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 $i:M\to H$. 



Multiple of identity plus compact
OK, let me try too. It is going to be a somewhat long story. WLOG, $\|T\|\le 1$.


**Step 1:** It is enough to show that for every finite-dimensional subspace $E$ and every $\delta>0$, there exists a unit vector $v\in E^\perp$ and $z\in\mathbb C$ such that $\|Tv-zv\|\le\delta$.


**Proof:** Suppose that the claim holds. Choose a square summable sequence $\delta\_j>0$. Construct by induction a sequence of pairwise orthogonal unit vectors $v\_j$ and numbers $z\_j$ such that $\|Tv\_j-z\_jv\_j\|\le\delta\_j$. Passing to a subsequence, we can ensure that $z\_j\to z$ and, moreover, $\sum\_j|z\_j-z|^2<+\infty$. Then $T-zI$ is a Hilbert-Schmidt operator on $\operatorname{span}(v\_j)$.


**Step 2:** Suppose that $1>\delta>0$ and $a\_j$ is a sufficiently long (i.e., of length $\ge N(\delta)$) sequence of positive numbers such that $\delta^2 a\_j\le a\_{j+1}\le a\_j$. Then there exist $m<n$ such that if we multiply $a\_j$ by an (automatically non-decreasing) geometric progression so that the resulting products $a\_j'$ will satisfy $a\_m'=a'\_n=1$, then we shall have $\sum\_{k=m}^{n-1} a\_k'\ge \delta^{-2}$.


**Proof:** Consider the points $P\_k=(k,\log a\_k)$, $k=0,\dots,N$ and take the convex hull of rays going from those points up. It will be bounded by two vertical rays and several slanted intervals. Now, if there is a slanted interval between $P\_m$ and $P\_n$ with $n-m>\delta^{-2}$, we are done (because for the corresponding modified sequence $a\_k'$, we'll have $a\_k'\ge 1$ for $m\le k\le n$). Otherwise, since the slopes of slanted intervals are squeezed between $2\log\delta$ and $0$ and increasing, we can find $2\delta^{-2}$ subsequent intervals that form an almost straight piece, provided that $N$ is large enough. Then if $P\_m$ is the left endpoint of the left of those intervals and $P\_n$ is the right endpoint of the right of those intervals, we have $n-m\ge 2\delta^{-2}$ and $a\_k'\ge\frac 12$ for $k$ between $m$ and $n$.


**Step 3:** Let $E$ be a finite-dimensional subspace. Let $\delta>0$ and let $N$ be large enough to guarantee the conclusion of Step 2. The conditions $x,Tx,T^2x,\dots, T^Nx\in E^\perp$ define a closed subspace of finite codimension in our infinite-dimensional Hilbert space $H$. Let $x$ be a unit vector in that subspace. Consider the sequence $x\_k=T^kx$, $k=0,\dots, N$ and the numbers $a\_k=\|x\_k\|^2$. If we have $a\_{j+1}<\delta^2a\_j$, then the normalized vector $x\_j$ is what we are looking for with $z=0$. Otherwise choose $m<n$ as in Step 2. Let $q^2\ge 1$ be the ratio of the corresponding geometric progression and $r^2$ be its $m$-th term. Then for the vectors $y\_k=rq^{k-m}x\_k$, we have $\|y\_m\|=\|y\_n\|=1$ and $y\_{k+1}=qTy\_k$. Let $|\zeta|=1$. Consider the vector
$$
y=y\_\zeta=\sum\_{k=m}^{n-1}\zeta^{-k}y\_k.
$$ 
Then $\|Ty-\zeta q^{-1}y\|=\frac 1q\|\zeta^{-n}y\_n-\zeta^{-m}y\_m\|\le 2$. However, the average of $\|y\_\zeta\|^2$ is $\sum\_{k=m}^{n-1}\|y\_k\|^2\ge \delta^{-2}$, so we can use the normalized $y\_\zeta$ with appropriately chosen $\zeta$ (with $2\delta$ instead of $\delta$).


It is the last step that uses the Hilbert space structure in a really essential way (though the previous steps made some limited use of it too), which makes me wonder what happens in an arbitrary Banach space.


