A question about quasi-nilpotent operators Let $X$ be an infinite dimensional Banach space, and let $Q\in B(X)$ be a bounded quasi-nilpotent operator ($\sigma(Q)=\{0\}$). I am trying to prove that for every $\epsilon >0$ we can find an infinite dimensional subspace $Y\subset X$ such that the restriction $\left. Q\right\vert _{Y}:Y\rightarrow X$ of $Q$ to $Y$ is such that $\left\Vert \left.Q\right\vert _{Y}\right\Vert <\epsilon $. 
Any help please ?
Thank you.
 A: This is a partial answer if we deal with a Hilbert space $\mathcal{H}$ :
If $\dim \left( \ker \left( Q\right) \right) =\infty $, then we take $%
Y=$ $\ker \left( Q\right) $ and hence $\left\Vert Q_{|Y}\right\Vert =0$
. Else, if $\dim \left( \ker \left( Q\right) \right) <\infty $ then it
is easy to see that for every closed subspace $V\subset \mathcal{H}$ such
that $co\dim V<\infty $ and for each $n\in 
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
$, $\dim \left( V\cap Q^{n}\left( V\right) \right) =\infty $ since $%
Q^{n}\left( V\right) $ is infinite dimensional. Then for every $n\in 
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
$ we can find a non zero vector $v_{n}\in V$ such that $\left\Vert
v_{n}\right\Vert =1$ and $Q^{i}\left( v_{n}\right) \in V\backslash \left\{
0\right\} $ for all $i\in \left\{ 1,\ldots ,n\right\} $. Suppose $%
Q_{|V}:V\rightarrow \mathcal{H}$ is bounded below, then there is some $a>0$
such that $\left\Vert Qv\right\Vert \geq a\left\Vert v\right\Vert $ for all $%
v\in V$. Then, for our sequence $\left( v_{n}\right) _{n\in 
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\mathbb{N}
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}$ we have $\left\Vert Q^{n}v_{n}\right\Vert \geq a\left\Vert
Q^{n-1}v_{n}\right\Vert \geq \cdots \geq a^{n}\left\Vert v_{n}\right\Vert
=a^{n}$ so, $\left\Vert Q^{n}v_{n}\right\Vert ^{\frac{1}{n}}\geq a$, wich
yealds $0=\underset{n\rightarrow \infty }{\lim }\left\Vert
Q^{n}v_{n}\right\Vert ^{\frac{1}{n}}$ $\geq a$, which is a contradiction
with the fact that $Q$ is quasi-nilpotent. Hence $Q_{|V}:V\rightarrow 
\mathcal{H}$ is not bounded below.
Let $\epsilon >0$, $S^{1}$ be the unit sphere in $\mathcal{H}$ and define an
orthonormal sequence $\left( y_{n}\right) _{n\in 
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\mathbb{N}
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}$ as follows :%
\begin{eqnarray*}
\exists y_{1} &\in &\left( \mathcal{N}\left( Q\right) \right) ^{\bot
}:\left\Vert y_{1}\right\Vert =1\text{ and }0<\left\Vert Qy_{1}\right\Vert
<2^{-1}\epsilon  \\
\exists y_{2} &\in &\left( \mathcal{N}\left( Q\right) +span\left\{
y_{1}\right\} \right) ^{\bot }:\left\Vert y_{2}\right\Vert =1\text{ and }%
0<\left\Vert Qy_{2}\right\Vert <2^{-2}\underset{v\in span\left\{
y_{1}\right\} \cap S^{1}}{\min }\left\Vert Qv\right\Vert  \\
