# The difference between Lebesgue and Hardy spaces

Are there any known inequalities of the following type for $f$ satisfying some conditions:

$$\|f\|_{H_p(\mathbb{R})} \le C\|f\|_{L_p(\mathbb{R})},$$

where $H_p$ denotes the real Hardy space and $p\le 1$ (otherwise the question would be trivial)?

Here is an example:

If $f$ is bandlimited, then

$$\|f\|_{H_p(\mathbb{R})} \sim \|f\|_{L_p(\mathbb{R})}.$$

Does a similar conclusion also hold if $f$ is sufficiently smooth?

Of particular interest to me is the following case:

Assume that $\varphi$ satisfies the inequality above. Is the same true for any function in the integer-shift-invariant space spanned by the integer shifts of $\varphi$?

• Are you talking about real or complex Hardy spaces? For complex Hardy spaces the $H^p$ norm is just (a multiple of) the $L^p$ norm of the boundary function, so the inequality is in fact an equality. For real Hardy spaces, it's a totally different question, but I'm not familiar with them. – Zen Harper Aug 4 '10 at 19:26
• It might be worth adding that the $H^p$ notation can mean real Hardy spaces, complex Hardy spaces, or Sobolev spaces, and they're different things, so a bit more explanation might be good. – Zen Harper Aug 4 '10 at 19:29
• I have edited the original question to address your comments. Thanks. – Philipp Aug 4 '10 at 19:38
• If we are talking about the same spaces, a characterization of the $H_1$ norm is $$\|f\|_{H_1} = \|f \|_{L^1}+\sum\|R_j f\|_{L^1}$$ where $R_j$ are the Riesz operators with symbol $\xi_j/|\xi|$. So when $p=1$ the Hardy norm is strictly stronger. – Piero D'Ancona Sep 12 '10 at 10:26
• I am talking about the real Hardy spaces which are defined e.g. by a Littlewood-Paley type decomposition. I am interested on results which rely on smoothness conditions on $f$. Of course, Hardy - and Lebesgue spaces are different for generic $f$. But if e.g. $f$ has bounded spectrum then the Hardy norm becomes comparable to the $L_p$ norm. I am asking to which extent this can be generalized by imposing smoothness conditions on $f$. – Philipp Sep 12 '10 at 10:53

Smoothness doesn't help much. Take any compactly supported smooth bump and put a few shifts of this bump far away to kill the first few moments. Then you get and $H^p$ function but the maximal operator will have a long stretch of $1/x$, which will give you a huge $H^p$ norm.
Your last question, as posed, also has negative answer: take this first example and add an $H^p$ function $f$ that satisfies the inequality and such that its norm is large but the norm of $f(x)-f(x+m)$ is small and $m$ is large (a stretch of a smooth atom renormalized properly would fit). Then the sum will satisfy the inequality because its norm sees only the good part but its difference with the shift by $m$ won't because its norm sees only the bad part.
• Hi Fedja, first let me thank you for your nice answer(s). As far as I see it, your example could also be made to work with a bandlimited bump function. The larger the support of the spectrum, the larger the $H_p$ norm can get in comparison to the $L_p$ norm. What if I impose that $$\label{1} |\hat f(\xi)| \leq (1+|\xi|)^{-N}$$ for some very large $N$. This is somewhat similar to requiring that $\mbox{supp }\hat f \subset [-1,1]$ where the desired inequality holds by some classical results. Could such an inequality hold over all $f$ with (\ref{1})? – Philipp Sep 12 '10 at 15:37
• The bump example exploits the bad behavior of $\widehat f$ at $0$, not at $\infty$. You seem to forget that the Paley-Littlewood decomposition has infinitely many terms with extra-short supports of the FT, not just with extra-long ones. Your words "band limited" should really mean "with spectrum bounded away from both $0$ and $\infty$". – fedja Sep 12 '10 at 16:42
• Thanks again and sorry for my misunderstanding. So how about if we require that $$|\hat f(\xi)| \leq \min(1,|\xi\|^N)(1 + |\xi|)^{-N} ?$$ My essential question is: Do we really need the compact support in the frequency domain or can we replace it with decay conditions? – Philipp Sep 12 '10 at 16:54
• Let us replace $\le$ by $\lesssim$. – Philipp Sep 12 '10 at 17:10