Why do atoms belong to the real Hardy space $\mathcal{H}^{1}(\mathbb{R}^n).$

Question

Define atom $a\in L^1(\mathbb{R}^n)$ associated to the ball $B_r(x_0)$ such that it satisfies

• $\text{supp}(a)\subset B_r(x_0).$
• $|a|\leq 1/|B_r(x_0)|$ and so $||a||_{L^1}\leq 1.$
• $\int_{B_r(x_0)} a dx = 0.$

Recall that $f\in \mathcal{H}^1(\mathbb{R}^n)$ if $f\in L^{1}(\mathbb{R}^n)$ and $M_{\varphi}f\in L^1(\mathbb{R}^n)$ where $$M_{\varphi}f (x) = \sup_{t>0} |\varphi_t*f|(x)$$ where $\varphi$ is the standard mollifier $0\leq \varphi\in C^{\infty}_c(\mathbb{R}^n)$ and $\phi_t(x) = t^{-n}\phi(x/t).$

Thus in order to show that an atom $a\in \mathcal{H}^1(\mathbb{R}^n)$ we need to show that $M_{\varphi} a\in L^{1}(\mathbb{R}^n)$ since $a\in L^{1}(\mathbb{R}^n).$ We can estimate the convolution as follows: \begin{align*} (\phi_t*a)(x) = \int_{B_r(x_0)} a(y)\phi_t(x-y) dy &= \int_{B_r(x_0)} a(y)[\phi_t(x-y)-\phi_t(x)] dy\\ &= t^{-n}\int_{B_r(x_0)} a(y)\left[\phi\left(\frac{x-y}{t}\right)-\phi\left(\frac{x}{t}\right)\right] dy\\ &= t^{-n} \int_{B_r(x_0)} a(y)\int_{0}^{1}\frac{d}{ds}\left[\phi\left(\frac{x-sy}{t}\right)\right] ds dy\\ &= t^{-n} \int_{0}^{1} \int_{B_r(x_0)} a(y) \nabla \phi\left(\frac{x-sy}{t}\right) \frac{-y}{t} dy ds \\ &\leq t^{-n-1} ||a||_{L^1} ||\nabla \phi||_{L^{\infty}} |y|\\ &\leq C t^{-n-1} ||a||_{L^1} (r+|x_0|) \end{align*} where the above estimate holds when $|x-y|<t$ since the support of $\phi$ is in $B_1(0).$ In this case we have $$t>|x-y|>|x|-(r+|x_0|)$$ since $|y|<|y-x_0|+|x_0|<r+|x_0|$ as $y\in B_r(x_0).$

I guess to conclude the argument we can work with large enough $x$ such that, $|x|>1+r+|x_0|$ then we get $$(\phi_t*a)(x) \leq C \frac{1}{||x|-(r+|x_0|)||^{n-1}}$$ which is integrable and thus we are done. Is this reasoning correct?

Answer 1

Your proof looks fine for me. Actually, as far as I know $f\in L^1$ is not required to $f$ be in $H^1$. It is more likely a result ($H^1\subset L^1$). In general setting $f$ just needs to be a tempered distribution, that is, $f\in\mathcal{S}^*$ where $\mathcal{S}$ is a Schwartz space. And for such $f$ if $M_\phi f\in L^1$ for some $\phi\in\mathcal{S}$ (not only $C_0^\infty$) then $f$ is said to be $H^1$.

Your reasoning is very appropriate, and let me just say that you can extend the result in $L^p$ with some control over $H^1$ norm:

Theorem. Let $B$ some open ball in $\mathbb{R}^n$. Denote $L_{B,0}^p\subset L^p$ the family of functions that are compactly supported in $B$, and has zero integration, i.e. $$f\in L_{B,0}^p\Longleftrightarrow f\in L^p,\,\text{supp}(f)\subset\subset B,\,\int f=0.$$ Then for any $1<p$, $L^p_{B,0}\hookrightarrow H^1$ with respect to the $L^p$ norm. More precisely, $$\|f\|_{H^1}\lesssim_{n,p}|B|^{1-\frac{1}{p}}\|f\|_{L^p}.$$

Of course atoms satisfy the above condition, if it is in $L^p$. The proof of this theorem is similar to your argument, but you need to verify first that $M_\phi f\lesssim_{n,p} Mf$, where $Mf$ is classical Hardy-Littlewood maximal function and from the maximal inequality you get $Mf\in L^p$ so is $M_\phi f\in L^p$. This nice bound does not hold anymore in $L^1$ case since the maximal inequality is only applicable for $p>1$.