\exists y_{2} &\in &\left( \mathcal{N}\left( Q\right) +span\left\{
y_{1},y_{2}\right\} \right) ^{\bot }:\left\Vert y_{3}\right\Vert =1\text{
and }0<\left\Vert Qy_{3}\right\Vert <2^{-3}\underset{v\in span\left\{
y_{1},y_{2}\right\} \cap S^{1}}{\min }\left\Vert Qv\right\Vert  \\
&&\vdots  \\
\exists y_{n} &\in &\left( \mathcal{N}\left( Q\right) +span\left\{
y_{1},y_{2},\ldots ,y_{n-1}\right\} \right) ^{\bot }:\left\Vert
y_{n}\right\Vert =1\text{ and }0<\left\Vert Qy_{n}\right\Vert <2^{-n}%
\underset{v\in span\left\{ y_{1},y_{2},\ldots ,y_{n-1}\right\} \cap S^{1}}{%
\min }\left\Vert Qv\right\Vert  \\
&&\vdots 
\end{eqnarray*}
The choice of $y_{n}$ is possible since $Q$ is not bounded below on the
closed finite codimensional subspace $\left( \mathcal{N}\left( Q\right)
+span\left\{ y_{1},y_{2},\ldots ,y_{n-1}\right\} \right) ^{\bot }$ and $%
\mathcal{N}\left( Q\right) \cap \left( \mathcal{N}\left( Q\right)
+span\left\{ y_{1},y_{2},\ldots ,y_{n-1}\right\} \right) ^{\bot }=\left\{
0\right\} $.
It is clear that the closed subspace $Y=\overline{span\left\{ y_{n}:n\in 
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%BeginExpansion
\mathbb{N}
%EndExpansion
\right\} }$ is in $\left( \mathcal{N}\left( Q\right) \right) ^{\bot }$ and
is infinite dimensional
Let $y\in Y$ be such that $\left\Vert y\right\Vert =1$. Writing $y=\underset{%
i=1}{\overset{\infty }{\sum }}a_{i}y_{i}$, then we have for sufficientely
great $n\in 
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
$ :
$\frac{\underset{i=1}{\overset{n}{\sum }}a_{i}y_{i}}{\left\Vert \underset{i=1%
}{\overset{n}{\sum }}a_{i}y_{i}\right\Vert }\in span\left\{
y_{1},y_{2},\ldots ,y_{n}\right\} \cap S^{1}$
On the other hand, since for each $n\in 
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
$, $\left\Vert Qy_{n}\right\Vert <2^{-n}\epsilon $ we find :
\begin{eqnarray*}
\underset{v\in span\left\{ y_{1},y_{2},\ldots ,y_{n}\right\} \cap S^{1}}{%
\max }\left\Vert Q\left( v\right) \right\Vert  &\leq &\underset{i=1}{\overset%
{n}{\sum }}\underset{v\in span\left\{ y_{1},y_{2},\ldots ,y_{n}\right\} \cap
S^{1}}{\max }\left\Vert \left\langle v,y_{i}\right\rangle Qy_{i}\right\Vert 
\\
&\leq &\underset{i=1}{\overset{n}{\sum }}\underset{v\in span\left\{
y_{1},y_{2},\ldots ,y_{n}\right\} \cap S^{1}}{\max }\left\vert \left\langle
v,y_{i}\right\rangle \right\vert \left\Vert Qy_{i}\right\Vert  \\
&\leq &\underset{i=1}{\overset{n}{\sum }}\left\Vert Qy_{i}\right\Vert  \\
&\leq &\underset{i=1}{\overset{n}{\sum }}2^{-i}\epsilon  \\
&\leq &\epsilon 
\end{eqnarray*}
Then :
\begin{eqnarray*}
\left\Vert Qy\right\Vert  &=&\left\Vert \underset{n\rightarrow \infty }{\lim 
}Q\left( \frac{\underset{i=1}{\overset{n}{\sum }}a_{i}y_{i}}{\left\Vert 
\underset{i=1}{\overset{n}{\sum }}a_{i}y_{i}\right\Vert }\right) \right\Vert 
\\
&\leq &\epsilon 
\end{eqnarray*}
And we are done.
