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Let $\psi \in L^2( \mathbb{R})$ and suppose that it satisfies the admissibility condition $$ \int_{-\infty}^{\infty} \frac{|\widehat{\psi}(\omega)|^2}{|\omega|}d\omega = C_{\psi} < \infty $$ where $ \widehat{\psi}$ denotes the fourier transform of $\psi$.

Then my textbook in Fourier analysis (and any other book I've seen) says that this implies that $$ \widehat{\psi}(0) = \int_{-\infty}^{\infty} \psi(x) dx = 0 .$$

So I can see why this is the case, since plugging in $\omega = 0$ in the first expression would make the integral infinite, however I'm not sure how to prove this rigorously. Could someone give me a hint on how to do that? I think we also can assume that $ \psi$ is continuous.

Thanks

Henrik Finsberg
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    If $\widehat{\psi}(0)\neq 0$, then the integral would be divergent (because so is $\int_0^1\frac 1t\mathrm dt$). – Davide Giraudo Aug 02 '13 at 11:29
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    Equivalently: ...because so is $\int\limits_{-\varepsilon}^\varepsilon\frac1{|t|}\mathrm dt$ for every positive $\varepsilon$. – Did Aug 02 '13 at 11:31

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For integrability at $\omega=0$, the integrand must be $\in o\left(1/w^{1-\epsilon}\right)$, for all $\epsilon > 0$. Thus, $|\hat{\psi}(\omega)| \in o\left(\omega^{\epsilon/2}\right)$ as $\omega \to 0$, which implies that $|\hat{\psi}(0)|=0$.

Ron Gordon
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