Let $L:D \rightarrow H$ be a closed, self adjoint operator, with two additional properties: (a) closed image. (b) finite dimensional kernel.
Is it true that $\dim(\ker L^*) = \dim (\operatorname{coker} L)$?
The reason I asked this is because when $D=H$, and $L$ is in fact bounded the result is true. What I have is written below.
If $L$ were a bounded self adjoint operator with $D=H$, then $H= \operatorname{im}(L) \oplus \operatorname{im} (L)^\perp =\operatorname{im}(L) \oplus \ker(L^*)$
(I) $\ker (L^*) \subseteq \operatorname{im} (L)^\perp$
(II) If $x \in \operatorname{im} L^\perp \cap \ker L$, $$ \langle x , Ly \rangle = 0 \Rightarrow \langle L^*x,y \rangle = 0$$
for all $y \in D$, hence $x \in \ker L^*$.
So we may write $H \supseteq \ker(L^*) \oplus \operatorname{im}(L)$ a dense subset of $H$.