Seeking an example of Schwartz function $f$ such that $ \int_{\bf R}\left|\frac{f(x-y)}{y}\right|\ dy=\infty$

by Jack   Last Updated January 16, 2018 15:20 PM

In the introduction section of Hilbert transform in Grafakos's Classical Fourier Analysis (3rd) (Section 5.1.1), it is said that

for Schwartz functions $f$, the integral $$ \int_{\bf R}\frac{f(x-y)}{y}\ dy\tag{1} $$ may not converge absolutely for any real number $x$.

This motivates the use of principal value integrals: $$ \lim_{\epsilon\to 0}\int_{|y|\geq\epsilon}\frac{f(x-y)}{y}\ dy. $$

Could anyone give an example such that (1) does not converge absolutely?

Answers 2

If the Schwartz function $f$ is $0$ at $x,$ then the integral in question converges absolutely. Thus the Schwartz functions $f$ for which absolute convergence of the integral fails for every $x$ are precisely the Schwartz functions that are positive everywhere, or negative everywhere.

January 14, 2018 23:35 PM

Thanks to @Calvin Khor and @zhw's helpful comments, I have an answer. One can take for instance a Schwartz function $f$ such that $f(t)=1$ for $-1 \leq t \leq 1$. Then the integral $\int_{\bf R} \frac{f(t)}{t}\ dt$ will not be absolutely convergent regardless of what $f$ is doing outside of the interval $-1 \leq t \leq 1$.

January 16, 2018 15:06 PM

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