Is it true that $T$ is surjective IFF $T-K$ is surjective?

Question

Let $E$ be an infinite-dimensional real Banach space. Let $\mathcal L(E)$ be the space of bounded linear operators on $E$ and $\mathcal K(E)$ its subspace consisting of compact operators.

Let $T\in \mathcal L(E)$ and $K \in \mathcal K(E)$ such that $T$ is injective. In solving Brezis' exercise 6.18.10, I'm thinking of

$T$ is surjective IFF $T-K$ is surjective.

Could you elaborate on whether above statement is true or not? Thank you so much for your help!

What if $T$ is the identity and $K$ is a projector onto some finite dimensional subset? — Didier, Jul 15 '23 at 08:55
@Didier You are right! Could you post your comment as an answer so that I can accept it? — Akira, Jul 15 '23 at 09:05

score 1 · Accepted Answer · answered Jul 15 '23 at 09:07

Assume one has a decomposition $E = V\oplus W$ with $V$ a finite dimensional subspace. Let $K$ be the projector onto $V$ in the direction of $W$ (which is a compact operator) and $T$ the identity map (which is a bounded operator). Then for any $u = v + w \in E$, $(T-K)(u) = u-v = w \in W$, and therefore, $T$ has range in $W$ and cannot be surjective.

Is it true that $T$ is surjective IFF $T-K$ is surjective?

1 Answers1