Holder continuity using Sobolev imbedding We assume for any $V\subset \subset U$ and $1<p<\infty$ 
$||u||_{W^{2,p}(V)}\le C(||\Delta u||_{L^p(U)}+||u||_{L^1(U)})$ for some $C=C(V,U,p)$. 
Given, $B=\{x∈R^3,|x|<1/2\}$ and we suppose $u∈H^1(B)$ is a weak solution of $-\Delta u+cu=f$ for some $c(x){\in L^q(B)}, 3/2<q<2$ and $f\in C^\infty (B)$. How would I show that $u$ is Hölder continuous inside $B$?

First notice that, by Hölder's inequality, we have $cu\in L^p$ for $1/p=1/q+1/6$ (since $u\in L^6$ by the Sobolev embedding). Next your inequality gives $u\in W^{2,p}$ for this same $p$. Applying Sobolev embedding again, we get $\nabla u \in L^{p*}$ where as usual $1/p^* =1/p-1/n$. If we substitute we get
$$
\frac{1}{p^*} = \frac{1}{q} +\frac{1}{6} -\frac{1}{3} = \frac{1}{q}-\frac{1}{3} < \frac{1}{3},
$$
where the first equality is the expression we got for $p$ at the start, and the last inequality is because $q>3/2$. This means that $\nabla u \in L^{p^*}$ with $p^*>3=n$, so that Morrey's inequality gives $u$ Hölder continuous (in the interior of $B$).
