Suppose $\phi$ is a weak solution of $\Delta \phi = f \in \mathcal{H}^1$. Then $\phi\in W^{2,1}$

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!

Solution 1:

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.