Laplace equation in 2D for a continuous source: is there a continuous solution?

Question

Given the Laplace equation on $\mathbb{R}^2$ $$ \Delta u = f, $$ where $f$ is a continuous function on $\mathbb{R}^2$, can we find a solution $u$ that is in fact also continuous?

Given the fact that the fundamental solution to this equation reads $\log(\sqrt{x^2+y^2})$, it is easy to conclude that $$ g = \log\left(\sqrt{x^2+y^2}\right) \ast f $$ is a solution, if the convolution integral converges that is. Unfortunately, for a generic continuous function, we cannot be sure of this. Is there another way of finding a continuous solution?

A slightly weaker solution would be a continous function $h$ such that the distribution defined by $u$ can be expressed as $$ \langle u, \phi \rangle = \int_{\mathbb{R}^2} h \phi. $$ Is it perhaps possible to conclude that at least such a function $h$ exists?

Answer 1

After some reading, I think I know how to answer this:

if $f \in C^0(\mathbb{R}^2)$, then $f\in L_{2,loc}$, so $f\in H^0_{loc}$. By the regularity theorems on elliptic operators we know now that $u \in H^2_{loc}$ (see e.g. Grubb chapter 6, theorem 6.29). But $H^2_{loc} \subset C^0$, so $u$ itself must be continuous.

So $u$ fullfills the requirements, although the relation with the convolution integral is unclear. My guess would be that the convolution integral only specifies a solution when it converges, so this cannot be used for all possible $f$. However, the elliptic operator theory is strong enough to prove that $u$ must be continuous.

Answer 2

If you know your solution is growth at rate $\log(|x|)$ as $x\to\infty$, then it has to agree with the solution generated by fundamental solution up to a constant.

However, the weakest condition I know for $f$ so that you could have a solution is that $f$ at least be Holder continuous for some order $0<\alpha<1$ and $f$ is integrable, i.e., at least $L^1(R^2)$. This result hold for general $N\geq 2$. But to my knowledge, I don't know any result which hold for barely continuous $f$.