in $\mathbb{R}^n$.

First, claim 3 of The Gaussian Smoothing Weight: Normalization, Derivatives, Exponential Tilting, Moments, and First-Order Remainder (exponential tilting, with $m=q$, $\eta=s$ and the point $a$, so that $Z_{a}(v)=a\cdot v/s$ and $\kappa_{a}/2=\lVert a\rVert^{2}/(2s)$) gives

$$g_{s}(v)\,\exp\bigl(a\cdot v/s\bigr)=\exp\bigl(\lVert a\rVert^{2}/(2s)\bigr)\,g_{s}(v-a).$$

Since $0\le\lVert a\rVert^{2}/(2s)$, claims 1 and 4 of Basic Properties of the Exponential Function ($\exp(0)=1$ and $\exp$ strictly increasing) give $1\le\exp\bigl(\lVert a\rVert^{2}/(2s)\bigr)$, and as $g_{s}(v-a)>0$ we obtain $g_{s}(v-a)\le g_{s}(v)\exp(a\cdot v/s)$.

Second, choose $k_{*}\in[q]$ with $|v_{k}|\le|v_{k_{*}}|$ for every $k\in[q]$ (a largest of finitely many real numbers), let $\theta=1$ if $0\le v_{k_{*}}$ and $\theta=-1$ otherwise, so that $\theta v_{k_{*}}=|v_{k_{*}}|$, and put $b=(\theta q)\,e_{k_{*}}$. Then $b\cdot v=q\,|v_{k_{*}}|$, and since $|a_{k}|\le\lVert a\rVert\le1$ for every $k$ by Step 0(b),

$$a\cdot v=\sum_{k=1}^{q}a_{k}v_{k}\le\sum_{k=1}^{q}|v_{k}|\le q\,|v_{k_{*}}|=b\cdot v .$$

As $0<s$ and $\exp$ is increasing, $\exp(a\cdot v/s)\le\exp(b\cdot v/s)$. Tilting once more, now with the point $b$, for which $\lVert b\rVert=q$ by Step 0(b), gives $g_{s}(v)\exp(b\cdot v/s)=\exp\bigl(q^{2}/(2s)\bigr)\,g_{s}(v-b)=C_{s}\,g_{s}(v-b)$. Altogether

$$g_{s}(y-z)=g_{s}(v-a)\le C_{s}\,g_{s}(v-b).$$

Third, $v-a=(v-b)+(b+(-1)a)$, so by the triangle inequality and homogeneity of Step 0(b), $\lVert v-a\rVert\le\lVert v-b\rVert+\lVert b\rVert+\lVert a\rVert\le\lVert v-b\rVert+q+1$. Since also $1\le\lVert v-b\rVert+q+1$, both $g_{s}(y-z)$ and $\lVert y-z\rVert\,g_{s}(y-z)$ are at most $C_{s}\,(\lVert v-b\rVert+q+1)\,g_{s}(v-b)=C_{s}\,G(v-b)$. Finally $v-b$ equals $(y_{0}-qe_{k_{*}})-z$ if $\theta=1$ and $(y_{0}+qe_{k_{*}})-z$ if $\theta=-1$, so $C_{s}\,G(v-b)$ is one of the nonnegative summands defining $\Gamma(z)$ and is therefore at most $\Gamma(z)$. This proves the first two inequalities of (E). The third follows from the first and $|H(z)|\le M$; the fourth follows from (C1) of Step 0(c), $|H(z)|\le M$ and the second inequality.

Step 4 (Claim 2: the partial derivatives exist and are given by the kernel). Fix $0<s\le1$, $i\in[q]$ and $y_{0}\in\mathbb{R}^{q}$, and let $\Gamma$ be the function of Step 3 for these $s$ and $y_{0}$. Let $U$ be the open interval with endpoints $-1$ and $1$, and define $f:U\times\mathbb{R}^{q}\to\mathbb{R}$ by $f(t,z)=g_{s}\bigl((y_{0}+te_{i})-z\bigr)\,H(z)$. We check the hypotheses of Differentiation under the Integral Sign for the measure space $(\mathbb{R}^{q},\mathcal{B}(\mathbb{R}^{q}),\lambda_{q})$.

