# Convolution of two Schwartz functions is Schwartz

I am trying to show directly (i.e., not using the Fourier transform) that if $S=S(\mathbb{R}^n)$ is the class of Schwartz functions then $f,g \in S$ implies $f*g \in S$.

The definition I have is $f \in S$ iff for all $N$, $\|f\|_N:=\sup_{x \in \mathbb{R}^d, |\alpha|,|\beta| \leq N} |x^\beta\partial_x^\alpha f(x)|<\infty$. I have shown that $\|f\|_N<\infty$ is equivalent to: for all $|\alpha|,|\beta| \leq N$, there exists a constant $C_N$ such that $|\partial_x^\alpha f(x)| \leq \frac{C_N}{\prod_{i=1}^n (1+|x_i^{\beta_i}|)}$.

I also know that $\partial_x^\alpha (f*g)(x) = \int_{\mathbb{R}^n} \partial_x^\alpha f(x-t) g(t) \ dt$. Since $\partial_x^\alpha f \in S$, I believe it suffices to show that for all $f,g \in S$, all $N$, and all $|\beta| \leq N$, we have $|(f*g)(x)| \leq \frac{C_N}{\prod_{i=1}^n (1+|x_i^{\beta_i}|)}$. I can prove this in the case $\mathbb{R}^n=\mathbb{R}$ as follows: Let $C_N$ be such that for all $b \leq N$, $|f(x)|,|g(x)| \leq \frac{C_N}{1+|x|^b}$. Then

\begin{align*} \left| \int_{\mathbb{R}} f(x-t)g(t) \ dt \right| & \leq \int_{\mathbb{R}} \left| f(x-t)g(t)\right| \ dt\\ &= \int_{|x-t| \geq |x|/2} \left| f(x-t)g(t)\right| \ dt + \int_{|x-t| < |x|/2} \left|f(x-t)g(t)\right| \ dt\\ &\leq \sup_{|x-t| \geq |x|/2} |f(x-t)|\int_{\mathbb{R}}|g(t)| \ dt + \sup_{|t| \geq |x|/2} |g(t)| \int_{\mathbb{R}} |f(x-t)| \ dt\\ &\leq \sup_{|x-t| \geq |x|/2}\frac{C_N}{1+|x-t|^b} \|g\|_{L^1} + \sup_{|t| \geq |x|/2}\frac{C_N}{1+|t|^b} \|f\|_{L^1}\\ &\leq \frac{C_N(\|g\|_{L^1}+\|f\|_{L^2})}{1+(|x|/2)^b}\\ & \leq \frac{2^NC_N(\|g\|_{L^1}+\|f\|_{L^2})}{1+|x|^b}. \end{align*} (Note that $|x-t|<|x|/2$ implies $|t|>|x|/2$.)

However, I am struggling with how to split up the integral in $\mathbb{R}^n$. In order to do the step $\frac{C_N}{\prod_{i=1}^n (1+|x_i-t_i|^{\beta_i})} \leq \frac{C_N}{\prod_{i=1}^n (1+(|x_i|/2)^{\beta_i})}$ I need $|x_i-t_i| \geq |x_i|/2$ for all $i$. But in order to do the step $\frac{C_N}{\prod_{i=1}^n (1+|t_i|^{\beta_i})} \leq \frac{C_N}{\prod_{i=1}^n (1+(|x_i|/2)^{\beta_i})}$ I need $|t_i| \geq |x_i|/2$ for all $i$. However these two sets do not encompass all of $\mathbb{R}^n$ (in particular, we miss all $t$ such that $|x_i-t_i| < |x_i|/2$ for some $i$ and $|t_j| < |x_j|/2$ for some $j$). I have tried integrating over one copy of $\mathbb{R}$ at a time, but as soon as we replace $f$ and $g$ by their upper bounds in the innermost integral, they are no longer guaranteed to have finite $L^1$ norm, so I cannot continue. (Note: I can show the inequality holds for $N=1$, i.e., $x^\beta = x_i$ for some $i$. Is that enough?) Any suggestions would be greatly appreciated.

One can proceed as follows: For $N > 0$ and $x \in \mathbb{R}^{n}$ \begin{align*} (1 + |x|)^{N}|f \star g(x)| &\leq \int_{\mathbb{R}^{n}} (1 + |x|)^{N}|f(x-t)||g(t)|\,dt\\ &\leq \int_{\mathbb{R}^{n}} (1 + |x-t|)^{N}(1 + |t|)^{N}|f(x-t)||g(t)|\,dt\\ &\leq ||f||_{N}\int_{\mathbb{R}^{n}}(1+|t|)^{N}|g(t)|\,dt\\ &\leq C_{N+2}||f||_{N}||g||_{n+N+1}\int_{\mathbb{R}^{n}}\frac{1}{(1 + |t|)^{n+1}}\,dt \end{align*} Here we have made use of the fact that $(1+ |x|) \leq (1+ |x-t|)(1+ |t|)$.