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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.

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.

Christmas Bunny
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2 Answers2

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$(f*g)' = (f')*g = f*(g')$. 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 $D = \|f\|_1+\|g\|_1$.

Stephen Montgomery-Smith
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  • Yep. I also show it here http://en.wikipedia.org/wiki/Convolution#Differentiation. But then we need to show $f'*g\in S$, too. – Christmas Bunny Nov 23 '13 at 00:48
  • I don't totally remember the definition of Schwartz class. But it is something like any number of derivatives of $f$ decay nicely at infinity. Now all you need to show is that $f*g$ decays nicely at infinity. And that should be about as hard as showing that $fg$ decays nicely at infinity. – Stephen Montgomery-Smith Nov 23 '13 at 00:51
  • @StephenMontgomeryu-Smith: you need $C^\infty$ and $f^{(m)}(x)x^n \to 0$ as $|x| \to \infty$ for all $m,n \in \mathbb{N}$ (I consider $0$ a natural number here). – Stefan Smith Nov 23 '13 at 01:25
  • So you can conclude that $|f(x)|,|g(x)| \le C_n/(1+|x|^n)$ for any $n \ge 0$ where $C_n$ depends upon $n$. Hence, you should be able to show $|f*g(x)| \le D_n / (1+|x|^n)$ for any $n \ge 0$, just by writing out the definition of convolution, and making crude estimates for the integral. – Stephen Montgomery-Smith Nov 23 '13 at 01:37
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    We have $f\ast g=\mathscr{F}(\mathscr{F}^{-1}(f)\mathscr{F}^{-1}(g))$ by the Fourier inversion theorem – Alexander Grothendieck Nov 23 '13 at 02:07
  • I added lots of details in the hope that the person who downvoted my answer will change their mind. – Stephen Montgomery-Smith Nov 23 '13 at 22:40
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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.

Vladhagen
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  • Montgomery-Smith's proof below is much less reliant on outside theorems of course, but many textbooks on Schwartz spaces have the theorems I have cited (i.e. Jones' book Lebesgue Integration on Euclidean Spaces.) – Vladhagen Nov 23 '13 at 05:49
  • That's also the book I'm using! :) – Christmas Bunny Nov 23 '13 at 07:53