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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.

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  • $\begingroup$ 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. $\endgroup$ Nov 14, 2015 at 14:51
  • $\begingroup$ Not an answer, but somewhat related: mathoverflow.net/questions/455725/… $\endgroup$ Oct 4, 2023 at 9:30

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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 $\log(2+|f(x)|+|x|)$.

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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.

My guess is that you need to assume more decay at infinity for this to work, see this post.

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