# Elliptic regularity for two dimensional domains

Suppose $\Omega$ is a smooth bounded domain in $R^2$. I am interested in the regularity of solutions to $$-\Delta u(x) = f(x) \mbox{ in } \Omega$$ with $u=0$ on $\partial \Omega$.

If $f \in L^1(\Omega)$ then one just misses $u \in C(\Omega)$. There was a result of Wente that said something like if $f= \nabla a \cdot \nabla^\perp b$ (where $a$ and $b$ have certain regularity assumptions, but not really enough to see the right hand side is better than $L^1$) then $u \in C(\Omega)$. I believe there is also a result that says something like if $f$ in a certain Hardy space (I am not familiar with these spaces) then one also has $u$ continuous.

QUESTION. I recall someone mentioning a version similar to the above. They had said if $f(x) ={\rm div}(F(x))$ where $F \in W^{1,1}(\Omega, R^2)$ then $u \in C(\Omega)$. So my question is. Is this correct or not ?

Thanks

## 1 Answer

It is true, but you must use the fact that $W^{1,1}$ is embedded in the Lorentz space $L^{2,1}$, see Helein's book, Harmonic maps, conservation laws and moving frames, theorem 3.3.10, you will find all the material about Hardy, Lorentz spaces in chapter 3 and more generally this book is just awsome!!

Then using the fact the gradient of the Green function $G$ is in $L^{2,\infty}$ because like $\frac{1}{\vert x\vert}$. Then you have $u=\nabla G * F$ is continuous.

• Is it actually necessary to use Lorentz spaces? That is surprising, if true. – Tommi Oct 19 '15 at 6:20
• Yes it is. If u=ln(ln(r)), then \nabla u \in L^2 but conclusion fails – Paul Oct 19 '15 at 6:46
• @Paul. Thank you very much for the answer. I guess at some point i should learn these more advanced spaces. I will look at the references you mentioned. thanks – Math604 Oct 19 '15 at 20:19