# Proof that the convolution of a tempered distribution with some $\varphi\in\scr{S}$ is a $C^\infty$ function

Let $u\in\mathscr{S}'(\mathbb{R}^n)$ be a tempered distribution, and let $\varphi\in\mathscr{S}(\mathbb{R}^n)$ be Schwartz class, then we consider $\varphi*u\in\scr{S}'$ by setting $$(\varphi*u)(\psi)=u(\tilde{\varphi}*\psi)$$ where $\tilde{\varphi}(x)=\varphi(-x)$, for all $\varphi\in\scr{S}$, and now it is not too hard to check that this is itself a tempered distribution. However, I'm struggling with the following proof that $\varphi*u$ is actually $C^\infty$.

Let $\psi\in\scr{S}(\mathbb{R}^n)$, so we have that \begin{align} (\varphi*u)(\psi)=u(\tilde{\varphi}*\psi) &= u\left(\int_{\mathbb{R}^n}\tilde{\varphi}(\cdot-y)\psi(y)\,\mathrm{d}y \right) \\ &= u\left(\int_{\mathbb{R}^n}\tau^y(\tilde{\varphi})(\cdot)\psi(y)\,\mathrm{d}y \right) \\ &= \int_{\mathbb{R}^n}u(\tau^y(\tilde{\varphi}))\psi(y)\,\mathrm{d}y \end{align}

and I'm told that the last step is justified by the continuity of $u$ and the fact that the Riemann sums converge in the $\mathscr{S}$ topology, though this isn't proven.

I suspect that the proof of that last statement will be very long and tedious, so I'm more curious about the general techniques that are necessary to prove it. Is there any specific sequence of test sets in $\mathbb{R}^n$ over which we can conveniently integrate, such that we have convergence w.r.t. $x$? My first naive attempt was to define $\Lambda_k=[-k,k]^n\cap\frac{1}{k}\mathbb{Z}^n$ but I was unable to prove that this would give us the necessary convergence.

• Issues of moving continuous linear maps inside vector-valued integrals occur everywhere. I think this is best addressed via Gelfand-Pettis/weak integrals, rather than using too much the specifics of various situations. The only specific here would be to follow Schwartz in compactifying $\mathbb R^n$ and extending the integrand continuously to that compactification, so that we have a continuous, compactly-supported vector-valued integral (values in quasi-complete, locally convex space), so there exists a weak integral, whose basic charactization is that interchange. Would this be of interest? Commented May 7, 2017 at 20:14
• Yes, that does seem potentially useful. Do you have a reference in mind for the Gelfand-Pettis integral? Commented May 7, 2017 at 20:31
• The originals by Gelfand and by Pettis are from the 1930s... Schwartz' compactification of $\mathbb R^n$ is from his c. 1950 work. Various of my functional analysis notes at www.math.umn.edu/~garrett/m/fun/ address these things quite explicitly, for example. Commented May 7, 2017 at 20:35
• If $\varphi$ is Schwartz and $u$ is a tempered distribution then $\frac{\partial^\alpha}{\partial x^\alpha}[\varphi \ast u] =u \ast\frac{\partial^\alpha \varphi }{\partial x^\alpha}$ which is well-defined since $\frac{\partial^\alpha \varphi }{\partial x^\alpha}$ is again Schwartz. Thus $\varphi \ast u$ is infinitely differentiable. Commented May 8, 2017 at 18:23

