associative law for convolution of distribution with test functions

Question

Let ${\mathcal D} = C^\infty_c$, the space of test functions (smooth functions $\Bbb R \to \Bbb C$ with compact support). Let $\mathcal D^*$ be the space of distributions (continuous linear functionals ${\mathcal D} \to {\Bbb C}$).

I am trying to fill in details of a proof in Richards & Youn's Theory of Distributions: A Nontechnical Introduction that

$\qquad$ If $S \in \mathcal D^*$ and $\varphi, \psi \in {\mathcal D}$, then $S * (\varphi * \psi) = (S * \varphi) * \psi)$.

My sticking point is this: I need to show that $\rho_n \to \rho$ in $\mathcal D$, where

$\qquad \rho(x) = \int_{-\infty}^{+\infty} \varphi(v) \psi(x-v) dv \quad$ (This defines the convolution $\rho := \varphi * \psi.)$ $\qquad \rho_n(x) = \sum_{m=-\infty}^{+\infty} \varphi (\frac m n)\psi(x - \frac m n) {\frac 1 n}$.

By "convergence in $\mathcal D$" is meant that, for each $k = 0, 1, 2, \dots$, the sequence of derivatives $\rho_n^{(k)}$ converges uniformly to the derivative $\rho^{(k)}$.

Because $\text{support }\varphi \subseteq [-M, M]$ for some positive integer $M$, the ostensibly infinite limits can be replaced by finite ones. Observe that $\rho_n$ is a Riemann sum for $\rho$, where $\rho_n$ uses the partition ${\mathscr P}_n = \{ \frac m n \}$ consisting of equally spaced points separated by distance $\frac 1 n$. As $\rho$ necessarily exists and as $\text{mesh } {\mathscr P}_n = {\frac 1 n} \to 0$, one has pointwise convergence $\rho_n(x) \to \rho(x)$ for each $x \in \Bbb R$.

Any test function is uniformly continuous, and $\varphi, \psi, \rho,$ and $\rho_n$ are all test functions. Richards & Youn state:

$\qquad \rho_n \to \rho$ uniformly because of that uniform continuity.
$\qquad$ Can anyone provide details as to why that should be so?

Some facts that might be helpful: $\text{support }\psi \subseteq [-N, N]$ for some positive integer $N$. Therefore $\text{support }\rho, \text{ support }\rho_n \subseteq [-M-N, M+N]$.

(I don't need to prove uniform convergence of the $k$th derivatives, $k>0$. Such a proof would be virtually identical to that for uniform convergence of $\rho_n$ to $\rho$.)

Answer 1

$$|\rho(x)-\rho_n(x)|=\left|\sum_m{\int_{m/n}^{(m+1)/n}{(\varphi(v)\psi(x-v)-\varphi(m/n)\psi(x-m/n))\,dv}}\right| \leq \sum_m{\int_{m/n}^{(m+1)/n}{|\varphi(v)\psi(x-v)-\varphi(v)\psi(x-m/n)+\varphi(v)\psi(x-m/n)-\varphi(m/n)\psi(m/n)|\, dv}}$$

Thus

$$|\rho(x)-\rho_n(x)| \leq \sum_m{\|\psi\|_{\infty}\int_{m/n}{(m+1)/n}{|\varphi(v)-\varphi(m/n)|\,dv}+\|\psi’\|_{\infty}\int_{m/n}^{(m+1)/n}{|\varphi(v)||v-m/n|\,dv}}$$.

Hence

$$|\rho(x)-\rho_n(x)| \leq \|\psi\|_{\infty}\|\varphi’\|_{\infty}\sum_{|m| \leq M}{\int_{m/n}^{(m+1)/n}{|v-m/n|\,dv}}+1/n\|\psi’\|_{\infty}\|\varphi\|_{L^1} \leq C/n$$

where $C$ is a constant depending on the first-order seminorms of $\varphi,\psi$ and on the support of $\varphi$ (assumed to be in $(-(M-1),M-1)$).

Answer 2

I didn't quite understand Mindlack's answer, but it did suggest some critical steps which I've been able to exploit to form my own answer:

Write $$\rho(x) := \int_{-M}^M {\varphi(v) \psi(x-v) \, dv} = \sum_{m=-Mn+1}^{Mn} \int_{(m-1)/n}^{m/n} {\varphi(v) \psi(x-v) \, dv},$$
$$\rho_n(x) := \sum_{m=-Mn+1}^{Mn} {\varphi({\frac m n}) \psi(x-{\frac m n}) {\frac 1 n}} = \sum_{m=-Mn+1}^{Mn} \int_{(m-1)/n}^{m/n} {\varphi({\frac m n}) \psi(x-{\frac m n}) \,dv}.$$

