# associative law for convolution of distribution with test functions

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 $$\rho$$.)

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

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 $$(-(M-1),M-1)$$).

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 $$\mathcal D$$.