Fletcher Hays

2022-06-22

Let ${H}_{1}$ and ${H}_{2}$ be Hilbert spaces, let $T\in B\left({H}_{1},{H}_{2}\right)$. Suppose that $\mathrm{Ker}T$ is finite-dimensional and that $\mathrm{Im}T$ is closed in ${H}_{2}$. Prove that $\mathrm{Ker}\left(T+K\right)$ is finite-dimensional for each $K\in K\left({H}_{1},{H}_{2}\right)$.

1. Define Hilbert space as direct sum of two complemented subspaces
2. For compact operator use Hilbert Schmidt decomposition

Main idea is to prove that intersection of Spectrum of such \$T and compact K is finite

Stevinivm

Expert

By assumptions the operator is invertible. Therefore for any $w\perp \mathrm{ker}T$ we have
$‖Tw‖\ge c‖w‖$
for a constant $c>0.$

Assume by contradiction, that $\mathrm{ker}\left(T+K\right)$ is infinite dimensional. Then there exists an infinite orthonormal system ${v}_{n}$ in $\mathrm{ker}\left(T+K\right).$ As $K$ is compact, then $K{v}_{n}\to 0.$ Thus $T{v}_{n}\to 0.$ We have

and
$‖T{v}_{n}‖=‖T{u}_{n}‖\ge c‖{u}_{n}‖$
hence ${u}_{n}\to 0.$ The sequence ${w}_{n}$ has an accumulation point, therefore the sequence ${v}_{n}$ has an accumulation point as well. This leads to a contradiction.

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