Proof of Differentiating a Convolution through the Kernel

Throughout, sums, differences and scalar multiples of points of $\mathbb{R}^n$ are those of Euclidean Space $\mathbb{R}^n$ is a Real Vector Space and Difference, Dot Product, and Orthogonality in $\mathbb{R}^n$, $0$ denotes the origin, $\lambda_n$ is Lebesgue measure on the Borel $\sigma$-algebra $\mathcal{B}(\mathbb{R}^n)$, and $\bar B(x,r)$ is the closed ball in $(\mathbb{R}^n,d)$. Let $e_i\in\mathbb{R}^n$ have $i$th coordinate $1$ and all other coordinates $0$, and for $z\in\mathbb{R}^n$ and a real number $s$ let $z[s]$ be the point whose $i$th coordinate is $s$ and whose $k$th coordinate is $z_k$ for every $k\ne i$, so that $z[z_i+t]=z+te_i$ for every real $t$. By claims 1, 2 and 5 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, $\lVert e_i\rVert=1$ and hence

$$d(z+te_i,z)=\lVert te_i\rVert=|t|\qquad(z\in\mathbb{R}^n,\ t\in\mathbb{R}).\tag{$*$}$$

Also, for $y,w\in\mathbb{R}^n$ the triangle inequality (claim 6 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$) applied to $y=(y+w)+(-w)$ together with claim 5 there gives

$$\lVert y+w\rVert\ \ge\ \lVert y\rVert-\lVert w\rVert.\tag{$**$}$$

Claim 1. The set $\mathbb{R}^n$ is Euclidean open, and by clauses 1 and 3 of C^k Maps on a Euclidean Open Set the hypothesis that $\rho$ is of class $C^1$ on $\mathbb{R}^n$ says exactly that $\rho$ is continuous at every point of $\mathbb{R}^n$, that the partial derivative of $\rho$ with respect to the $i$th variable exists at every point of $\mathbb{R}^n$, and that the function $\partial_i\rho:\mathbb{R}^n\to\mathbb{R}$ is continuous at every point of $\mathbb{R}^n$. By claim 1 of Euclidean Continuity Agrees with Metric Continuity for Real-Valued Functions (taken with $E=\mathbb{R}^n$) the last statement is exactly continuity of $\partial_i\rho$ on $\mathbb{R}^n$ as a map from $(\mathbb{R}^n,d)$ to $(\mathbb{R},d)$.

Now let $y\in\mathbb{R}^n$ with $\lVert y\rVert>\delta$ and put $r=\lVert y\rVert-\delta>0$. For every real $t$ with $|t|<r$ we get from $(**)$ and $(*)$ that $\lVert y+te_i\rVert\ge\lVert y\rVert-|t|>\delta$, so $\rho(y+te_i)=0$ by hypothesis on $\rho$; taking $t=0$ also gives $\rho(y)=0$. Hence every difference quotient $\bigl(\rho(y[y_i+t])-\rho(y)\bigr)/t$ with $0<|t|<r$ equals $0$, and Partial Derivative on a Euclidean Open Set gives $\partial_i\rho(y)=0$.

Thus $\partial_i\rho$ satisfies exactly the hypotheses imposed on the kernel in Convolution of a Continuous Function with a Compactly Supported Continuous Kernel, with the same $\delta$. Since the set $\Omega^{\delta}$ depends only on $\Omega$ and $\delta$, the convolution $f*(\partial_i\rho)$ is a real-valued function on $\Omega^{\delta}$.

Claim 2. By claim 2 of The $\delta$-Interior of an Open Subset of $\mathbb{R}^n$ is Open the set $\Omega^{\delta}$ is open in $(\mathbb{R}^n,d)$, hence Euclidean open by Euclidean Openness Agrees with Metric Openness on $\mathbb{R}^n$. If $\Omega^{\delta}$ is empty there is nothing further to prove, so fix $x\in\Omega^{\delta}$.

