TITLE: Multiple of identity plus compact
QUESTION [15 upvotes]: Is there an example of a bounded operator $T\in\mathcal{B}(H)$, where $H$ is a separable complex Hilbert space,  such that no restriction to an infinite dimensional closed subspace is multiple of identity plus compact?  
Edit: What I mean by "restriction" is $T_{|M}:M\to H$, where $M$ is an infinite dimensional subspace of $H$, and by "multiple of identity" I mean a complex multiple of the inclusion map $i:M\to H$.

REPLY [6 votes]: I think this is essentially Theorem 3.1 in:
A. Brown, C. Pearcy, Compact restrictions of operators, Acta. Sci. Math (32) 1971, 271-282 (link)
In fact, something a bit stronger and in more generality is true (see Proposition 3.2 in this paper):
Proposition:  Let $X$ is a  Banach space and $T\in\mathcal{B}(X)$ such that there exists $\lambda\in\partial\sigma(T)\setminus\sigma_p(T)$. Then there exists $e\in X$ such that for any $\epsilon>0$ we can find $Y$ an infinite dimensional and infinite co-dimensional subspace of $X$ with the property that $(\lambda I-T)_{|Y}:Y\to Y+[e]$, and $(\lambda I-T)_{|Y}$ is compact and has norm less than $\epsilon$.