I'm reading the following proof that boils down to the following argument.
$A: H \rightarrow H$ compact, with $H$ Hilbert space. $B: H\rightarrow H$ bounded, invertible. $0$ not an eigenvalue of $A+B$.
Then, $A+B$ is a Fredholm operator, and since $0$ is not an eigenvalue of $A+B$, then the kernel is trivial. Therefore, $A+B$ is invertible.
My first guess is that this is false, as just injectivity is not enough, we require the Fredholm operator to be of index 0 for invertibility. (Ex. the shift operator)
But maybe a Fredholm obtained in this way is forcefully of index 0?
Best Answer
Since $B$ is invertible and $A$ is compact we have that $A+B$ is Fredholm and $ind(A+B)=ind(B)=0$. Since $ \dim (ker(A+B))=0$ we have that $ codim (Im(A+B))=0$, hence $A+B$ is surjective.
Conclusion: $A+B$ is bijective.