I have the following, possibly trivial question, regarding checking for the essential spectrum:
Given a self-adjoint bounded operator $T\in \mathcal{L}(\mathcal{H})$, does $\dim \Big(\text{Im}\big(\chi_{[a-\epsilon,a+\epsilon]}(T) \big) \Big)=\infty$ imply that $\sigma_{ess}(T)\cap [a-\epsilon,a+\epsilon] \neq \emptyset$? Is it true that $\dim \Big(\text{Im}\big(\chi_{[a-\epsilon,a+\epsilon]}(T) \big) \Big)=\infty$ if and only if $\sigma_{ess}(T)\cap [a-\epsilon,a+\epsilon] \neq \emptyset$?
I have the following argument which I am unsure of.
I know that for such a $T$ as above, $\lambda \in \sigma_{ess}(T)$ if and only if there exists a sequence $\psi_n\in \mathcal{H}$ such that $\Vert \psi_n\Vert\equiv 1$, $\Vert (T-\lambda I)\psi_n\Vert \to 0$ and $\psi_n$ converge weakly to $0\in \mathcal{H}$. Using an iterated partition of $[a-\epsilon,a+\epsilon]$, I can find a nested sequence of shirnking closed intervals, $I_{n+1}\subseteq I_n \subseteq [a-\epsilon,a+\epsilon]$ and $Leb(I_n)\to 0$. By Cantor's intersection theorem and the criterion I mentioned, there exists $\lambda_0 \in \cap I_n$ such that $\lambda_0\in \sigma_{ess}(T)$.
The other direction is easier, since discrete spectrum corresponds to isolated eigenvalues with finite dimensional eigenspaces.
Can anyone help me verify this fact or this argument? I confused myself on this.
I think your argument is correct, not much else to say.
Another way to see this which is on the same page is to consider the fact that the discrete spectrum is, well, discrete, and thus $E := \sigma_\mathrm{dis}(T) \cap [a - \varepsilon, a + \varepsilon]$ is necessarily finite.
Indeed, that intersection remains discrete, yet if it were infinite, then by compactness of $[a - \varepsilon, a + \varepsilon]$ you could find a limit point, which would be absurd.
$E$ being finite, and the ranks of the Riesz projectors (the fancy name for the spectral projections) being finite for elements of $E$, you would have $\dim \Big(\text{Im}\big(\chi_{[a-\epsilon,a+\epsilon]}(T) \big) \Big) < \infty$ if there were no elements from the essential spectrum in our interval, thus $\sigma_{\mathrm{ess}}(T)\cap [a-\epsilon,a+\epsilon] \neq \emptyset$ needs to hold.