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I'm trying to prove the statement in the title in as simple a way as possible. It is Theorem 3.2.9 in Helein's book "Harmonic maps, conservation laws, and moving frames", although it is not proved there. The statement is as follows.

Suppose $\phi\in\mathbb{R}^m$ is a weak solution of $\Delta \phi = f \in \mathcal{H}^1$, where $\mathcal{H}^1$ is the standard Hardy space on $\mathbb{R}^m$. Then $$ \Big\lVert\frac{\partial^2\phi}{\partial x^\alpha \partial x^\beta}\Big\rVert_{L^1(\mathbb{R}^m)} \le C\lVert f \rVert_{\mathcal{H}^1(\mathbb{R}^m)}. $$

My idea is to use convolution with the kernel of the Laplacian, and then differentiate, estimate in $L^1$ and somehow interpolate between the $\mathcal H^1$ and $BMO$ norms. Then since the kernel of the Laplacian is in $BMO$, I am finished. However there are two problems with my proof: I don't know how to prove that one can interpolate a convolution between $\mathcal H^1$ and $BMO$ (atomic decomposition?) and I don't know how to prove that the kernel of the Laplacian is in $BMO$.

Does anyone have either a better way to prove this theorem, or a way to fix up my proof? Thanks!

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Glen: I just wanted to point out that you can bump your question to the front page by editing it (of course, you should do this with reasonable restraint). I think this would get noticed more than advertising it on the chat (where there are at most 3 users a day). Also, if you have a heavy-handed solution, could you give some indication of what it is? I'd certainly be interested in seeing it. – t.b. Apr 12 '11 at 9:23
@Theo Buehler Yeah, I am aware of that method of 'bumping' questions, but is this not considered poor behaviour? I wouldn't want to go against site policy. My solution is just to flesh out what I wrote in the question, and solve the two problems. The first is easy. The second is the reason I see it as heavy handed: the interpolation of a convolution between $\mathcal{H}^1$ and $BMO$ is a result from Stein, "Harmonic Analysis", which I do not understand at the moment. Atomic decomposition is the right idea, but the theorem is just a black box for me at the moment. – Glen Wheeler Apr 12 '11 at 9:46
Maybe when I understand it a little better I will be able to provide a good answer. – Glen Wheeler Apr 12 '11 at 9:46
Thanks for the info, I appreciate it. In fact, if you add actual content (such as your first comment), I don't think anyone would construe this as poor behavior. I think you're reasonable enough to avoid exaggerating. Moreover, there is of course the bounty system. – t.b. Apr 12 '11 at 9:54
up vote 7 down vote accepted

Let me take a try:

Recall that the Riesz transform is bounded from $\mathcal H^1$ to $\mathcal H^1$ (and from $L^2$ to $L^2$).

We have the inequality

$$\|\partial_i \partial_j u \|_{L^2} \leq \| \Delta u \|_{L^2}.$$

This is because $R_i R_j \Delta u = \partial_i \partial_j u$ where $R_i$ is the $i$-th Riesz transform. Now because the Riesz transform is also $\mathcal H^1$ bounded we have

$$\|\partial_i \partial_j u \|_{\mathcal H^1} \leq \| \Delta u \|_{\mathcal H^1}.$$

But the $L^1$ norm is dominated by the $\mathcal H^1$ norm so

$$\|\partial_i \partial_j u \|_{L^1} \leq \| \Delta u \|_{\mathcal H^1}.$$

Further note that

$$t f'(s) = f(t + s) - f(s) - \int_0^1 f''(s+r)(t - r) \, dr$$

and a similar statement holds for partial derivatives. This implies that we can control the first derivative by the second and the function itself. This should give the result as asked in the title.

Alternatively we could use Mikhlin's multiplier theorem for $\mathcal H^1$ for the second part.

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Hi Jonas, thanks for the answer. It looks like you've perhaps made a typo in the second inequality? Should that not be the Hardy space there? Actually, I'm confused about the second inequality. Could you please write a bit more clarification there? – Glen Wheeler Apr 12 '11 at 15:19
@Glen: I have added some information. Please tell me if it is enough or if you think it is wrong. – Jonas Teuwen Apr 12 '11 at 15:30
Oh sigh, I read the $H^1$ as the Sobolev space $W^{1,2}$ before the edit. Yes! Its correct, and it is much shorter than what I originally proposed. Thanks again for the answer. – Glen Wheeler Apr 12 '11 at 15:33
You're welcome. – Jonas Teuwen Apr 12 '11 at 15:34
@Jonas: Very nice! (I voted this up before but I was distracted by a phone call, that's why I write only now) – t.b. Apr 12 '11 at 16:17

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