I have a (densely defined) operator of the form $$D:=\frac d {dt} + L + h:L^2(\mathbb{R}\times Y)\dashrightarrow L^2(\mathbb{R}\times Y)$$ where $L:C^\infty(Y)\to L^2(Y)$ is self-adjoint, elliptic differential of order 1 with $C^\infty$-coefficients and $h:C^\infty(Y)\to L^2(Y)$ extends as a bounded operator on $L^2_j(Y)$ for $j=-1,0,1$. Also $dim(Y)>2$ and $Y$ compact so that we can assume Sobolev embedding $L^2_1\to L^2$ is continuous and Rellich $L^2_1\subset \subset L^2$.
Claim: $D:L^2\to L^2_{-1}$ is invertible iff the $L^2$-adjoint $D^*:=-\frac d {dt} + L + h^*:L^2_1\to L^2 $ is invertible.
I would like to prove this but I have some problems already with the statement.
- First of all why the $L^2$-adjoint of $D$ is even defined over $L^2_1$? We know that $D:L^2\to L^2_{-1}$ is continuous, thus the $(L^2,L^2_{-1})$-adjoint will be a continuous map $L^2_1\to L^2$, but $D^*$ is the $L^2$-adjoint = $(L^2,L^2)$-adjoint which is different a priori.
- If $D^{\dagger}$ is the $(L^2,L^2_{-1})$-adjoint then is standard that if $D^\dagger$ is invertible also $D$ is invertible: $D^\dagger$ closed $\Rightarrow$ $D$ closed, $\ker D^\dagger = 0\Rightarrow D$ surjective, and $D^\dagger$-surjective $\Rightarrow$ $\ker D = 0$. However again, we are dealing with the adjoint with respect to different inner-products.
Probably I am confused due to the dual $L^2_{-1}$ because I'm not used to it. If I am correct, the hypothesis imply that $h(L^2_j(Y))\subset L^2_j(Y)$ so that $h$ is compact and $L+h:L_1^2(Y)\to L^2(Y)$ is Fredholm.