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 $H$.

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

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