# weak solutions need to have local integrability condition?

I am currently studying Poisson and Laplace equations. This is just a small question that has been causing me some confusion, and I would like some clarification before I resume my study. For example, when we write $\triangle u =0$ in the sense of distributions and u is also a distribution, are we assuming that $u$ is locally integrable. By this, I mean, are we assuming that $u$ is a function that is in $L^1_{loc}$? Thanks very much.

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I don't think that the local integrability is necessarily an a priori assumption in an arbitrary equation when interpreting it in the sense of distributions. It just turns out that any locally integrable function also defines a distribution. However, when we interpret the Poisson equation $\Delta u = f$ in the sense of distributions, the solution is usually locally integrable, at least when $f \in L^p$. Notice that the class of locally integrable functions is HUGE. For instance it contains all the $L^p$ for $1 \le p \le \infty$.
True, you could of course consider solutions in the entire class of distributions. However, this is pointless unless you want the data $f$ in a larger class than say $\bigcup L^p$. I am just not sure what the existence theorem would look like for an arbitrary distribution $T$, instead of $T = T_f$ for some $f \in L^p$... – user12014 Jan 3 '12 at 9:51