Injectivity removes the $L^p$ term from the global $W^{2,p}$ estimate

Contradiction setup and a bounded sequence. Suppose the inequality fails: then for every integer $j\ge 1$ there is $u_j\in W^{2,p}(\mathrm{\Omega })\cap W_0^{1,p}(\mathrm{\Omega })$ with $\mathrm{\parallel }{u}_j{\mathrm{\parallel }}_{W^{2,p}}=1$ and $\mathrm{\parallel }L{u}_j{\mathrm{\parallel }}_{L^p}<1\mathrm{/}j$; equivalently, after rescaling, a sequence with $\mathrm{\parallel }{u}_j{\mathrm{\parallel }}_{W^{2,p}}=1$ and $\mathrm{\parallel }L{u}_j{\mathrm{\parallel }}_{L^p}\to 0$. For each $j$ the $W^{1,p}$ norm is bounded by the $W^{2,p}$ norm up to constants, so $(u_j)$ is bounded in $W^{1,p}(\mathrm{\Omega })$, and by [F3] the domain $\mathrm{\Omega }$ is a bounded extension domain for $W^{1,p}$.

An $L^p$-convergent subsequence. By [F2] applied to the bounded sequence $(u_j)$, in each of the three cases $p<n$, $p=n$, $p>n$ there is a subsequence, relabelled $(u_j)$, converging in $L^p(\mathrm{\Omega })$ to some $w\in L^p(\mathrm{\Omega })$. In the case $p<n$ the admissible exponents form the interval $[1,p^{\ast })$ and $q=p$ is admissible; in the case $p=n$ every finite $q$ is admissible and $q=n=p$; in the case $p>n$ every $1\le q<\mathrm{\infty }$ is admissible and $q=p$.

Cauchy in $W^{2,p}$ via the a priori estimate. Apply [F1] to the differences $u_j-u_k\in W^{2,p}(\mathrm{\Omega })\cap W_0^{1,p}(\mathrm{\Omega })$: $\mathrm{\parallel }{u}_j-u_k{\mathrm{\parallel }}_{W^{2,p}}\le C_0(\mathrm{\parallel }L(u_j-u_k){\mathrm{\parallel }}_{L^p}+\mathrm{\parallel }{u}_j-u_k{\mathrm{\parallel }}_{L^p})\le C_0(\mathrm{\parallel }L{u}_j{\mathrm{\parallel }}_{L^p}+\mathrm{\parallel }L{u}_k{\mathrm{\parallel }}_{L^p}+\mathrm{\parallel }{u}_j-u_k{\mathrm{\parallel }}_{L^p})\mathrm{.}$ The first two terms tend to $0$ by construction, and the third by the $L^p$ convergence of step 2.1; hence $(u_j)$ is Cauchy in $W^{2,p}(\mathrm{\Omega })$, which is complete by Integer-order Sobolev spaces are Banach, and converges to some $v\in W^{2,p}(\mathrm{\Omega })$ with $v$ equal to the $L^p$-limit $w$ of step 2.1.

The limit is a vanishing strong solution. The space $W^{2,p}(\mathrm{\Omega })\cap W_0^{1,p}(\mathrm{\Omega })$ is closed in $W^{2,p}(\mathrm{\Omega })$ by [F4], so $v$ belongs to it; the boundedness of $L:W^{2,p}(\mathrm{\Omega })\to L^p(\mathrm{\Omega })$ and $L{u}_j\to 0$ in $L^p$ give $Lv=0$ almost everywhere. By the injectivity hypothesis $v=0$.

Contradiction. Applying [F1] to $u_j$ and using the normalization, $1=\mathrm{\parallel }{u}_j{\mathrm{\parallel }}_{W^{2,p}}\le C_0(\mathrm{\parallel }L{u}_j{\mathrm{\parallel }}_{L^p}+\mathrm{\parallel }{u}_j{\mathrm{\parallel }}_{L^p})⟶0$ because $\mathrm{\parallel }L{u}_j{\mathrm{\parallel }}_{L^p}\to 0$ and $\mathrm{\parallel }{u}_j{\mathrm{\parallel }}_{L^p}\to \mathrm{\parallel }v{\mathrm{\parallel }}_{L^p}=0$ by steps 2.1 and 4.1. This contradiction shows that the failure assumed in step 1.1 is impossible, that is, there is $C$ with $\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{2,p}(\mathrm{\Omega })}\le C\mathrm{\parallel }Lu{\mathrm{\parallel }}_{L^p(\mathrm{\Omega })}$ for all $u\in W^{2,p}(\mathrm{\Omega })\cap W_0^{1,p}(\mathrm{\Omega })$.

Conclusion. Assume that the only strong solution of $Lw=0$ in $W^{2,p}(\mathrm{\Omega })\cap W_0^{1,p}(\mathrm{\Omega })$ is $w=0$. Then the compactness of the Sobolev embedding upgrades the a priori estimate [F1] to the pure $L^p$ estimate displayed in the statement, the constant absorbing the $L^p$ term through the contradiction argument. No symmetry, self-adjointness or spectral hypothesis on $L$ is used, and the only choice principle invoked is the Axiom of Choice, including its inherited uses in the a priori estimate, completeness, Rellich--Kondrachov and extension theorems.