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Let $\mathcal H^1(\mathbb R^n)$ be the real Hardy space (as in Stein's "Harmonic Analysis", Chapter 3). It is well known that $\mathcal H^1(\mathbb R^n)\subset L^1(\mathbb R^n)$ and its elements have zero mean.

I would like to know: does a function $f\in L^1\cap L^\infty(\mathbb R^n)$ with $\int_{\mathbb R^n} f(x)\,dx=0$ belong to $\mathcal H^1(\mathbb R^n)$ assuming it decays fast enough at infinity? For example, is polynomial decay $$ |f(x)|\leq (1+|x|)^{-M} $$ for some $M>0$ large enough a sufficient condition, together with the zero mean condition? If not, what about exponential decay, etc...? A Related question would be: is any Schwartz function with zero mean also in $\mathcal H^1(\mathbb R^n)$?


What I know is that any function in $L^q(\mathbb R^n)$, $q>1$ with compact support and zero mean belongs to $\mathcal H^1(\mathbb R^n)$ (and the $L^q$ assumption can be relaxed to $L\log L$). I had a quick look at Stein's book and Grafakos' book, but I did not find anything like what I am looking for. I thought that if no condition like that exists, it must be something well known.

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    $\begingroup$ (replying to a deleted comment). I was thinking about using the atomic decomposition characterization of H1, the problem though is that I don't know how to chop the function in pieces so that all the pieces have zero mean. Maybe it's easy, I am not sure, but for the moment I don't know how to do that in general $\endgroup$ Commented Oct 2, 2023 at 23:16

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Edit. For the sake of improving the quality of the post, I modified the proof to make it work for all $M>n$ after Prof. Tao’s comments (the previous version was admittedly way too loose).

In the end I just tried by hand, as I probably should have done from the beginning. The answer is yes. More precisely, if $M>n$, any function $f$ satisfying $$ |f(x)|\leq\frac {C}{1+|x|^M} $$ belongs to the Hardy space $\mathcal H^1(\mathbb R^n)$ if and only if $\int_{\mathbb R^n} f(x)\,dx=0$, with $\mathcal H^1$ norm bounded by $C$. This is of course sharp in the constant $M$. It would be reasonable to expect that a similar proof works for weighted $L^p$ spaces or similar, as soon as there is enough regularity and a way of controlling the amount of mass of the function at infinity, but I will refrain from guessing.


Proof

We fix a function $\phi\in\mathscr S(\mathbb R^n)$ with non-zero mean. We know that $f$ belongs to $\mathcal H^1(\mathbb R^n)$ if and only if the maximal function defined as $$ M_\phi f(x):=\sup_{t>0}|\phi_t*f(x)| $$ is in $L^1(\mathbb R^n)$, where $\phi_t(x):=\frac{1}{t^n}\phi(\frac{x}{t})$. In that case, we have $$ \|f\|_{\mathcal H^1(\mathbb R^n)}\lesssim_{n,\phi} \|M_\phi f\|_{L^1(\mathbb R^n)}. $$ We then want to try to find an upper bound on $|\phi_t*f(x)|$ that does not depend on $t$. By linearity, we assume $C=1$.

First bound

Here we use the zero mean condition: $$ |\phi_t*f(x)|=\left|\int_{\mathbb R^n} t^{-n}\phi(t^{-1}(x-y))f(y)\,dy\right|= $$ $$ =\left|\int_{\mathbb R^n} t^{-n}\left[\phi(t^{-1}(x-y))-\phi(t^{-1}x)\right]f(y)\,dy\right|, $$ since $f$ has zero mean. Then one can estimate the previous integral with $$ \leq\left|\int_{\mathbb R^n} t^{-(n+1)}|y|\int_0^1|\nabla\phi(t^{-1}(x-sy))|\,ds\,f(y)\,dy\right|, $$ which, since $\phi$ has bounded gradient, can be estimated by $$ \lesssim_{n,\phi} t^{-(n+1)}\|\,|\cdot|f\,\|_{L^1(\mathbb R^n)}\lesssim_{n,M,\phi} t^{-(n+1)}. $$