(i) For $t\in U$, Step 1 with $y=y_{0}+te_{i}$ shows that $z\mapsto f(t,z)$ is integrable, and by Step 2 its integral is $H_{s}(y_{0}+te_{i})$.

(ii) Fix $z\in\mathbb{R}^{q}$ and $t\in U$, and put $w=(y_{0}+te_{i})-z$. For real $h$ we have $(y_{0}+(t+h)e_{i})-z=w+he_{i}$, the point obtained from $w$ by replacing its $i$th coordinate $w_{i}$ by $w_{i}+h$ (Step 0(a)). By Step 0(c) the partial derivative of $g_{s}$ with respect to the $i$th variable exists at $w$, with value $\partial_{i}g_{s}(w)$; by Partial Derivative on a Euclidean Open Set (on the open set $\mathbb{R}^{q}$) this means that for every positive $\varepsilon'$ there is a positive $\delta$ such that $\bigl|\bigl(g_{s}(w+he_{i})-g_{s}(w)\bigr)/h-\partial_{i}g_{s}(w)\bigr|<\varepsilon'$ whenever $0<|h|<\delta$. Given a positive $\varepsilon$, apply this with $\varepsilon'=\varepsilon/(|H(z)|+1)$; multiplying the difference quotient by $H(z)$ shows that for $0<|h|<\delta$ with $t+h\in U$,

$$\Bigl|\frac{f(t+h,z)-f(t,z)}{h}-\partial_{i}g_{s}(w)\,H(z)\Bigr|=|H(z)|\,\Bigl|\frac{g_{s}(w+he_{i})-g_{s}(w)}{h}-\partial_{i}g_{s}(w)\Bigr|<\varepsilon .$$

Hence, by Derivative at an Interior Point, the function $t'\mapsto f(t',z)$ on $U$ is differentiable at $t$, with derivative $D_{1}f(t,z)=\partial_{i}g_{s}\bigl((y_{0}+te_{i})-z\bigr)\,H(z)$.

(iii) For $t\in U$ the point $y=y_{0}+te_{i}$ satisfies $\lVert y-y_{0}\rVert=|t|<1$ by Step 0(b), so the last inequality of (E) gives $|D_{1}f(t,z)|\le M\,s^{-1}\,\Gamma(z)$ for every $z$, and $M\,s^{-1}\,\Gamma$ is integrable by Step 3 and Linearity and Monotonicity of the Lebesgue Integral §integrable.

By Differentiation under the Integral Sign, taken at $t=0$, the function $z\mapsto D_{1}f(0,z)=\partial_{i}g_{s}(y_{0}-z)\,H(z)$ is measurable, that is Borel, and integrable, and the function $F:U\to\mathbb{R}$, $F(t)=H_{s}(y_{0}+te_{i})$, is differentiable at $0$ with

$$F'(0)=\int_{\mathbb{R}^{q}}\partial_{i}g_{s}(y_{0}-z)\,H(z)\,dz=:J_{i}(y_{0}).$$

Now let $\varepsilon$ be positive, let $\delta$ be the positive number provided by Derivative at an Interior Point for $F$ at $0$ and $\varepsilon$, and let $\delta'$ be the lesser of $\delta$ and $1$. Every real $h$ with $0<|h|<\delta'$ lies in $U$, and since $F(0)=H_{s}(y_{0})$,

$$\Bigl|\frac{H_{s}(y_{0}+he_{i})-H_{s}(y_{0})}{h}-J_{i}(y_{0})\Bigr|<\varepsilon ,$$

where $y_{0}+he_{i}$ is the point obtained from $y_{0}$ by replacing $y_{0,i}$ by $y_{0,i}+h$. By Partial Derivative on a Euclidean Open Set, on the open set $\mathbb{R}^{q}$ (open by claim 1 of Euclidean Space is Open in Itself, and $C^k$ Maps are Continuous), the partial derivative of $H_{s}$ with respect to the $i$th variable exists at $y_{0}$ with value $J_{i}(y_{0})$. Such a value is unique: if $L$ and $L'$ both qualify, then for every positive $\varepsilon$ the difference quotient lies within $\varepsilon$ of both for all small $h\ne0$, so $|L-L'|<2\varepsilon$, whence $L=L'$. Thus $\partial_{i}(H_{s})(y_{0})=J_{i}(y_{0})$. As $y_{0}$ was arbitrary, for every $y\in\mathbb{R}^{q}$ the function $z\mapsto\partial_{i}g_{s}(y-z)\,H(z)$ is Borel and integrable and $\partial_{i}(H_{s})(y)=\int_{\mathbb{R}^{q}}\partial_{i}g_{s}(y-z)\,H(z)\,dz$, which is the displayed formula of claim 2.

