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Question

I have a question concerning the proof of following proposition:

Proposition: Let $\phi\in S(\mathbb{R})$ be given. Then $H\phi \in L^1(\mathbb{R})$ if and only if $\int_{\mathbb{R}}\phi(x)dx=0$.

Where H is the Hilbert Transform ($Hf:=\frac{1}{x}\int_{\mathbb{R}}\frac{f(y)}{x-y}dy$)

Proof: Suppose first that $\int_{\mathbb{R}}\phi(x)dx=0=\hat{\phi}(0)=0$. We then use the Fourier Inversion Formula to compute: \begin{equation} H\phi(x)=\int_{\mathbb{R}}\widehat{H\phi}(\xi)e^{2\pi i x \xi}d\xi \end{equation} Indeed, this first equality is made rigorously because, for $\phi\in S(\mathbb{R})$, we have $H\phi \in L^2(\mathbb{R})$....The proof continues.

My question: why is $H\phi \in L^2(\mathbb{R})$?

That's one of the basic results - the Fourier transform of any $L^1$ function is continuous. Trivial from dominated convergence. — David C. Ullrich, Jul 16 '16 at 13:16
You really shouldn't edit questions in a way that makes previous answers and comments look like nonsense. My previous comment was an answer to a question that has disappeared, so it looks ridiculous. Anyway, regarding the current question: Plancherel. — David C. Ullrich, Jul 16 '16 at 13:38
Ok @DavidC.Ullrich I'm sorry for this. - But Plancherel just say that the $L^2-$norm of the fourier Transform is the $L^2$-norm of the function itself ... how can I use this fact here? — MorganeMaPh, Jul 16 '16 at 13:47
It's $\hat{H\phi}(\xi)=\int_{\mathbb{R}}H\phi(x)e^{-2\pi i x \cdot \xi}dx$ isn't it? — MorganeMaPh, Jul 16 '16 at 13:56
Only if $H\phi\in L^1$. There's a simple expression for the FT of $H\phi$ in terms of $\hat\phi$, presumably coming up later... — David C. Ullrich, Jul 16 '16 at 14:13
Can I get a reference for this proposition? I would like to see this proof. — user593295, Mar 19 '21 at 05:55

score 1 · Answer 1 · answered Jul 16 '16 at 14:10

This is an answer to the current version of the question, which asks why $H\phi\in L^2$ if $\phi$ is a Schwarz function. This is clear from Plancherel.

Except that doesn't work if we don't know yet what the Fourier transform of $H\phi$ is; could be the current proof is leading up to that. One can prove it directly from the definition as a convolution:

At least formally, ignoring irrelevant constant factors, $H\phi=\phi*K$, where $$K(t)=\frac1t.$$ Write $K=K_1+K_2$, where $$K_1=K\chi_{[-1,1]}.$$

Now $K_2\in L^2$, so $$||\phi*K_2||_2\le||K_2||_2||\phi||_1<\infty.$$

Define $$\psi(x)=\sup_{|t|\le1}|\phi'(x+t)|.$$You cann use the fact that $\phi$ is a Schwarz function to show that $\psi\in L^2$. On the other hand, $$\int_{-1}^1\frac{\phi(x-t)}{t}\,dt=\int_0^1\frac{\phi(x-t)-\phi(x+t)}{t}\,dt$$ shows that $$|\phi*K_1|\le c\psi.$$

But actually, why do we need $H\phi$ to be in $L^2$ for the step to be rigorous? — MorganeMaPh, Jul 16 '16 at 14:35

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