A question about quasi-nilpotent operators

Question

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.

Answer 1

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 $\left( v_{n}\right) _{n\in %TCIMACRO{\U{2115} }% %BeginExpansion \mathbb{N} %EndExpansion }$ 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 $\left( y_{n}\right) _{n\in %TCIMACRO{\U{2115} }% %BeginExpansion \mathbb{N} %EndExpansion }$ 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 $Y=\overline{span\left\{ y_{n}:n\in %TCIMACRO{\U{2115} }% %BeginExpansion \mathbb{N} %EndExpansion \right\} }$ 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.