Let $X$ be a Banach space and $K:X\to X$ be a compact operator. If $I+K$ is injective then it is onto and hence the inverse $(I+K)^{-1}$ is bounded. What kind of qualitative or quantitative assumptions on $K$ are necessary in order to find an explicit bound on the operator norm of $(I+K)^{-1}$?

In such generality, there is probably no answer (or there are too many), so let me be more specific and explain the situation I've encountered this. The Banach space is either $L^\infty$ or a weighted $L^2$ with weight $(1+|x|)^s$ on $\mathbb{R}^n$, these cases are all of interest. Call $H_0$ the operator $-\Delta$, $V\ $ the multiplication by a function $V(x)$, and $H=H_0+V$. This should all be properly defined but I leave the definitions aside so to accommodate as many additional assumptions as necessary. A standard procedure in spectral theory is to express the resolvent operator $R(z)=(H-z)^{-1}$ as a perturbation of $R_0(z)=(H_0-z)^{-1}$ through the formula
$$ R_0(z)=R(z)(I+VR_0(z)). $$
Now the main step: assume we know that $VR_0$ is a compact operator and $I+VR_0$ is injective (this typically amounts to requiring that $H$ has no eigenvalues, or resonances). Then we can invert $I+VR_0$ and write
$$ R(z)=R_0(z)(I+VR_0(z))^{-1} $$
which is a very powerful formula, but, alas!, not quantitative. Is there a way to bound the norm of the inverse based on **any** reasonable assumption on $H_0$ and $V$? Notice that the operator $K=VR_0$ is very explicit, one can write an explicit integral kernel and extract any kind of detailed information on it. Additionally, one can make any sort of assumption on the perturbed operator $H=H_0+V$ (e.g., about the distance of the eigenvalues from the continuous spectrum etc.)

EDIT: I might add that in the example the main point is to obtain a uniform estimate for the norm of $R(z)$ as $z$ approaches the real axis, which contains the spectrum of $H$. Otherwise it might seem odd to look for an estimate on $R(z)$, which is already a well defined and bounded operator on $L^2$, with known bounds. But this procedure gives a reasonable definition of $R(z)$ also for real $z$, certainly not as a bounded operator on $L^2$ but on different spaces.