I am convinced I have seen results along the lines of: if $ u \ge 0$ is an $H_0^1(\Omega)$ solution of $$\Delta u = u^{q1}$$ in $\Omega$ with $ u=0$ on $ \partial \Omega$ (here $\Omega$ is a smooth bounded domain in $R^N$ and $ q=2^*$, then $u$ is smooth. Any idea how one proves the regularity result? Since $ q=2^*$ it appears the standard iteration method fails. Maybe one uses some $ \epsilon$ regularity approaches (which I don't know). Or maybe I am wrong and the result is false. thanks
1 Answer
The result is true. As you suggest, the starting regularity assumption $u \in H^1_0$ is critical in the sense that, if you additionally knew $u \in L^p$ with $p>2^*$, then you could bootstrap to smoothness (at least, when $2^*$ is an integer, as in dimension $N=3$, for which $2^*=6)$. In the case $p=2^*$, the key observation is that $H^1_0$ solutions to $\Delta u = Vu$ do satisfy higher integrability in the critical case $V \in L^{N/2}$. That is, such solutions belong to all $L^p$, $p < +\infty$. The idea is to split the potential $V$ into a bounded part and a part which is small in $L^{N/2}$ and run Moser's iteration. The argument is given in the book Elliptic Partial Differential Equations by Qing Han and Fanghua Lin, see Theorem 4.4, p. 76.

$\begingroup$ i will take a look in teh book. Thanks for the comment. $\endgroup$– Math604May 13 at 21:15