Dense subspaces of the Hardy space $H^1$

Question

Let $H^1$ be the Hardy space on $\mathbb{R}^n$, defined e.g. as the set of $u\in L^1$ such that $Ru\in L^1$, where $R$ is the Riesz transform on $\mathbb{R}^n$. It seems to me that simple functions with bounded support and average 0 are dense in $H^1$ (simple functions = finite linear combinations of characteristic functions). But I grew suspicious since I can not find any mention of this fact in the literature. If true this would be very useful, e.g. it implies a short proof that the complex interpolation between $H^1$ and $L^\infty$ is an $L^p$ space.

So, can anyone confirm/disprove/point at the relevant literature?

Answer 1

It seems to be true. Due to the atomic decomposition, it is enough to approximate in $H^1$ any "$(1,\infty)$"-atom, that is, a function $a$ such that $$\mathrm{supp}\, a \subset Q,\quad \|a\|_{L^\infty}\leq |Q|^{-1}, \quad \int_Q a=0$$ for a cube $Q$.

In order to do this, we can cut $Q$ into $N=k^n$ smaller equal cubes $Q_1, Q_2, \ldots, Q_N$ for a large $k$. We put $$ b_N(x)=\sum_{j=1}^N \Big( \frac{1}{|Q_j|}\int_{Q_j}a(y)\,dy\Big) \chi_{Q_j}(x). $$ Put $g_N=a-b_N$. Due to Lebesgue differentiation theorem we have $$ \lim\limits_{N\to \infty}\int |g_N(x)|^2dx = 0. $$

Then for big $N$ the following holds: $$ \mathrm{supp}\, g_N\subset Q, \quad \|g_N\|_{L^2}\leq \varepsilon |Q|^{-1/2}, \quad \int_Q g_N = 0. $$

It means that $\varepsilon^{-1} g_N$ is a $(1,2)$-atom. Hence, $\|g_N\|_{H^1}\leq C\varepsilon$.

The atomic decomposition in $H^1$ is described in the book "Weighted norm inequalities and related topics" by Garcia-Querva and Rubio de Francia. Another reference is "Modern Fourier Analysis" by Grafakos.