# Criterion for Membership in Hardy Space $H^{1}(\mathbb{R}^{n})$

Let $H^{1}(\mathbb{R}^{n})$ denote the real Hardy space (I am agnostic about the choice of characterization). It is known that if $f:\mathbb{R}^{n}\rightarrow\mathbb{C}$ is a compactly supported function (say in a ball $B$) such that $\int f=0$ and

$$\int_{\mathbb{R}^{n}}|f(x)|\log^{+}|f(x)|dx<\infty$$

then $f\in H^{1}(\mathbb{R}^{n})$ and moreover, we have an estimate of the sort

$$\|f\|_{H^{1}}\lesssim |B|\|f\|_{L\log L(dx/|B|)}$$

See Lemma 3.10 in these notes for details. One can also show this result by means of a Calderon-Zygmund decomposition for a truncated Hardy-Littlewood maximal function to produce an atomic decomposition for $f$.

Suppose we now consider measurable functions $f$, not necessarily compactly supported, such that $\int f=0$ and

$$\int_{\mathbb{R}^{n}}\left|f(x)\right|\log\left||2+|f(x)|\right|dx<\infty$$

Is it true that $f\in H^{1}(\mathbb{R}^{n})$? I don't have my intuition for the answer to this question at the moment. A corresponding result for functions $f\in L^{p}(\mathbb{R}^{n})$, where $1<p<\infty$, with $\int f=0$ fails. In one dimension, take

$$f(x):=\dfrac{\text{sgn}(x)}{|x|^{(1+\epsilon)/p}}\chi_{(-1,1)^{c}}(x),$$

where $\epsilon>0$ is sufficiently small so that $(1+\epsilon)/p<1$. Since $f\notin L^{1}(\mathbb{R})$, we see that $f\notin H^{1}(\mathbb{R})$. The problem here is that we can have $f\in L^{p}\setminus L^{1}$. However, by considering the factor $\log(2+|f|)$, we have the estimate

$$\int|f|\leq(\log|2|)^{-1}\int|f|\log|2+|f||,$$ which addresses this issue.

• There are two issues here - the change of factor from $\log^+ |f|$ to $\log (2+ |f|)$ and dropping the assumption that $f$ is compactly supported. The first change concerns the size of the sets $\{|f| > M\}$ for large $M$. It is inessential since the ratio of these factors is bounded above and away from 0 for large arguments. The second change is the essential one. It concerns the size of the sets $\{|f| < \epsilon\}$ for small $\epsilon$. And since the $H^1$ bound that you are citing contains the size of the support explicitly, I doubt that your conjecture is true. Nov 14, 2015 at 14:51
• Not an answer, but somewhat related: mathoverflow.net/questions/455725/… Oct 4, 2023 at 9:30

I am not sure but I would think that it should be $$\int_{\mathbb R^n} f(x)\,dx=0$$ and $$\int_{\mathbb R^n} |f(x)| \log(2+|f(x)|+|x|)\,dx<\infty.$$ I mean $$\log(2+|f(x)|)$$ should be $$\log(2+|f(x)|+|x|)$$.
The answer is no, if I am not mistaken. The reason is that the function $$\Phi(f)=\int |f(x)||\log(2+|f(x)|)|dx$$ defines an Orlicz space in which the functions with zero mean are a closed subspace. So, a condition like the one you want would imply a continuous embedding of this closed (Banach) subspace into $$\mathcal H^1(\mathbb R^n)$$. But this is not true for the same reason why the subspace of $$L^1$$ of function with zero mean are not included in $$\mathcal H^1$$: consider a sequence of functions $$f_n(x)=g(x)-g(x-ne_1)$$, where $$g$$ is a non-negative test function. This is a bounded sequence in the Orlicz space, but not in the Hardy space.