Room to spare, and a bound for $f$. By claim 1 of The $\delta$-Interior of an Open Subset of $\mathbb{R}^n$ is Open there is a real $\eta>0$ with $\bar B(x,\delta+\eta)\subseteq\Omega$. Write $K=\bar B(x,\delta+\eta)$ and $B=\bar B(0,\delta+\eta)$. By claim 2 of A Closed Euclidean Ball is Convex and Compact both $K$ and $B$ are compact, and both are nonempty (they contain their centres). Since $f$ is continuous on $\Omega$ and $K\subseteq\Omega$, the restriction of $f$ to $K$ has the continuity property required in Extreme Value Theorem on a Compact Subset of a Metric Space, so there are $z_-,z_+\in K$ with $f(z_-)\le f(z)\le f(z_+)$ for every $z\in K$. Choose a real number $M\ge0$ with $f(z_+)\le M$ and $-f(z_-)\le M$; then $|f(z)|\le M$ for every $z\in K$ by claim 6 of Properties of the Absolute Value in an Ordered Field. By claim 3 of Balls Have Positive Lebesgue Measure and Bounded Sets Have Finite Lebesgue Measure, $B\in\mathcal{B}(\mathbb{R}^n)$ and $C:=\lambda_n(B)$ is a real number with $C\ge0$.

A mean value identity for $\rho$. Let $y\in\mathbb{R}^n$ and let $t$ be a nonzero real number. Put $\varrho=|t|+1$ and $J=\{s\in\mathbb{R}:y_i-\varrho<s<y_i+\varrho\}$, an open interval, and let $\psi:J\to\mathbb{R}$ be given by $\psi(s)=\rho(y[s])$. Fix $s_0\in J$ and put $\varrho_0=\varrho-|s_0-y_i|>0$. Since the domain of $\rho$ is all of $\mathbb{R}^n$, the number $\varrho_0$ is an admissible radius at the point $y[s_0]$ in the sense of claim 1 of Slice Function and the Partial Derivative, and the associated slice function of $\rho$ at $y[s_0]$ in the $i$th variable is $s\mapsto\rho(y[s])$ on the interval $I_0=\{s:|s-s_0|<\varrho_0\}$, because $(y[s_0])[s]=y[s]$. As $\partial_i\rho(y[s_0])$ exists by claim 1, claim 2 of Slice Function and the Partial Derivative says that this slice function is differentiable at $s_0$ with derivative $\partial_i\rho(y[s_0])$. The slice function is the restriction of $\psi$ to $I_0\subseteq J$, and $I_0$ contains all points $s_0+u$ with $|u|<\varrho_0$; since differentiability at $s_0$ only constrains difference quotients over increments $u$ with $|u|$ smaller than a threshold that we may shrink to $\varrho_0$, $\psi$ is itself differentiable at $s_0$ with $\psi'(s_0)=\partial_i\rho(y[s_0])$.

Apply Mean Value Theorem on an Open Interval to $\psi$ on $J$ with the two points $y_i$ and $y_i+t$ of $J$ (they lie in $J$ because $|t|<\varrho$), taking $a$ to be the smaller and $b$ the larger of them. The conclusion is that $\psi(b)-\psi(a)=\psi'(c)(b-a)$ for some $c$ with $a<c<b$. If $t>0$ then $a=y_i$ and $b=y_i+t$, so this reads $\psi(y_i+t)-\psi(y_i)=\psi'(c)\,t$; if $t<0$ then $a=y_i+t$ and $b=y_i$, so it reads $\psi(y_i)-\psi(y_i+t)=\psi'(c)\,(-t)$, which is the same identity. Writing $\theta=c-y_i$, so that $y[c]=y+\theta e_i$ and $\theta$ lies strictly between $0$ and $t$, hence $|\theta|<|t|$, we obtain

$$\rho(y+te_i)-\rho(y)=\psi(y_i+t)-\psi(y_i)=\partial_i\rho(y+\theta e_i)\,t .$$

Uniform continuity of $\partial_i\rho$. By claim 1, $\partial_i\rho$ is continuous on $\mathbb{R}^n$ and vanishes at every $y$ with $\lVert y\rVert>\delta$, so it is compactly supported by claim 2 of Compact Support on $\mathbb{R}^n$ Means Vanishing Outside a Bounded Set, and therefore uniformly continuous on $\mathbb{R}^n$ by A Continuous Compactly Supported Function on $\mathbb{R}^n$ is Uniformly Continuous.

