Question

I stumbled across the book Second Oder Elliptic Equations and Elliptic Systems by Yah-Ze Chen which appears to contain an answer to my question. It is available on google books here

http://books.google.com/books?id=eQcbiPQPweQC&pg=PA49&dq=strong+solution+dirichlet+problem+Lp&hl=en&ei=byKiTeXXO6GG0QGG7dGgBQ&sa=X&oi=book_result&ct=result&resnum=1&ved=0CCcQ6AEwADgK#v=onepage&q=strong%20solution%20dirichlet%20problem%20Lp&f=false

To save time for those who are interested, here is the relevant argument:

For large $\lambda > 0$ we want to show that $L_{\lambda} = L - \lambda I$ is injective on $W_0^{1,p}$.

Claim: Let $L^T_{\lambda}$ be the transpose of $L^{\lambda}$ with respect to the paring that defines weak solutions. Then we claim that $L^T_{\lambda}$ inective on $W_0^{2,p}$ implies that $L_{\lambda}$ is injective on $W_0^{1,p}$

Proof: Suppose that $L^T_{\lambda}$ is injective on $W_0^{2,p}$. Then, by an argument contained in the original post above, for every $f \in L^p(\Omega)$ we can find $u \in W_0^{2,p}(\Omega)$ such that $L^T_{\lambda}u = f$. Now, suppose that $L_{\lambda}v = 0$ for some $v \in W_0^{1,p}$. After an integration by parts and the definition of weak solution, we see that $\varphi \in W_0^{2,q}$ implies that

$\int_{\Omega}uL^T_{\lambda}\varphi = 0$.

Now choose $\Omega'' \subset\subset \Omega' \subset\subset \Omega$ and a bump function $\rho$ identically one in $\Omega''$ with support in $\Omega'$. $\rho\text{sgn}(u)$ is in $L^q$, and we can find $g \in W_0^{2,q} $ such that $L^T_{\lambda}g = \rho\text{sgn}(u)$. Plugging this $g$ into the above equality gives

$\int_{\Omega''}|u| = -\int_{\Omega\setminus\Omega''}\rho |u|$

Due to the arbitrariness of $\Omega''$, this implies that $\int_{\Omega} |u| = 0$ and hence $u$ is $0$ a.e.

Claim: For $\lambda$ large enough, $L_{\lambda}$ is injective on $W_0^{1,p}$. W_0^{2,p}$.

Proof: Suppose $L_{\lambda}u = 0$ for $u \in W_0^{2,p}$. Let $\tilde{\Omega} = \Omega \times (-1,1)$, and $\tilde{\Omega'} = \Omega \times (-1/2,1/2)$. Let $(x,t)$ be the coordinates on $\Omega \times (-1,1)$. Then define $v(x,t) = \cos(\sqrt{\lambda}t)u(x)$. Let $\hat{L_{\lambda}} = L_{\lambda} + \partial_t^2$. We have $\hat{L_{\lambda}}v = 0$. The strong solution estimates give

$\vert\vert v\vert\vert_{W^{2,p}(\tilde{\Omega'})} \leq C\vert\vert v\vert\vert_{L^p(\tilde{\Omega})} \leq C\vert\vert u\vert\vert_{L^p(\Omega)} \Rightarrow $

$\vert\vert \partial_t^2v\vert\vert_{L^p(\tilde{\Omega'})} \leq C\vert\vert u\vert\vert_{L^p(\Omega)} \Rightarrow $

$\lambda\vert\vert u\vert\vert_{L^p(\Omega)}(\int_{-1/2}^{1/2}|\cos(\sqrt{\lambda}t|^p)^{1/p} \leq C\vert\vert u \vert\vert_{L^p(\Omega)} \Rightarrow$

$\lambda^{1 - \frac{1}{2p}}\vert\vert u\vert\vert_{L^p(\Omega)}(\int_{-1/2}^{1/2}|\cos t|^p)^{1/p} \leq C\vert\vert u\vert\vert_{L^p(\Omega)}$

Now taking $\lambda$ large enough implies that $u = 0$ almost everywhere.