Step 5 (Claim 2: continuity of $H_{s}$ and of its partial derivatives). Fix $0<s\le1$, $i\in[q]$ and $y_{0}\in\mathbb{R}^{q}$, and let $\Gamma$ be the function of Step 3 for these $s$ and $y_{0}$. Let $\Phi$ denote either $H_{s}$ or $\partial_{i}(H_{s})$.

(a) Sequences. Let $(y_{m})_{m\in\mathbb{N}}$ be a sequence in $\mathbb{R}^{q}$ with $\lVert y_{m}-y_{0}\rVert\le1$ for every $m$ and with $(\lVert y_{m}-y_{0}\rVert)_{m}$ converging to $0$. For each $z$, Step 0(b) gives $d_{E}(y_{m}-z,\,y_{0}-z)=\lVert(y_{m}-z)-(y_{0}-z)\rVert=\lVert y_{m}-y_{0}\rVert$, which converges to $0$; by the sequential continuity of $g_{s}$ and $\partial_{i}g_{s}$ (Step 0(c)), $g_{s}(y_{m}-z)\to g_{s}(y_{0}-z)$ and $\partial_{i}g_{s}(y_{m}-z)\to\partial_{i}g_{s}(y_{0}-z)$, and multiplying by the fixed number $H(z)$ preserves these convergences, since $|H(z)c_{m}-H(z)c|=|H(z)|\,|c_{m}-c|$. The functions $z\mapsto g_{s}(y_{m}-z)\,H(z)$ and $z\mapsto\partial_{i}g_{s}(y_{m}-z)\,H(z)$ are Borel by Steps 1 and 4, and by (E) they are dominated, for every $m$, by the integrable functions $M\,\Gamma$ and $M\,s^{-1}\,\Gamma$ respectively. By claim 3 of Dominated Convergence Theorem, together with Steps 2 and 4, $(\Phi(y_{m}))_{m}$ converges to $\Phi(y_{0})$.

(b) Continuity at $y_{0}$. We show that $\Phi$ is continuous at $y_{0}$. Suppose not. Then there is a positive $\varepsilon$ such that for every positive $\delta$ some $x\in\mathbb{R}^{q}$ satisfies $\sum_{k=1}^{q}(x_{k}-y_{0,k})^{2}<\delta^{2}$ and $\varepsilon^{2}\le(\Phi(x)-\Phi(y_{0}))^{2}$. Let $(h_{m})_{m\in\mathbb{N}}$ be a sequence of positive reals converging to $0$, which exists by Existence of a Sequence of Positive Real Numbers with Limit Zero, and for each $m$ choose such an $x$, called $y_{m}$, for $\delta$ the lesser of $h_{m}$ and $1$. By Step 0(b), $\lVert y_{m}-y_{0}\rVert^{2}<\delta^{2}$, so $\lVert y_{m}-y_{0}\rVert<\delta$, whence $\lVert y_{m}-y_{0}\rVert\le1$ and $0\le\lVert y_{m}-y_{0}\rVert<h_{m}$; thus $(\lVert y_{m}-y_{0}\rVert)_{m}$ converges to $0$. Moreover $\varepsilon\le|\Phi(y_{m})-\Phi(y_{0})|$ for every $m$. This contradicts part (a). Hence $\Phi$ is continuous at $y_{0}$.

Step 6 (Claim 2: class $C^{1}$ and the bound). Let $0<s\le1$. By Steps 4 and 5, applied at every point and for every $i\in[q]$, the function $H_{s}$ is continuous at every point of $\mathbb{R}^{q}$, its partial derivative with respect to each variable exists at every point, and each $\partial_{i}(H_{s})$ is continuous at every point. By clause 1 of C^k Maps on a Euclidean Open Set, read through its clause 3, $H_{s}$ is of class $C^{1}$ on $\mathbb{R}^{q}$, which is the class of Differential Calculus and Convexity on Euclidean Open Sets: Standing Notation §derivatives.

