Let $A$ be a bounded linear operator on some separable Hilbert space $\mathcal{H}$. We have the following iff condition for $A$ to have closed range: $\exists C>0$ s.t. $\forall\varphi\in\ker(A)^\perp$ we have $\|A\varphi\|\geq C \|\varphi\|$.
This apparently also means that zero is not, or is an isolated point of the spectrum of $A$, that is, $0\notin\sigma_{\mathrm{ess}}(A)$, and since $\sigma_{\mathrm{ess}}(A)^c$ is an open set, this implies that we have an open ball about zero in the essential resolvent set of $A$.
We also have the following criterion for a point $\lambda\in\mathbb{C}$ to be in the resolvent set of $A$, if $A$ is self-adjoint: $\exists\varepsilon>0$ s.t. $\forall\varphi\in\mathcal{H}\setminus\{0\}$, $\|(A-\lambda)\varphi\|\geq\varepsilon\|\varphi\|$
These two conditions look extremely similar. It almost feels like the closed range condition allows us to make conclusions in the neighborhoud of zero just by checking what happens at zero, at least when operators are self-adjoint.
Can anyone explain the relationship between these two conditions? Is there something about when $A$ is not self-adjoint?
The first condition implies that $A$ has closed range and it implies that $A : \ker(A)^\perp\to R(A)$ can be continuously inverted. We have to factor out $\ker A$ first.
The second condition is stronger, because it also implies (and requires) that $A$ is injective. Hence $A^{-1}:R(A)\to X$ exists and is continuous. It does not imply that $A$ is surjective.
Your criterion is not the definition of resolvent set: to have $\lambda\in \rho(A)$ the operator $\lambda I-A$ has to be continuously invertible. For example, the right-shift operator $S$ on $l^2$ satisfies $\|Sx\|_{l^2}= \|x\|_{l^2}$, however $0$ is not in the resolvent set because $S$ is not surjective.