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 ?



1 Answer 1


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.

  • $\begingroup$ Is it actually necessary to use Lorentz spaces? That is surprising, if true. $\endgroup$
    – Tommi
    Oct 19, 2015 at 6:20
  • $\begingroup$ Yes it is. If u=ln(ln(r)), then \nabla u \in L^2 but conclusion fails $\endgroup$
    – Paul
    Oct 19, 2015 at 6:46
  • $\begingroup$ @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 $\endgroup$
    – Math604
    Oct 19, 2015 at 20:19

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