For the bound fix $i\in[q]$ and $y\in\mathbb{R}^{q}$. By (C1) and $|H(z)|\le M$, $|\partial_{i}g_{s}(y-z)\,H(z)|\le M\,s^{-1}\,\lVert y-z\rVert\,g_{s}(y-z)$ for every $z$. Claim 3 of Translation and Reflection Invariance of Lebesgue Measure on $\mathbb{R}^n$, with $a=y$ applied to the integrable function $x\mapsto\lVert x\rVert\,g_{s}(x)$ of Step 0(c), shows that $z\mapsto\lVert y-z\rVert\,g_{s}(y-z)$ is integrable with $\int_{\mathbb{R}^{q}}\lVert y-z\rVert\,g_{s}(y-z)\,dz=\int_{\mathbb{R}^{q}}\lVert x\rVert\,g_{s}(x)\,dx\le\sqrt{q\,s}$. By Step 4, domination (Step 0(d)) and Linearity and Monotonicity of the Lebesgue Integral §integrable,

$$|\partial_{i}(H_{s})(y)|\le M\,s^{-1}\int_{\mathbb{R}^{q}}\lVert y-z\rVert\,g_{s}(y-z)\,dz\le M\,s^{-1}\sqrt{q\,s}.$$

Finally $\sqrt{q}\,\sqrt{s}$ is nonnegative with square $q\,s$, so $\sqrt{q\,s}=\sqrt{q}\,\sqrt{s}$ by the uniqueness in Existence and Uniqueness of the Nonnegative Square Root; and $s=(\sqrt{s})^{2}$ with $\sqrt{s}>0$, so $s^{-1}\sqrt{s}$ is the multiplicative inverse of $\sqrt{s}$, which is $s^{-1/2}$. Hence $|\partial_{i}(H_{s})(y)|\le M\sqrt{q}\,s^{-1/2}$. This proves claim 2.

Step 7 (Claim 3: derivatives through the function). Suppose $H\in C^{1}_{b}(\mathbb{R}^{q})$, and let $0<s\le1$ and $i\in[q]$. By Bounded Continuously Differentiable Functions with Bounded Partial Derivatives on Euclidean Space §bounded, $H$ is of class $C^{1}$ on $\mathbb{R}^{q}$, the partial derivative $\partial_{i}H(x)$ exists at every $x$, and $\partial_{i}H$ is bounded: there is a nonnegative real $M_{i}$ with $|\partial_{i}H(x)|\le M_{i}$ for every $x$. By clause 1 of C^k Maps on a Euclidean Open Set, read through its clause 3, $\partial_{i}H$ is continuous at every point in the sense of Continuity at a Point for Maps Between Euclidean Spaces.

Borel. Let $(x^{k})_{k\in\mathbb{N}}$ be a sequence in $\mathbb{R}^{q}$ and $x\in\mathbb{R}^{q}$ with $d_{E}(x^{k},x)\to0$, and let $\varepsilon$ be positive. Take the positive $\delta$ given by continuity of $\partial_{i}H$ at $x$ for $\varepsilon$, and $N$ with $d_{E}(x^{k},x)<\delta$ for every $k\ge N$. For such $k$, Step 0(b) gives $\sum_{j=1}^{q}(x^{k}_{j}-x_{j})^{2}=d_{E}(x^{k},x)^{2}<\delta^{2}$, hence $(\partial_{i}H(x^{k})-\partial_{i}H(x))^{2}<\varepsilon^{2}$ and so $|\partial_{i}H(x^{k})-\partial_{i}H(x)|<\varepsilon$. Thus $\partial_{i}H$ is sequentially continuous, hence measurable with respect to $\mathcal{B}_{q}$ and $\mathcal{B}(\mathbb{R})$ by claim 3(a) of The Borel Sigma-Algebra of a Euclidean Space as a Product, and Measurability of Projections, Sequentially Continuous Maps, and Open and Closed Sets, and $\mathcal{B}_{q}=\mathcal{B}(\mathbb{R}^{q})$ by claim 5 there; so $\partial_{i}H$ is Borel and bounded by $M_{i}$. Hence claim 1, applied with $\partial_{i}H$ and its bound $M_{i}$ in place of $H$ and $M$ (Steps 1 and 2 used only that $H$ is Borel and bounded by $M$), shows that for every $y\in\mathbb{R}^{q}$ the functions $x\mapsto\partial_{i}H(y-x)\,g_{s}(x)$ and $z\mapsto g_{s}(y-z)\,\partial_{i}H(z)$ are Borel and integrable with equal integrals, so that the smoothing $(\partial_{i}H)_{s}$ is defined, with $(\partial_{i}H)_{s}(y)=\int_{\mathbb{R}^{q}}\partial_{i}H(y-x)\,g_{s}(x)\,dx$ and $|(\partial_{i}H)_{s}(y)|\le M_{i}$.

