Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

the standard intuition for Lebesgue spaces $L^p(\mathbb R^n)$ for $p \in [1,\infty]$ are measurable functions with certain decay properties at infinity or at the singularities.

In particular, a typical $L^p$ function is a function like $|x|^{-\alpha}$ with $\alpha > \frac{n}{p}$ for $p$ < $\infty$, or simply $1$ for $p = \infty$. Furthermore, functions which are almost everywhere absolute-value-dominated by an $L^p$ function are elements of $L^p$, too. This is useful for approximation arguments, as the pointwise error has just to be dominated by an $\epsilon$-multiple of another $L^p$ function.

In contrast to that, hardy spaces seem to be less intuitive due to cancellation properties. Hence I wonder:

  1. How do "typical" $H^1$ functions look like?
  2. In particular, what can you typically "do" with Hardy functions?

For example, I guess convolution arguments take a more prominent role in approximation arguments than perturbation arguments do, but I am not sure about that.

The hardy space $H^1$ shall be defined via

$f \in H^1 $ if $f \in L^1_{\mathrm{loc}}$ and $Mf \in L^1$, where

$$Mf(x) := \sup \limits_{B_r(x_0) \ni x, \phi \in \mathcal L(B_r(x_0))} \int f \phi dx$$

$$\mathcal{L}(B_r(x_0)) = \left\{ \phi \in C(B_r(x_0)) s.t. |\phi(x)| < \dfrac{\max(r-|x-x_0|,0)}{r|B_r(x_0)|} , \mathrm{Lip}(\phi) < \frac{1}{r|B_r(x_0)|} \right\}$$

share|improve this question
    
I have always found the real Hardy spaces very confusing; it seems to me that they are vaguely related to the usual (complex) Hardy spaces (which are spaces of analytic functions); in particular, many results in one space have a natural translated version in the other space; but the analogy is quite hard for simpletons like me to understand. Still, maybe knowing about the Hardy spaces of analytic functions could be helpful? –  Zen Harper Jan 19 '11 at 1:13

1 Answer 1

In many ways $H^1$ is just a natural substitute for $L^1$.

A typical $H^1$ function is a $1$-atom, i.e. a function $\phi\in L^1(\mathbb R^n)$ such that the support of $\phi$ is contained in some ball $B(a,r)$, the bound $$\sup|\phi(\cdot)|\leq \frac{1}{m(B(a,r))}$$ holds true, and $$\int_{\mathbb R^n} \phi(x)dx=0.$$

In fact, following a suggestion by Fefferman, one can use $1$-atoms to characterize the whole Hardy space $H^1$. More precisely, the following is true.

Theorem. A function $f$ belongs to $H^1(\mathbb R^n)$ if and only if there exist $1$-atoms $\phi_k$ and complex numbers $c_k$ such that $$f=\sum_{k=1}^{\infty}c_k\phi_k,$$ where the convergence is in $L^1$ and $\sum_{k}|c_k|<\infty.$

One should note that the mean-value condition in the definition of $1$-atom plays a key role here. If it is removed from the definition, the above description with sums would result in the standard Lebesgue space $L^1$.

$H^1$ is a good testing field to study various classes of multiplier transformations. For instance, the singular integral operators of the form $$Tf=\lim_{\epsilon\to0}\int_{|y|>\epsilon}\frac{K(x-y)}{|x-y|^n}f(y)dy$$ where $K$ is a smooth homogeneous function of degree $0$ such that $\int_{|y|=1}K(y)dy=0$, extend to bounded operators on $H^1$. In particular, the Riesz transforms are bounded on $H^1$. Moreover, an integrable function $f$ belongs to $H^1$ if and only if its Riesz transforms $R_jf$ are also in $L^1$.

share|improve this answer
    
@Andrey This is a great answer, but I'm just a bit confused as to why the $c_k$ are complex? –  Glen Wheeler Apr 19 '11 at 9:38
    
@GlenWheeler I think that this is to cover the case of complex-valued functions. If the functions $f$ and $\phi$ are real, then the coefficients $c_k$ can be taken to be real. –  Jean Van Schaftingen Jul 10 at 12:23

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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