I am trying to find a proof for the following problem:
Let $X,Y$ be Banach spaces $A,B:X \rightarrow Y$ are bounded linear operators $Ran(A)$ is closed, and $\dim(\mathrm{Ker}(A))$ or $\dim(Y/\mathrm{Ran}(A))$ is finite (or at least one of them). If $\left \| Bx\right \|<\left \| Ax\right \|$ for all $x \in X$ then
$\dim(Y/\mathrm{Ran}(A+B))\le\dim(Y/\mathrm{Ran}(A))$
Can you give me some hints on how to get a grip on the codimensions for this?
Thank you.
Regarding my attempts: It flows easily that $\mathrm{Ker}(A)=\mathrm{Ker}(A+B)$ hence we can assume the kernel is empty or else we factorize $X$ with it. Than thanks to the closeness of $\mathrm{Ran}(A)$ there exists a continuous $A^{-1}$ on $\mathrm{Ran}(A)$, so we have a continuous bijection from $\mathrm{Ran}(A)$ to $\mathrm{Ran}(A+B)$. However this does not help if $\dim(Y)=\infty$
Also it is enough to consider the case where $\dim(Y/\mathrm{Ran}(A))$ is finite. In this case $Y/\mathrm{Ran}(A)\cong(Y/\mathrm{Ran}(A))^*\cong \mathrm{Ran}(A)^0 $ the last element is the annihilator. But this is just as hard to use as the cokernel in my opinion.