Differentiation. Fix $y\in\mathbb{R}^{q}$, let $U$ be the open interval with endpoints $-1$ and $1$, and define $f:U\times\mathbb{R}^{q}\to\mathbb{R}$ by $f(t,x)=H\bigl((y+te_{i})-x\bigr)\,g_{s}(x)$. (i) For $t\in U$, Step 2 with $y+te_{i}$ in place of $y$ shows that $x\mapsto f(t,x)$ is integrable with integral $H_{s}(y+te_{i})$. (ii) Fix $x$ and $t\in U$ and put $w=(y+te_{i})-x$; then $(y+(t+h)e_{i})-x=w+he_{i}$, the partial derivative of $H$ with respect to the $i$th variable exists at $w$, and exactly as in Step 4(ii), with the constant factor $g_{s}(x)$ in place of $H(z)$, the function $t'\mapsto f(t',x)$ on $U$ is differentiable at $t$ with $D_{1}f(t,x)=\partial_{i}H\bigl((y+te_{i})-x\bigr)\,g_{s}(x)$. (iii) $|D_{1}f(t,x)|\le M_{i}\,g_{s}(x)$ for all $t\in U$ and $x$, and $M_{i}\,g_{s}$ is integrable by Step 0(c). By Differentiation under the Integral Sign the function $F(t)=H_{s}(y+te_{i})$ on $U$ is differentiable at $0$, with

$$F'(0)=\int_{\mathbb{R}^{q}}\partial_{i}H(y-x)\,g_{s}(x)\,dx=(\partial_{i}H)_{s}(y),$$

the last equality being the first form of claim 1, applied with $\partial_{i}H$ and $M_{i}$ in place of $H$ and $M$. Exactly as at the end of Step 4, the partial derivative of $H_{s}$ with respect to the $i$th variable exists at $y$ with value $F'(0)$, and by uniqueness of that value $\partial_{i}(H_{s})(y)=(\partial_{i}H)_{s}(y)$. As $y$ was arbitrary, $\partial_{i}(H_{s})=(\partial_{i}H)_{s}$. This proves claim 3.

Step 8 (Claim 4: pointwise convergence). Let $y\in\mathbb{R}^{q}$ be a point at which $H$ is continuous in the sense of Continuity at a Point for Maps Between Euclidean Spaces, and let $\varepsilon$ be a positive real number. The choices are made in the order $\varepsilon$, then $r$, then $s_{0}$, then $s$.

Choice of $r$. By Continuity at a Point for Maps Between Euclidean Spaces, taken with $n=q$, $m=1$, $E=\mathbb{R}^{q}$, $f=H$ (whose single coordinate function is $H$) and $a=y$, and applied with $\varepsilon/2$ in place of $\varepsilon$, there is a positive real $r$ such that for every $x\in\mathbb{R}^{q}$ with $\sum_{k=1}^{q}(x_{k}-y_{k})^{2}<r^{2}$ we have $(H(x)-H(y))^{2}<(\varepsilon/2)^{2}$. Let $x\in\mathbb{R}^{q}$ with $\lVert x-y\rVert<r$. Since $0\le\lVert x-y\rVert$, we get $\lVert x-y\rVert^{2}<r^{2}$, and $\sum_{k=1}^{q}(x_{k}-y_{k})^{2}=\lVert x-y\rVert^{2}$ by Step 0(b); hence $(H(x)-H(y))^{2}<(\varepsilon/2)^{2}$, that is, $|H(x)-H(y)|^{2}<(\varepsilon/2)^{2}$, and as both $|H(x)-H(y)|$ and $\varepsilon/2$ are nonnegative, $|H(x)-H(y)|<\varepsilon/2$. Thus $|H(x)-H(y)|<\varepsilon/2$ for every $x\in\mathbb{R}^{q}$ with $\lVert x-y\rVert<r$.