A convenient choice of Riemann sum in this case is the following: Let $h>0$ and consider the Riemann sum for $(\phi*\psi)(x)$ given by $$\tag{1} R_h(x):=\sum_{k\in\mathbb{Z}^n} \phi(x-hk)\psi(hk)h^n.$$ As $h\rightarrow 0$, one finds $R_h\rightarrow \phi*\psi$ with respect to most reasonable topologies. I took this parametrization from Lemma 4.1.3 of the classic 'The Analysis of Linear Partial Differential Operators 1' by Hörmander. It looks quite close to what you had in mind!
First, for fixed $h>0$, to show that the series (1) converges in $C_b(\mathbb{R}^n)$, the space of bounded continuous functions with supremum norm, note that $\lvert\phi(x-hk)\rvert\leqslant \lVert \phi \rVert_{L^\infty(\mathbb{R}^n)}$, while there is a constant $C>0$ such that $\psi(hk)\leqslant C(1+ h^2 \lvert k\rvert^2)^{-n}$. But then \begin{align*} \sum_{k\in\mathbb{Z}^n} \sup_{x\in\mathbb{R}^x}\lvert \phi(x-hk)\psi(hk)h^n \rvert & \leqslant C\lVert \phi\rVert_{L^\infty(\mathbb{R}^n)}\sum_{k\in\mathbb{Z}^n} \left(\frac{h}{1+h^2\lvert k \rvert^2}\right)^n \\ & \leqslant C\lVert \phi\rVert_{L^\infty(\mathbb{R}^n)}\left(h+2\sum_{k=1}^\infty \frac{h}{1+ h^2 k^2 }\right)^n \\ & \leqslant C2^n\lVert \phi\rVert_{L^\infty(\mathbb{R}^n)}\left(h + \int_{-\infty} ^{\infty}\frac{dt}{1+ t^2 }\right)^n < \infty, \end{align*} which ensures convergence of (1) in $C_b(\mathbb{R}^n)$. Now argue that convergence also happens in $\mathscr{S}(\mathbb{R}^n)$ (still for fixed $h>0$).
Next, we argue that $R_h\rightarrow \phi*\psi$ in $C_b(\mathbb{R}^n)$ as $h\rightarrow 0$. Let $$C_{k,h}=\{x\in\mathbb{R}^n \, | \, \forall j\in\{1,\ldots,n\}:\lvert x_j - k_j \rvert \leqslant h/2 \}$$ be the cube with side length $h>0$ centered at $k\in\mathbb{R}^n$. Let $f^y(x)=f(y-x)$ and let $f_h=\sum_{k\in\mathbb{Z}^n} f(k) \cdot 1_{C_{k,h}}$. Then we see that $$R_h(x)=\int (\phi^x)_h \psi_h \, dy.$$ Noting that \begin{align*} R_h(x)-\phi*\psi(x)&=\int (\phi^x)_h \psi_h \, dy - \int \phi^x \psi \, dy\\ &= \int ((\phi^x)_h-\phi^x) \psi_h \, dy + \int \phi^x (\psi_h-\psi) \, dy, \end{align*} we have $$\lVert R_h - \phi*\psi\rVert_{L^\infty(\mathbb{R}^n)} \leqslant \lVert \phi_h - \phi\rVert_{L^\infty(\mathbb{R}^n)} \lVert \psi_h\rVert_{L^1(\mathbb{R}^n)} + \lVert \phi\rVert_{L^\infty(\mathbb{R}^n)} \lVert \psi_h - \psi\rVert_{L^1(\mathbb{R}^n)}.$$ Since $\phi$ is uniformly continuous, $\lVert \phi_h - \phi\rVert_{L^\infty(\mathbb{R}^n)}\rightarrow 0$ as $h\rightarrow 0$, so it remains to show that $\psi_h\rightarrow \psi$ in $L^1(\mathbb{R}^n)$. Since $\psi_h\rightarrow \psi$ pointwise, it suffices to find a dominating function $m\in L^1(\mathbb{R}^n)$ with $\lvert\psi_h(x)\rvert\leqslant m(x)$ for, say, $0\leqslant h \leqslant 1$. Consider $m(x)=\max_{k\in C_{x,1}} \lvert \psi(k)\rvert$. Then we clearly have $\lvert \psi_h(x) \rvert \leqslant m(x)$ when $0\leqslant h \leqslant 1$. Furthermore, there is a constant $C'>0$ such that $$\lvert\psi(x)\rvert \leqslant C'(1+n+\lvert x\rvert^2)^{-n}.$$ thus, if $k\in C_{x,1}$, then $$\lvert\psi(k)\rvert \leqslant C'(1+n+\lvert k\rvert^2)^{-n} \leqslant C'(1+\lvert x\rvert^2)^{-n},$$ since $\lvert x- k\rvert^2\leqslant n$. We conclude that $$m(x)=\max_{k\in C_{x,1}} \lvert \psi(k)\rvert \leqslant C'(1+\lvert x\rvert^2)^{-n},$$ and therefore $m\in L^1(\mathbb{R}^n)$.
Finally, finish the argument by showing that $R_h\rightarrow \phi*\psi$ in $\mathscr S(\mathbb{R}^n)$ as well.