Let $S$ be the Schwartz class. Show that if $f,g\in S$, then $fg\in S$ and $f*g\in S$, where $*$ denotes convolution.

Question

Let $S$ be the Schwartz class. Show that if $f,g\in S$, then $fg\in S$ and $f*g\in S$, where $*$ denotes convolution.

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To differentiate $fg$, we may apply Leibniz's rule ( http://en.wikipedia.org/wiki/General_Leibniz_rule ). And then maybe induct on the order of the derivative.

Is there something useful that can be applied to differentiating $f*g$? I guess there's a product rule for convolution. But after using product rule, there's still a convolution sign.

Answer 1

$(f*g)' = (f')*g = f*(g')$. Just differentiate under the integral.

So it is sufficient to show that if $f$ and $g$ are Schwartz, then $f*g(x)/|x|^n \to \infty$ as $x \to \pm\infty$. Let's just do the case $+\infty$. Since $f$ and $g$ are $C^\infty$, they are bounded on any compact set, and hence $|f(x)|,|g(x)| \le C_n /(1+|x|^n)$ where $C_n$ depends upon $n$.

Then $$ |f*g(x)| = \left|\int_{-\infty}^\infty f(x-y) g(y) \, dy\right| \\ \le \left|\int_{-\infty}^{x/2} f(x-y) g(y) \, dy\right| + \left|\int_{x/2}^\infty f(x-y) g(y) \, dy\right| \\ \le \sup_{y\le x/2}|f(x-y)| \int_{-\infty}^\infty |g(y)| \, dy + \sup_{y\ge x/2}|g(y)| \int_{-\infty}^\infty |f(x-y)| \, dy \\ \le D C_n /(1+|x/2|^n) ,$$ where $D = \|f\|_1+\|g\|_1$.

Answer 2

You can use the fact that $\widehat{f*g} = \hat f \hat g$. If you know that $f,g \in S$ implies that $\hat f$ and $\hat g$ are in $S$, then use the Fourier inversion formula to get $f * g (x)= (2\pi) ^{-n}~~\widehat{\widehat{f*g}}(-x)$.

Since $\widehat{f*g} = \hat f \hat g$ and you know that the pointwise product of two functions in $S$ is in $S$, $\widehat{f*g}$ is in $S$, implying that $\widehat{\widehat{f*g}}$ is in the Schwartz space $S$.

All told, we get that $f * g$ is equal to a constant times a function in $S$.

Notice that everything in $S$ lies in $L^1$, so use of the Inversion formula is valid.