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 $H\phi \in L^2(\mathbb{R})$?
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 $$|\phi*K_1|\le c\psi.$$