The estimate. Let $\varepsilon>0$ be a real number and put $\varepsilon'=\varepsilon/\bigl(1+M(1+C)\bigr)>0$. By uniform continuity there is a real $\sigma>0$ such that $|\partial_i\rho(y')-\partial_i\rho(y)|<\varepsilon'$ whenever $y,y'\in\mathbb{R}^n$ satisfy $d(y',y)<\sigma$. Put $\tau=\min\{\eta,\sigma\}>0$ and fix a real $t$ with $0<|t|<\tau$; write $x_t=x+te_i$.

(i) $x_t\in\Omega^{\delta}$. If $z\in\bar B(x_t,\delta)$ then $d(z,x_t)\le\delta$, so by the triangle inequality for the metric $d$ (Euclidean Distance is a Metric on $\mathbb{R}^n$) and by $(*)$, $d(z,x)\le d(z,x_t)+d(x_t,x)\le\delta+|t|<\delta+\eta$, so $z\in K\subseteq\Omega$. Hence $\bar B(x_t,\delta)\subseteq\Omega$.

(ii) Three integrands. Define $u,v,w:\mathbb{R}^n\to\mathbb{R}$ by

$$u(y)=f(x-y)\,\rho(y+te_i),\qquad v(y)=f(x-y)\,\rho(y),\qquad w(y)=f(x-y)\,\partial_i\rho(y)$$

for those $y$ with $x-y\in\Omega$, and $u(y)=v(y)=w(y)=0$ for all other $y$. Then $v$ is the function $h_x$ of Convolution of a Continuous Function with a Compactly Supported Continuous Kernel for the kernel $\rho$ and $w$ is the corresponding function for the kernel $\partial_i\rho$ (legitimate by claim 1), so both are integrable by claim 1 of The Convolution Integrand is Continuous, Compactly Supported and Integrable and

$$(f*\rho)(x)=\int_{\mathbb{R}^n}v\,d\lambda_n,\qquad \bigl(f*(\partial_i\rho)\bigr)(x)=\int_{\mathbb{R}^n}w\,d\lambda_n.$$

Let $p:\mathbb{R}^n\to\mathbb{R}$ be the function $h_{x_t}$ of Convolution of a Continuous Function with a Compactly Supported Continuous Kernel at the point $x_t$, which lies in $\Omega^{\delta}$ by (i); thus $(f*\rho)(x_t)=\int_{\mathbb{R}^n}p\,d\lambda_n$ and $p$ is integrable, again by claim 1 of The Convolution Integrand is Continuous, Compactly Supported and Integrable. Since $x_t-(y+te_i)=x-y$ for every $y$, the definitions give $p(y+te_i)=u(y)$ for every $y\in\mathbb{R}^n$. Hence, by claim 3 of Translation and Reflection Invariance of Lebesgue Measure on $\mathbb{R}^n$ applied with the point $te_i$, the function $u$ is integrable and

$$\int_{\mathbb{R}^n}u\,d\lambda_n=\int_{\mathbb{R}^n}p\,d\lambda_n=(f*\rho)(x_t).$$

