3
$\begingroup$

It is well-known (see Grafakos' Classical Fourier Analysis, Exercise 5.1.12) that if $f$ is a real valued $L^p(\mathbb R)$ function and $1<p<2$ , then we have the following inequality: $$ \|Hf\|_{L^p(\mathbb R)}\leq A_p\|f\|_{L^p(\mathbb R)}, $$ where $A_p=\tan \frac {\pi}{2p}$. Moreover, this bound is sharp.

A nice and elegant proof and also be found in https://faculty.missouri.edu/~grafakosl/preprints/pichorides.pdf.

However, I was wondering if this bound is also true for complex valued functions $f$. The proof above used the fact that if $f$ is real valued, then the real part and the imaginary part of the integral $$ \frac i \pi\int_{-\infty}^\infty \frac {f(t)}{z-t}dt $$ converges to $f(x)$ and $Hf(x)$ respectively, as $y\to 0^+$, where $z=x+iy$. For complex valued $f$, this is no longer true.

$\endgroup$

1 Answer 1

1
$\begingroup$

I have found the answer myself; it is a result of Theorem 5.5.1 of Grafakos’ Classical Fourier Analysis. The result we need is that if a linear operator maps real valued functions to real valued functions and has real L^p-L^p operator norm bounded by A, then its complex operator norm is also bounded by A.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.