I have a question related to a certain elliptic operator on $R^N$ but I think i can clarify my confusion if I just consider the Laplacian $\Delta$ on the unit ball in $R^N$.
If $ 1 <p< \infty$ we know that $ \Delta:W_0^{2,p}(B) \rightarrow L^p(B)$ is an isomorphism (one to one, onto, continuous, inverse continous). Here I am using some non standard notation in the sense that by $W^{2,p}_0(B)$ i mean the functions in $W^{2,p}(B)$ which are zero on the boundary. Now we know the above does not hold for $p=1,\infty$.
I am interested in having the target space to be $L^\infty(B)$. So, as well known counterexamples show, there are examples of $ f $ bounded such that the solution $ \Delta u =f $ in $B$ with $u=0$ are not in $W^{2,\infty}$. So the inverse isn't continuous.
So I would like to just try and define the appropriate space on the left using some graph norm type idea. So , for instance, pick $ N<p< \infty$ and set $$ X:=\{ u \in W^{2,p}_0(B): \Delta u \in L^\infty\} $$ and lets put the norm
$ \| u\| = \| \Delta u \|_{L^\infty} $.
So it appears $X$ is a Banach space and $ \Delta :X \rightarrow L^\infty$ should be an isomorpism. Is this correct?
If this is the case I realize that $X$ is slightly bigger than $ W^{2,\infty}_0$ (as defined above).
My functional analysis has decayed to zero and so any comments would be greatly appreciated.
thanks Craig
