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Let $f:\mathbb{R} \to \mathbb{R}$ such that $\int|f|<\infty$. Consider the Fourier transform $$\hat{f}(\xi)=\frac1{\sqrt{2\pi}}\int e^{-i\xi > x}f(x)dx.$$ Give necessary conditions on $f$ if $$\int|\hat{f}(\xi)|^2+\frac1{|\xi|}|\hat{f}(\xi)|^2d\xi<\infty.$$

Ideas: This implies that $\int_{\mathbb{R}}|\hat{f}(\xi)|^2<\infty$, and since the Fourier transform is a unitary transformation, $\int |f|^2<\infty$ also. (Technically it is a unitary transformation on the Schwarz space, but I believe it is still unitary considered everywhere it is defined.) I am having a bit more trouble with the second part. I know multiplication by $\frac1{|\xi|}$ should correspond to integration somehow...

Eric Auld
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    Actually, you could also write that as $f∈ L^2$ and $(-\Delta)^{-\frac{1}{2}} f∈ L^2$ if you want it as "somehow an integration". And then yes, the integral expression of $(-\Delta)^{-\frac{1}{2}}$ would give you the answer of Stephen Montgomery-Smith. This implies that your condition is equivalent to $f∈L^2$ and $f = \Delta^{1/2}g$ for some $g∈ L^2$. – LL 3.14 Apr 29 '20 at 22:20

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So you want to know when $|\xi|^{-1/2} \hat f(\xi) \in L^2$. Now the Fourier transform turns multiplication into convolution, and the Fourier transform of $|\xi|^{-1/2}$ is $C|x|^{-1/2}$ for some constant $C>0$. So your condition is equivalent to $f \in L^2$ and

$$\int \left|\int \frac{f(x-y)\,dy}{|y|^{1/2}}\right|^2 \, dx < \infty .\tag1$$

However, if $f \in L^2$, then $$\int \left|\int_{|y|<1} \frac{f(x-y)\,dy}{|y|^{1/2}}\right|^2 \, dx < \infty .$$ because $I_{|x|<1} |x|^{-1/2}$ is in $L^1$, and convolution takes $L^2\times L^1 \to L^2$.

So (1) is equivalent to $$\int \left|\int_{|y|>1} \frac{f(x-y)\,dy}{|y|^{1/2}}\right|^2 \, dx < \infty .$$

Probably more work could be done to simplify this further, but I can't think what it might be right now.

Stephen Montgomery-Smith
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