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?

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.

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 $-1 \leq t \leq 1$.

- ServerfaultXchanger
- SuperuserXchanger
- UbuntuXchanger
- WebappsXchanger
- WebmastersXchanger
- ProgrammersXchanger
- DbaXchanger
- DrupalXchanger
- WordpressXchanger
- MagentoXchanger
- JoomlaXchanger
- AndroidXchanger
- AppleXchanger
- GameXchanger
- GamingXchanger
- BlenderXchanger
- UxXchanger
- CookingXchanger
- PhotoXchanger
- StatsXchanger
- MathXchanger
- DiyXchanger
- GisXchanger
- TexXchanger
- MetaXchanger
- ElectronicsXchanger
- StackoverflowXchanger
- BitcoinXchanger
- EthereumXcanger