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

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

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

  • $\begingroup$ 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? $\endgroup$
    – Zen Harper
    Commented Jan 19, 2011 at 1:13
  • $\begingroup$ I think this post of mine is somewhat related. mathoverflow.net/questions/455725/… $\endgroup$ Commented Oct 8, 2023 at 15:35

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 $L^1$.

  • $\begingroup$ @Andrey This is a great answer, but I'm just a bit confused as to why the $c_k$ are complex? $\endgroup$ Commented Apr 19, 2011 at 9:38
  • $\begingroup$ @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. $\endgroup$ Commented Jul 10, 2014 at 12:23

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