Let $ T:X \rightarrow Y$ denote some linear operator and suppose we know its one to one (here $X$ and $Y$ are Banach spaces). I believe their is results that say $Ker(T^*)= (R(T))^\perp$ (where for the perp we are using the generalization people use in non Hilbert spaces). Here we are defining $T^*:Y^* \rightarrow X^*$ via the usual definition but now we replace the inner products with the duality pairings.

QUESTION. I would to try and use this to say something about the mapping $\Delta$ in the $L^p$ setting (I never learned the proof of the $L^p$ regularity results and now I need to it for a general second order operator and I can't perturb the laplacian).

So we set $ T(u):=\Delta u$ where $T:W_D^{2,p}(\Omega) \rightarrow L^p(\Omega)$ (here $ W^{2,p}_D = \{ u \in W^{2,p}(\Omega): u=0 \; on \; \partial \Omega\}$) where $ 1<p<\infty$. Now of course we know $T$ is onto and the inverse has an estimate (but this is want i want to prove via this adjoint theory).

Let $ g \in L^{p'}(\Omega)$ be such that $T^*(g)=0$ in $ (W_D^{2,p})^*$ and hence $$0= (T^*(g), u) = (g,T(u)) = \int_\Omega g (\Delta u) dx $$ for all $ u \in W^{2,p}_D$. Lets assume we know the $L^2$ theory of the Laplacian; which also tells us that for all $ f $ smooth there is a $u$ smooth such that $ \Delta u = f$ in $\Omega$ and $u=0$ on $ \partial \Omega$. This would be enough $u$ to show that $g=0$ in $L^{p'}(\Omega)$ and hence the kernel of $T^*$ is empty (hence we should have $T$ onto and I assume it probably tells us something about an estimate via Open Mapping Theorem).

thanks Craig

  • $\begingroup$ maybe I am cheating here: to go from $R(T)^\perp = \{0\}$ to $R(T)=Y$ maybe i am using $R(T)$ is closed (which we don't have...apriori... ??) $\endgroup$ – Math604 Oct 28 '18 at 13:23

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