# Elliptic pde L^p theory via adjoint theory

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 $$1. Now of course we know $$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).

DOES THIS ACTUALLY WORK OR IS THIS ALL NONSENSE?
thanks Craig

• 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... ??) – Math604 Oct 28 '18 at 13:23