essential numerical range of a self-adjoint operator

Question

If $T$ is a self-adjoint operator in $L(H)$ and the essential numerical range $W_e(T)$ of $T$ is the subset of $(–\infty, c]$, where $0\leq c＜1$. Then there exists a self-adjoint compact operator $L$ such that $$T\leq cI+L$$.

My question is that how to prove the existence of the self-adjoint compact operator.

If $W_e(T)\subset(–\infty, c]$, we have $\sigma_e(T)\subset W_e(T)\subset(–\infty, c]$. Thus $\sigma_e(T-cI)\subset(–\infty, c].$

We conclude that $\sigma(\pi(T-cI))\subset(–\infty, 0],$ where $\pi:L(H)\rightarrow L(H)/K(H)$, $K(H)$ is the set of all compact operators on $H$.

How to construct the compact operator $L$?

Answer 1

From your post we have that $\sigma(\pi(T-cI))\subset(\infty, 0]$, which combined with the fact that $\pi (T-cI)$ is self-adjoint, gives $\pi(T)-cI\leq 0$, or equivalently that $cI-\pi(T)\geq 0$.

We now use the following:

Lemma. Let $\varphi :A\to B$ be a surjective $^*$-homomorphism of C$^*$-algebras. Then, for every $b\geq 0$ in $B$, there is some $a\geq 0$ in $A$ such that $\varphi (a)=b$.

to conclude that there exists $S\geq 0$ in $L(H)$ such that $\pi (S)=cI-\pi(T)$. Consequently $L:= S-(cI-T)$ satisfies $\pi (L)=0$, so $L$ is compact. Also, since $S\geq 0$,we have $$ T =cI+L-S \leq cI+L. $$