As in the title - I would like to show that if $f$ is a smooth ($C^{\infty}$) function then for any distribution $u$ satisfying $$ \Delta u = fu$$ in the distributional sense we have, in fact, $u \in C^{\infty}$. Initially my problem was to show the same thing but for $u$ being a solution of $$ \Delta u = b \cdot \nabla u + f$$ with $b, f$ some given $C^{\infty}$ functions, but I got rid of the $\nabla u$ term by considering $$ v(x) := \exp (- \frac{1}{2} b \cdot x) u(x)$$ and observing that $v \in C^{\infty}$ if and only if $u \in C^{\infty}$. I'm not quite sure it helps - I was trying to go back to $\Delta w = g $ situation with some smooth $g$, but I seem to be stuck - I'd much appreciate some hints.
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$\begingroup$ Just an observation: As soon as $u\in L^p$ for some $p>1$ we get that $u\in L^\infty$ by an iteration of Young's inequality for the convolution with the Newtonian potential. From there it's easy to see that $u\in C^{1,\alpha}$, again by standard estimates for the NP, and from there smoothness follows by the Schauder estimates. $\endgroup$– Jose27Jan 16, 2015 at 5:37
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ok, I think I've figured it out:
in my course, while showing that distributional solutions of $$\Delta u = 0$$ are smooth we've shown that you can estimate (locally at least) $|| u||_{H^{k+1}} $ norm by the $|| u ||_{H^k}$ norm, and that works for any $k$ - negative as well. The proof can be also easily adapted to the case $$ \Delta u = f $$ with $f \in H^k$. Combining this with the fact that any distribution is locally of finite order, hence in some $H^{-n}$ space this lets us easily conclude - by induction we show that $u$ is in all $H^k$ spaces, hence is smooth.