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I want to proof the following (simple case/version of) Wiener's tauberian theorem: The (span of the) set of translates $\{f_a | a \in \mathbb R^n\}$, $f_a(x) = f(a+x)$, is dense in $L^2(\mathbb R^n)$ if and only if $\hat f(p) \neq 0$ a.e., where $\hat f$ denotes the Fourier transform of $f$.

My try: Now lets assume $\hat f(p) \neq 0$ a.e and assume $g$ is orthogonal to all translates $f_a$, i.e. $\int f_a(x) \overline{g(x)} dx = 0$. (I want to show that $g=0$).

Then $\int \hat f_a(p) \overline{\hat g(p)} dp = 0$ (by Parseval) and by the formula $\hat f_a(p) = e^{iap} \hat f(p)$ we get $\int e^{iap} \hat f(p) \overline{\hat g(p)} dp = 0$. Now by assumption $\hat f(p) \neq 0$ a.e, but that doesnt let me deduce that $\hat g(p) \neq 0$ a.e. (at first i thought so, but trivially its wrong, just integrate the product of an even and an odd function..) How can I argue here?

Edit: Just to confirm: is my understanding correct, that the converse direction is trivial: if the span of $\{f_a | a \in \mathbb R^n\}$ is dense, then $\{\hat f_a | a \in \mathbb R^n\} = \{e^{iap} \hat f | a \in \mathbb R^n\}$ is dense, since Fourier transform is in particular surjective on $L^2$ and this set trivially couldnt be dense if $\hat f$ were zero almost everywhere (i.e. = 0 in $L^2$).

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I think you're on a good way. Recall that the Fourier Transform $\mathcal{F}$ is a unitary operator mapping $L^{2}(\mathbb{R}^n)$ onto itself, and you have already shown that $$ \int e^{iap} \hat f(p) \overline{\hat g(p)} dp = 0 \quad \forall a \in \mathbb{R}^{n}$$

This can be written as $$\mathcal{F}[ \hat f \overline{\hat g}](-a) = 0 \quad \forall a \in \mathbb{R}^n$$

Now you can use that $\mathcal{F}$ is unitary and hence injective.

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  • $\begingroup$ Thanks. Isnt it the inverse Fourier transform because inside the integral its $e^{iap}$ rather than $e^{-iap}$. But i guess that doesnt make a difference. $\endgroup$ – Mekanik Jun 17 '15 at 9:47
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    $\begingroup$ It really does not make a difference. As you know that the first equation holds for all $a \in \mathbb{R}^n$, my conclusion is valid, maybe I should have written $-a$ in the second equation to make matters less confusing. Regardless, of course $\mathcal{F}^{-1}$ is injective as well, so you can apply the same argument using $\mathcal{F}^{-1}$ if it suits you better. $\endgroup$ – user159517 Jun 17 '15 at 11:27

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