# Two subsets of the spectrum, do they agree?

Let $H$ be a Hilbert space and $A:H\supset \text{dom}(A)\rightarrow H$ be a self-adjoint and unbounded operator defined on some dense subset $\text{dom}(A)\subset H$. Further, let $\sigma(A)\subset\mathbb{R}$ denote the spectrum of $A$. In my lecture, we defined two subsets of $\sigma(A)$ and denoted them by the same letter, namely:

$$1) \quad \mathcal{C}:=\lbrace \lambda\in\mathbb{R}: \text{Im}(T-\lambda\cdot I)\neq \overline{ \text{Im}(T-\lambda\cdot I)} \rbrace.$$

On the other hand, he showed $H=P\oplus P^{\perp}$, where $P$ denotes the closure of the subspace spanned by all eigenspaces of $A$. Moreover, because $A( P^{\perp})\subset P^{\perp}$, $A$ induces a unique operator $A_{\vert P^{\perp}}P^{\perp}\supset \text{dom}(A_{\vert P^{\perp}})\rightarrow P^{\perp}$ in $P^{\perp}$ with $\text{dom}(A_{\vert P^{\perp}}):=\text{dom}(A)\cap P^{\perp}$. Thus we defined

$$2)\quad \mathcal{C}:=\sigma(A_{\vert P^{\perp}}).$$

I am confused now. Both definitions look different, but we used the same letter as if both objects were the same. Are both sets equal, or are they indeed different?

I would appreciate if you help clarifying my confusion!

Assume that $\text{Im}(A-\lambda I)\neq \overline{ \text{Im}(A-\lambda I)}$. We may assume that $\lambda$ is not an eigenvalue, because even if it is, we can work on the restriction of $A$ to the orthogonal of $\ker(A-\lambda I)$. Then $A-\lambda I$ is not bounded below: because a bounded-below selfadjoint operator is injective, so it has dense range, and a bounded-below operator with dense range is surjective). So there exists a sequence $\{x_n\}$ with $\|x_n\|=1$ for all $n$, and such that $(A-\lambda I)x_n\to0$. If $v\in P$ is an eigenvector with nonzero eigenvalue $\mu\ne\lambda$, then $$\langle x_n,v\rangle=\frac1\mu\,\langle x_n,Av\rangle=\frac1\mu\,\langle Ax_n,v\rangle.$$ So $$\left(1-\frac\lambda\mu\right)\,\langle x_n,v\rangle\to0,$$ which implies $\langle x_n,v\rangle\to0$. This means that if we write $x_n=y_n+z_n$, with $y_n\in P$ and $z_n\in P^\perp$, we have $y_n\to0$. Thus $(A-\lambda I)z_n\to0$ with $\|z_n\|\to1$, implying that $\lambda\in\sigma(A|_{P^\perp})$.
If $\mu=0$ the reasoning is similar: $$0=\langle x_n,Av\rangle = \langle Ax_n,v\rangle,$$ so $\lambda \langle x_n,v\rangle\to0$, and $\lambda\ne\mu=0$.
Conversely, if $\lambda\in\sigma(A|_{P^\perp})$, we know that $\lambda$ is not an eigenvalue (because then it would be an eigenvalue for $A$). As $A|_{P\perp}$ is selfadjoint, any element of its spectrum that is not an eigenvalue is an approximate eigenvalue. So there exists a sequence $\{z_n\}$ with $\|z_n\|=1$ for all $n$, and $(A-\lambda I)z_n\to0$. This shows that $A-\lambda I$ is not bounded below, and so $$\text{Im}(A-\lambda I)\neq \overline{ \text{Im}(A-\lambda I)} .$$ This is because if $\text{Im}(A-\lambda I)=\overline{ \text{Im}(A-\lambda I)}$, the operator $(A-\lambda I)|_{ker(A-\lambda I)^\perp}$ is bijective onto its image, and thus invertible, thus bounded below.