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If $f,g\in L^1(\mathbb{R})$, it is not hard to show by definition that $$(\hat{f\ast g)}(t)=\hat{f}(t)\hat{g}(t).$$ But what about if $f,g\in L^2(\mathbb{R})$? The Fourier transform on $L^2(\mathbb{R})$ is defined in as an extension of the Fourier transform in the Schwartz class. Then it's hard to work with the definition...

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    $\begingroup$ You'll have to work with the Fourier transform on the space of tempered distributions since $f*g$ is an $L^\infty$ function. $\endgroup$ – Jose27 Dec 13 '13 at 7:40
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For $f,g\in L^2$, the convolution $f*g$ belongs to $C_0(\mathbb R)$. Taking the Fourier transform of a $C_0$ function is problematic: such transforms are generally not functions, but merely distributions (as Jose27 points out).

It is better to work from right to left. The product $\hat f \hat g$ is in $L^1$. Apply the inverse Fourier transform to it (essentially same as Fourier transform, but with opposite sign in the exponential, and maybe different normalization). The result is a $C_0$ function. You want to show that this function is $f*g$.

Argue by density. For $f,g\in \mathcal S$ the result is known. The map $(f,g)\mapsto f*g$ is a continuous map from $L^2\times L^2$ into $C_0$. Also, $(f,g)\mapsto \hat f \hat g$ is continuous from $L^2\times L^2$ to $L^1$. The inverse Fourier transform is continuous from $L^1$ to $C_0$. The conclusion follows.

In the preceding paragraph, $C_0$ can be replaced by $L^\infty$ throughout (the norm is the same), saving you the trouble of proving that $f*g$ vanishes at infinity.

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One might argue that this is only a partial argument.

I understand the question in the following way: 1) is $f \ast g$ well defined: Yes, certainly, and since an easy application of Cauchy-Schwartz yields $$ \| f \ast g\|_\infty \leq \|f\|_2 \|g\|_2$$ we certainly have a bounded function. Since $$ \| T_x(f \ast) g \|_\infty = \|(T_xf -f) \ast g\|_\infty \leq \|T_x f -f\|_2 \|g\|_2\to 0 \quad \mbox{for} \,\, |x| \to 0$$ because the shift is continuous in the $L^2$-norm (prove it for compactly supported continuous functions, then by density you get it for every $f \in L^2$) we find that $f \ast g$ is in fact uniformly contiuous. Using the density of compactly supported (continuous) fucntion in $L^2(R)$ once more and the fact that the convolution product of two such function is thus a compactly supported function implies that $f \ast g \in C^0(R)$, i.e. $$\lim_{x \to \infty} f \ast g (x) = 0 . $$

2) is the Fourier transform of $f \ast g$ defined (and there one has to say: only in the distributional setting, AND in this distributional setting there is an inverse to the Fourier transform, in the realm of tempered distributions!). This makes the (good) argument given above a (non-trivial) valid statement.

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