(iii) A single integral. By claim 2 of Linearity and Monotonicity of the Lebesgue Integral, applied twice to the integrable functions $u$, $v$ and $w$, the function $q=t^{-1}(u-v)-w$ is integrable, with

$$\int_{\mathbb{R}^n}q\,d\lambda_n=\frac{(f*\rho)(x_t)-(f*\rho)(x)}{t}-\bigl(f*(\partial_i\rho)\bigr)(x), \qquad\Bigl|\int_{\mathbb{R}^n}q\,d\lambda_n\Bigr|\le\int_{\mathbb{R}^n}|q|\,d\lambda_n .$$

(iv) A pointwise bound for $q$. Let $y\in\mathbb{R}^n$. If $x-y\notin\Omega$ then $u(y)=v(y)=w(y)=0$ and so $q(y)=0$. If $x-y\in\Omega$ and $\lVert y\rVert>\delta+\eta$, then $\lVert y\rVert>\delta$ gives $\rho(y)=0$ and, by claim 1, $\partial_i\rho(y)=0$, while $(**)$ and $(*)$ give $\lVert y+te_i\rVert\ge\lVert y\rVert-|t|>(\delta+\eta)-\eta=\delta$ and hence $\rho(y+te_i)=0$; so again $q(y)=0$. In the remaining case $x-y\in\Omega$ and $\lVert y\rVert\le\delta+\eta$; then $y\in B$, and $d(x-y,x)=\lVert -y\rVert=\lVert y\rVert\le\delta+\eta$ by claims 2 and 5 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, so $x-y\in K$ and $|f(x-y)|\le M$. By the mean value identity above there is a real $\theta$ with $|\theta|<|t|<\sigma$ and

$$q(y)=f(x-y)\Bigl(\frac{\rho(y+te_i)-\rho(y)}{t}-\partial_i\rho(y)\Bigr)=f(x-y)\bigl(\partial_i\rho(y+\theta e_i)-\partial_i\rho(y)\bigr),$$

and $d(y+\theta e_i,y)=|\theta|<\sigma$ by $(*)$, so $|q(y)|\le M\varepsilon'$ by claim 4 of Properties of the Absolute Value in an Ordered Field. Altogether

$$|q(y)|\le M\varepsilon'\,\mathbf{1}_{B}(y)\qquad\text{for every }y\in\mathbb{R}^n,$$

$\mathbf{1}_{B}$ being the indicator function of $B$.

(v) Integrating the bound. The function $|q|$ is measurable, being the sum of the positive and negative parts of the measurable function $q$ (see Integrable Function and the Lebesgue Integral and claim 1 of Linearity and Monotonicity of the Lebesgue Integral), and $M\varepsilon'\mathbf{1}_{B}$ is measurable and nonnegative because $B\in\mathcal{B}(\mathbb{R}^n)$. By the monotonicity and homogeneity of the nonnegative integral (claim 1 of Linearity and Monotonicity of the Lebesgue Integral) and by The Integral of an Indicator Function is the Measure of the Set,

$$\int_{\mathbb{R}^n}|q|\,d\lambda_n\ \le\ M\varepsilon'\,\lambda_n(B)=M\varepsilon' C .$$

Since $M\ge0$ and $C\ge0$ we have $MC\le M(1+C)<1+M(1+C)$, so $M\varepsilon' C<\varepsilon$.

(vi) Conclusion. Combining (iii), (iv) and (v): for every real $t$ with $0<|t|<\tau$ the point $x[x_i+t]=x+te_i$ lies in $\Omega^{\delta}$ and

$$\Bigl|\frac{(f*\rho)(x[x_i+t])-(f*\rho)(x)}{t}-\bigl(f*(\partial_i\rho)\bigr)(x)\Bigr|<\varepsilon .$$

As $\varepsilon>0$ was arbitrary and $\tau$ depends only on $\varepsilon$, Partial Derivative on a Euclidean Open Set (with $\tau$ in the role of the threshold called $\delta$ there, the symbol $\delta$ being already in use) shows that the partial derivative of $f*\rho$ with respect to the $i$th variable exists at $x$ and equals $\bigl(f*(\partial_i\rho)\bigr)(x)$. $\blacksquare$