Choice of $s_{0}$. Put $R=r/2$ and let $s_{0}$ be the lesser of $1$ and $\varepsilon\,r^{2}/(16\,M\,q+1)$; then $0<s_{0}\le1$. Let $A_{R}=\{x\in\mathbb{R}^{q}:R<\lVert x\rVert\}$, which is Borel by The Gaussian Kernels on Euclidean Space: Scaling, Derivatives up to Order Three, the Convolution Identity, Moments and Tails §moments; its indicator $\mathbf{1}_{A_{R}}$ is Borel by claim 1 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, and $\mathbf{1}_{A_{R}}\,g_{s}$ is Borel by claim 3 there.

Estimate. Let $0<s\le s_{0}$ and define $D(x)=H(y-x)\,g_{s}(x)-H(y)\,g_{s}(x)$. By Step 2, Step 0(c) and Linearity and Monotonicity of the Lebesgue Integral §integrable, $D$ is integrable with $\int_{\mathbb{R}^{q}}D\,d\lambda_{q}=H_{s}(y)-H(y)\int_{\mathbb{R}^{q}}g_{s}\,d\lambda_{q}=H_{s}(y)-H(y)$. Let $x\in\mathbb{R}^{q}$; as $g_{s}(x)>0$, $|D(x)|=|H(y-x)-H(y)|\,g_{s}(x)$. If $\lVert x\rVert<r$, then $\lVert(y-x)-y\rVert=\lVert(-1)x\rVert=\lVert x\rVert<r$ by Step 0(b), so $|D(x)|\le(\varepsilon/2)\,g_{s}(x)$. If $r\le\lVert x\rVert$, then $R<\lVert x\rVert$, so $\mathbf{1}_{A_{R}}(x)=1$, and $|H(y-x)-H(y)|\le2M$ gives $|D(x)|\le2M\,\mathbf{1}_{A_{R}}(x)\,g_{s}(x)$. In both cases, all terms being nonnegative,

$$|D(x)|\le\frac{\varepsilon}{2}\,g_{s}(x)+2M\,\mathbf{1}_{A_{R}}(x)\,g_{s}(x).$$

By Linearity and Monotonicity of the Lebesgue Integral §integrable and Linearity and Monotonicity of the Lebesgue Integral §nonnegative (monotonicity, additivity and homogeneity of the nonnegative integral), $\int g_{s}\,d\lambda_{q}=1$, and the tail bound of The Gaussian Kernels on Euclidean Space: Scaling, Derivatives up to Order Three, the Convolution Identity, Moments and Tails §moments,

$$|H_{s}(y)-H(y)|\le\int_{\mathbb{R}^{q}}|D|\,d\lambda_{q}\le\frac{\varepsilon}{2}+2M\int_{\mathbb{R}^{q}}\mathbf{1}_{A_{R}}\,g_{s}\,d\lambda_{q}\le\frac{\varepsilon}{2}+\frac{2M\,q\,s}{R^{2}}=\frac{\varepsilon}{2}+\frac{8M\,q\,s}{r^{2}} .$$

Since $s\le s_{0}\le\varepsilon\,r^{2}/(16Mq+1)$, we get $8Mqs/r^{2}\le8Mq\,\varepsilon/(16Mq+1)$, and $8Mq\,\varepsilon/(16Mq+1)<\varepsilon/2$ because $16Mq\,\varepsilon<(16Mq+1)\,\varepsilon$. Hence $|H_{s}(y)-H(y)|<\varepsilon$ for every $s$ with $0<s\le s_{0}$. This proves claim 4.