Does the Hilbert transform of Schwarz function decay far away

Question

The Hilbert transform $H$ of Schwarz functions can be defined as \begin{equation} Hf(x)=\int_{|y|<1}\frac{f(x-y)-f(x)}{y}dy + \int_{|y|>1} \frac{f(x-y)}{y}dy. \end{equation}

I would like to know whether \begin{equation} Hf(x)\to 0 \ \ \ \ \ as\ \ |x|\to\infty \end{equation} or not. If it is true, please show me how to prove this. If not, please give me the counter example. Thank you.

Answer 1

Note: the following is valid only for $f$ of compact support. As noted in mickeo's comment, functions in the Schwarz class do not in general have compact support.

Suppose that the support of $f$ is contained in $[-R,R]$, $R>0$.

1. If $|x|>R+1$ and $|y|<1$, then $f(x)=f(x+y)=0$, and the first integral is equal to $0$.
2. If $x>R+1$ and $|y|>1$, then $$ \Bigl|\int_{|y|>1} \frac{f(x-y)}{y}\,dy\Bigr|=\Bigl|\int_{x-R}^{R+x} \frac{f(x-y)}{y}\,dy\Bigr|\le\frac{\|f\|_\infty}{x-R}. $$
3. If $x<-R-1$ and $|y|>1$ a similar reasoning applies.

Proof for $f$ in te Schwarz class

There is a constant $C$ such that $$|f(x)|, |f'(x)|\le\frac{C}{1+x^2}\quad\forall x\in\mathbb{R}.$$ Then, by the mean value theorem,

$$ \Bigl|\int_{|y|<1} \frac{f(x-y)-f(x)}{y}\,dy\Bigr|\le\sup_{x-1\le\xi\le x+1}|f'(\xi)|\le\frac{C}{1+(x-1)^2}. $$ For the second integral we have $$ \Bigl|\int_{|y|>1} \frac{f(x-y)}{y}\,dy\Bigr|\le\int_{|y|>1} \frac{C}{1+(x-y)^2}\,dy, $$ which converges to $0$ as $x\to\infty$.