For the Riemann sum $\rho_n(x)$ we partitioned integration interval $[-M, M]$ (where $M$ is a positive integer) into subintervals $[{\frac {m-1} n}, {\frac m n}]$ of equal width $\frac 1 n$; the integrand is evaluated that the right endpoint $\frac m n$ of the subinterval. Then

$$|\rho(x)-\rho_n(x)| = \left| \sum_{m=-Mn+1}^{Mn} \int_{(m-1)/n}^{m/n} \left[ \varphi(v) (\psi(x-v) -\psi(x - {\frac m n})) + (\varphi(v)- \varphi({\frac m n})) \psi(x-{\frac m n}) \right] \right|$$
(we've added and subtracted the same term from the integrand), so use of the triangle inequality gives $$\le {\sum_{m=-Mn+1}^{Mn} \int_{(m-1)/n}^{m/n} \left( \left| \varphi(v) \right| \cdot \left| \frac {\psi(x-v) - \psi(x-{\frac m n})} {(x-v)-(x-{\frac m n})} \right| \cdot \left| {(x-v)-(x-{\frac m n})} \right| + \left| \frac {\varphi(v) - \varphi(\frac m n)} {v - \frac m n} \right| \cdot \left| v - \frac m n\right| \cdot \left| \psi(x - \frac m n) \right| \right) \,dv.}$$

Observe:

• Factor $|\varphi(v)|$ is dominated by supremum norm $||\varphi||_\infty$, while factor $\left| \psi(x - \frac m n)\right|$ is dominated by $||\psi||_\infty$.

• From Rudin's Principles of Mathematical Analysis: Suppose continuous vector-valued function ${\mathbf f}:[a,b] \to {\Bbb R}^d$ is differentiable on $(a,b)$. Then there exists a point $t \in (a,b)$ such that
  $$ \left| \frac {{\mathbf f}(b) - {\mathbf f}(a)} {b-a} \right| \le |{\mathbf f'}(t)|.$$ Consequently factor $\left| \frac {\varphi(v) - \varphi(\frac m n)} {v - \frac m n} \right|$ is dominated by $||\varphi'||_\infty$. (If $\varphi$ is complex-valued rather than real-valued, think of $\varphi(v)$ as an element of ${\Bbb R}^2.$) Similarly factor $\left| \frac {\psi(x-v) - \psi(x-{\frac m n})} {(x-v)-(x-{\frac m n})} \right|$ is dominated by $||\psi'||_\infty$.

• Factors $\left| {(x-v)-(x-{\frac m n})} \right|$ and $\left| v- {\frac m n} \right|$ in the integrand's two terms both equal ${\frac m n} - v$.

• $ \int_{(m-1)/n}^{m/n}({\frac m n} - v) \,dv = {\frac 1 {2n^2}}.$
• The summation $\sum_{m=-Mn+1}^{Mn}$ contains $2Mn$ terms.

Consequently $$\forall x: |\rho(x)-\rho_n(x) | \le {\frac {C_0M} n} \quad \text{where} \quad C_0 := ||\varphi||_\infty \cdot ||\psi' ||_\infty + ||\varphi'||_\infty \cdot ||\psi||_\infty.$$

Differentiating $k$ times with respect to $x$ the initial expressions for $\rho(x)$ and $\rho_n(x)$, we have more generally $$\rho^{(k)}(x) := \int_{-M}^M {\varphi(v) \psi^{(k)}(x-v) \, dv} = \sum_{m=-Mn+1}^{Mn} \int_{(m-1)/n}^{m/n} {\varphi(v) \psi^{(k)}(x-v) \, dv}$$
$$\rho_n^{(k)}(x) := \sum_{m=-Mn+1}^{Mn} {\varphi({\frac m n}) \psi^{(k)}(x-{\frac m n}) {\frac 1 n}} = \sum_{m=-Mn+1}^{Mn} \int_{(m-1)/n}^{m/n} {\varphi({\frac m n}) \psi^{(k)}(x-{\frac m n}) \,dv}.$$

An argument identical to the above but with $\psi^{(k)}$ replacing $\psi$ then gives $$|\rho^{(k)}-\rho_n^{(k)} | \le {\frac {C_k M} n} \quad \text{where} \quad C_k := ||\varphi||_\infty \cdot ||\psi^{(k+1)} ||_\infty + ||\varphi'||_\infty \cdot ||\psi^{(k)}||_\infty.$$

For each $k = 0,1,2,\dots$ we have $\rho_n^{(k)} \to \rho^{(k)}$ uniformly since the upper bound ${\frac {C_k M} n}$, which is independent of $x$, shrinks to zero as $n \to \infty$. Conclude $\rho_n \to \rho$ in $\mathcal D$.