# $\dim(\ker L^*) = \dim(\operatorname{coker} L)$

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 $$H$$.

• Is $H$ a hilbert space and $D$ a subspace? Is $D$ closed? – uniquesolution Apr 15 at 8:13
• $H$ is a Hilbert space, $D$ is a dense subspace. The definition of closed is as link in the notes by Tao. – CL. Apr 15 at 10:44