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Being far from analysis, I recently learned about the Invariant subspace problem and came up with the following (perhaps simple or well-known) question.

Let $H$ be a separable complex Hilbert space and $T:H\to H$ a bounded operator. Assume that the spectrum of $H$ is $\{0\}$, i.e. $T-\lambda I$ has a bounded inverse for every $\lambda\in\mathbb C\setminus\{0\}$. In finite dimensions, this would imply that $T$ is nilpotent ($T^n=0$ for some $n$). I wonder if there is something similar in the infinite dimensional case. The precise formulation I have in mind is the following.

It is easy to see that there is a maximal subspace $X\subset H$ such that $T(X)=X$. This is a purely set-theoretic fact, a sort of explicit construction is the intersection of the images of iterations of $T$ up to and beyond infinity (via transfinite induction).

Question: can it happen that $X\ne\{0\}$?

Here are some observations that I made:

  • $T$ cannot be onto. (I derived this from some random Wikipedia quotes so there are high chances of error; please correct me if I am wrong.) It follows that a nontrivial $X$ cannot be closed.

  • If there is an example, then there is one where $X$ is dense. Just take the closure of $X$ for $H$.

  • It is possible that $T(H)$ is dense. My example is the shift in $\ell^2(\mathbb Z)$ composed with a mutiplication by a positive function (sequence) that goes to zero at both ends. Again, please correct me if I am wrong.

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Bounded operators with spectrum $\\{0\\}$ are called quasinilpotent. I don't know much about them, but I guess that knowing the terminology will help you to find some references. –  Ian Morris May 15 '10 at 9:23
    
Incidentally, Apostol and Voiculescu proved that every quasinilpotent operator is a norm limit of nilpotent operators. See for example jstor.org/pss/2042882 –  Jonas Meyer May 17 '10 at 6:53

2 Answers 2

up vote 6 down vote accepted

Perhaps I am missing something. Isn't the example you mention quasinilpotent and and maps the finitely non zero sequences onto themselves?

I am not an expert on invariant subspaces, but I think it is widely believed among experts that the answer to the question is the same for quasinilpotent operators as for general operators.

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Sergei is asking for a large invariant subspace which may not be closed. I think that the existence of a large, dense $X$ such that $TX=X$ is not constrained by the ordinary invariant subspace problem, which asks for closed, proper invariant subspaces. –  Ian Morris May 15 '10 at 13:49
    
That is what I thought, Ian. The finitely non zero sequences $X$ satisfy $TX=X$ for the operator Sergei mentioned (indeed, for any right shift with all weights non zero, but you need the null condition at $\pm \infty$ to make it quasinilpotent). Am I misinterpreting Sergei's question? –  Bill Johnson May 15 '10 at 14:17
    
Sorry, ignore my comment. –  Ian Morris May 15 '10 at 16:37
    
Yes, this is exactly what I asked. It is simple as I suspected, I just did not see it. Thanks! –  Sergei Ivanov May 15 '10 at 17:23

Just some remarks. A natural example of a quasinilpotent operator, which may be of interest to you, is Tu(x):=∫0xu(s)ds on C[0,1], or on Lp[0,1], for 1≤p≤∞. In all these cases T is compact. However note that a noncompact example can be easily defined e.g. on a Hilbert sum of infinitely many Hilbert copies of a Hilbert space H, as the direct sum of infinitely many copies of a non-zero quasinilpotent T:H→H (that is, just taking {hj} to {Thj}).

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Another noncompact example is $\left( \begin{matrix} 0 & I \\ 0 & 0 \end{matrix} \right)$ acting on $H\oplus H$ if $H$ is infinite dimensional. (I'm just taking the Kronecker product in the reverse order.) –  Jonas Meyer May 17 '10 at 1:37

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