compact linear operator restricted to closed subspace of Hilbert space is one to one

Question

I am solving a problem:

For any closed subspace $V$ of Hilbert space $V$, the image$(I-K)(V)$ is a closed subspace of $H$. $K$ is a compact linear operator on $H$.

I think this is easy to prove only if I can prove that $I-K$ is one to one when restricted on $V$.

Can anyone please give me a hint to show that $I-K$ is one to one when restricted on $V$?

Thanks,

Answer 1

$L=I-K$ is a Fredlhom operator. This implies that its kernel is finite dimensional and the codimension of its image is finite. Let $U$ such that $H=Im(L)\oplus U$. Define $f:H/ker(L)\oplus U\rightarrow H$ by $f(x+u)=L(x)+u, x\in H/ker(L), u\in U$. $L$ is a bounded and bijective. The open mapping theorem implies that $f$ is invertible, thus closed, we deduce that $L(V)=f(V\oplus 0)$ is closed.

https://en.wikipedia.org/wiki/Fredholm_operator