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 $\rho$.)


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


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


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

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 $$\le {\sum_{m=-Mn+1}^{Mn} \int_{(m-1)/n}^{m/n} \left( \left| \varphi(v) \right| \cdot \left| \frac {\psi(x-v) - \psi(x-{\frac m n})} {(x-v)-(x-{\frac m n})} \right| \cdot \left| {(x-v)-(x-{\frac m n})} \right| + \left| \frac {\varphi(v) - \varphi(\frac m n)} {v - \frac m n} \right| \cdot \left| v - \frac m n\right| \cdot \left| \psi(x - \frac m n) \right| \right) \,dv.}$$


  • 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 $\mathcal D$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.