$L^1(\mathbb{R}^n)$ functions not in $\mathcal{H}^1(\mathbb{R}^n)$

Question

I am wondering how to imagine the Hardy space on $\mathcal{H}^1(\mathbb{R}^n)$ and in particular what sort of functions are in $L^1(\mathbb{R}^n)\backslash\mathcal{H}^1(\mathbb{R}^n)$. Furthermore, is it possible to find explicit examples of functions in $\partial (L^1(\mathbb{R}^n)\backslash\mathcal{H}^1(\mathbb{R}^n))$?

$\mathcal{H}^p(\mathbb{R}^n)=L^p(\mathbb{R}^n)$ for $p>1$, but what is the "problem"/ difference when $p=1$ ( or $p\leq 1)$?

Thanks!

Answer 1

Let $f(x) \in L^1(\mathbb R^n)$ satisfy $\int_{\mathbb R^n} f(x) \, dx = C \ne 0$. If you are using the maximal characterization of $H^1$, then $$ Mf(x) := \sup_{r>0} \frac1{w_nr^n} \int_{y\in B(x,r)} f(y) \, dy \ge \frac1{2^nw_n|x|^n} \int_{B(x,2|x|)} f(y) \, dy \sim \frac C{|2x|^n} \text{ as $x \to \infty$}.$$ Here $w_n = |B(0,1)|$. Hence $Mf \notin H^1$.

The problem is that the inequality $\|Mf\|_p \le c_p \|f\|_p$ is only true if $p>1$.

Also $\partial(L^1\setminus H^1) = L^1$ because $L^1\setminus H^1$ is dense in $L^1$. This isn't really a meaningful question, because $L^1$ and $H^1$ have different topologies.