Question about Hilbert transform(applying plancherel theorem) Let $f\in S(\mathbb{R})$(Schwartz function on real line). Then Hilbert transform $H$ of $f$ is defined by
$\displaystyle Hf(x)=\lim\limits_{t\rightarrow0}\int_{|y|>t}\frac{1}{y}f(x-y)\,dy$
One can show that this integral exists if $f$ is schwartz. It is known that $H$ is of strong type $(p,p)$ for $p>1$ and weak type $(1,1)$. (It is also known that $H$ cannot be of strong type (1,1).) In most books and references that I looked up so far, they prove weak $L^1$ bound and strong $L^2$ bound of $H$ and use Marcinkiewicz interpolation for finishing touch. Especially in their proof of strong $L^2$ bound, They argue as follows.
$\lVert Hf\rVert_2=\lVert (Hf)^\hat\ \rVert_2=\lVert f^\hat\ \rVert_2=\lVert f\rVert_2$
The first and third equality is by Plancherel's theorem, and the second equality is by the identity $(Hf)^\hat\ (\xi)=-i\rm{sgn}(\xi) f^\hat\ (\xi)$. My question is that,
how can I apply Plancherel's theorem on the first equality? Shouldn't I guarantee that $Hf$ is in $L^1\cap L^2$ first to apply it? As written earlier, there is an example where $Hf$ fails to be in $L^1. (f=\chi_{[0,1]})$. I am confused.
 A: The source of your confusion is a "density argument" which is often omitted.
The initial definition of $H$ is
$$Hf(x)=\lim_{t\rightarrow 0} \int_{|y|>t} \frac{f(x-y)}{y} dy\tag{1}$$
As you have said, a priori this makes sense only for functions $f$ having some nice properties, say $f$ Schwartz. So when we show that $H$ is bounded on $L^2$ using the calculation you have shown, we first do this only for Schwartz functions (as we don't even know yet what $Hf$ is supposed to mean for a general $f\in L^2$).
So say that we have established $\|Hf\|_2=\|f\|_2$ for $f$ Schwartz. 
Then we can use this to extend our definition of $H$ to all of $L^2$, since Schwartz functions are dense in $L^2$. Namely, for $f\in L^2$ arbitrary, choose $(f_n)_n$ Schwartz such that $\|f_n-f\|_2\rightarrow 0$. We define
$$Hf=\lim_{n\rightarrow\infty} Hf_n\tag{2}$$
where the limit is in the $L^2$ sense. Note that the limit exists, since $Hf_n$ is a Cauchy sequence (check it) and $L^2$ is complete. Also check that $Hf$ is well-defined. Using continuity of the norm and $(3)$ we can also extend our norm equality to all $f\in L^2$:
$$\|Hf\|_2=\lim_{n\rightarrow\infty} \|Hf_n\|_2 = \lim_{n\rightarrow\infty} \|f_n\|_2 = \|f\|_2$$