Second bound

We fix $N>0$. Now we look for a bound for all $t$ using the decay of $f$ and knowing that $\phi\lesssim \frac{1}{1+|x|^N}$. $$ |\phi_t*f(x)|=\left|\int_{\mathbb R^n} t^{-n}\phi(t^{-1}y)f(x-y)\,dy\right|\lesssim $$ $$ \lesssim \int_{\mathbb R^n} \frac{t^{-n}}{1+|t^{-1}y|^N}\frac{1}{1+|x-y|^M}\,dy.$$ Now we split the integral into the regions $|y|\leq \frac{|x|}{2}$ and its complementary:

a)

$$ \left|\int_{|y|\leq \frac{|x|}{2}} \frac{t^{-n}}{1+|t^{-1}y|^N}\frac{1}{1+|x-y|^M}\,dy\right|\leq $$

$$ \leq 2^M \int_{|y|\leq \frac{|x|}{2}} \frac{t^{-n}}{1+|t^{-1}y|^N}\frac{1}{1+|x|^M}\,dy\lesssim$$

$$ \lesssim \frac{1}{1+|x|^M}. $$

b)

$$ \int_{|y|\geq \frac{|x|}{2}} \frac{t^{-n}}{1+|t^{-1}y|^N}\frac{1}{1+|x-y|^M}\,dy\leq $$

$$ \leq \int_{|y|\geq \frac{|x|}{2}} \frac{t^{-n}}{1+|t^{-1}y|^N} dy=$$

$$ = \int_{|z|\geq \frac{|x|}{2t}} \frac{1}{1+|z|^N}\lesssim $$

$$ \lesssim \frac{1}{1+|\frac{x}{t}|^{N-n}}. $$

Hence, the function is bounded up to a constant by

$$ \max\left\{\frac{1}{1+|x|^M},\frac{1}{1+|\frac{x}{t}|^{N-n}}\right\} $$

Final estimate

Now, we take the minimum of the two bounds we obtained and then we take the supremum over $t$:

$$ \sup_{t>0}\min\left\{t^{-(n+1)}, \max\left\{\frac{1}{1+|x|^M},\frac{1}{1+|\frac{x}{t}|^{N-n}}\right\}\right\} $$

Using the property $a\wedge(b\vee c)=(a\wedge b)\vee(a\wedge c)$ one can bring the inner $\max$ outside and split the bound into the maximum of two separate quantities, the first of which yelds trivially $$ \frac{1}{1+|x|^M}, $$ and the second can be found explicitly simply by plugging $t$ such that $t^{-(n+1)}=\frac{1}{1+|x/t|^{N-n}}$, and can be estimated by $$ \frac{1}{1+|x|^{-\alpha}}, $$ with $$ \alpha=\frac{(N-n)(n+1)}{N+1}, $$ and we note that $\alpha>n$ (that is, the second bound is integrable in $x$) for $N$ large enough.

So, if $M>n$, then $M_\phi f$ belongs to $L^1(\mathbb R^n)$ and the claim is proved, namely $f\in\mathcal H^1(\mathbb R^n)$ with $$ \|f\|_{\mathcal H^1(\mathbb R^n)}\lesssim_{n,M} C. $$

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  • $\begingroup$ You can do much better in your second bound by extracting out the case where $|y| \leq |x|/2$, in which $|x-y|$ is comparable to $|x|$, and which will already be very good for handling the $t \ll |x|$ case. (See also my recent article terrytao.wordpress.com/2023/09/30/… .) $\endgroup$
    – Terry Tao
    Commented Oct 4, 2023 at 17:00
  • $\begingroup$ Also, the atomic decomposition approach will also work. For any radius $R>0$, let $f_R$ be the normalized truncated function $f_R = f \psi_R - \frac{\int f \psi_R}{\int f} \psi_R$, where $\psi_R$ is a bump function adapted to the ball $B(0,R)$ that equals 1 on $B(0,R/2)$. The mean $\int f \psi_R = - \int f (1-\psi_R)$ will decay nicely when $f$ has decay. Now decompose $f$ as a telescoping series $f_1 + \sum_{j=0}^\infty f_{2^{j+1}}-f_{2^j}$ and verify that all the summands are multiples of atoms. $\endgroup$
    – Terry Tao
    Commented Oct 4, 2023 at 17:02
  • $\begingroup$ Sorry, the denominator should be $\int \psi_R$ rather than $\int f$. $\endgroup$
    – Terry Tao
    Commented Oct 4, 2023 at 17:39
  • $\begingroup$ Thank you so much for your comments! I modified the proof to hide my terribly loose bound, although the atomic decomposition approach looks more intuitive for this problem (and probably easier to write down). $\endgroup$ Commented Oct 6, 2023 at 14:14

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