$(Tx,x) \geq ||x||^2$, $K$ compact operator. Then $T+K$ injective implies surjective

Question

I am trying to solve the following exercise.

Let $H$ be a Hilbert space and $T: H\to H$ be a bounded linear operator such that $$ (Tx,x) \geq ||x||^2 \quad \forall x \in H. $$

Let $K: H \to H$ be a compact linear operator. Prove that if $T+K$ is injective, then $T+K$ is surjective.

What I have done so far: I have proved that there exists $T^{-1}$ and it is a bounded linear operator.

Could you please help me to go further? Any hints or book suggestions with similar exercises are appreciated. Thank you

Answer 1

You have shown that $T$ is bijective. I will follow the hint by harfe. We have $$ T^{-1}(T+K) = I + T^{-1}K, $$ is injective. Since $T^{-1}$ is bounded and $K$ is compact $T^{-1}K$ is compact. By the Fredholm alternative $$ \mathrm{codim}\, R(I + T^{-1}K) = \dim N(I + T^{-1}K) = 0. $$ Thus $R(I + T^{-1}K) = H$. Applying $T$ we have $$ R(T+K) = H. $$

Answer 2

Since you are able to show that $T$ is invertible, $T+K$ is a Fredholm operator, look thereom 7 of this paper

https://users.math.yale.edu/users/mr2245/func2018Data/fredholm.pdf