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The following conjeture is stated here:

Every adjoint operator has a non-trivial closed invariant subspace.

Reference 11 where adjoint is supposedly defined can be found here. But I don't have access to the article.

So what's an "adjoint operator"?

Edit It is not "the adjoint of an operator" here because then the conjecture would be equivalent to:

Every operator has a non-trivial closed invariant subspace.

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    http://www.lmgtfy.com/?q=adjoint+operator – JMoravitz Jan 03 '15 at 03:47
  • I know the definition of the adjoint of an operator. But obviously that's not what is meant in the article since a conjecture like that would not make any sense. @JMoravitz –  Jan 03 '15 at 05:10
  • @student: I don't know a lot of functional analysis, but I'm not sure that your statement is equivalent. Is it the case that every Banach space is isomorphic to the dual of some Banach space? If it isn't, then perhaps not every operator is an adjoint operator, and the theorem is trying to point out a potentially important subtlety. – Eric Stucky Jan 03 '15 at 05:35
  • @EricStucky No, it seems... –  Jan 03 '15 at 06:15
  • @student: $$headdesk$$ I always forget to do obvious things like search MSE when I have questions I don't know the answer to… Thanks for the link :) – Eric Stucky Jan 03 '15 at 06:20
  • @EricStucky No, I thank you. Your comment was very helpful to me. –  Jan 03 '15 at 06:54

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For a Banach space $X$ , the dual space is the Banach space $X^*$ of all bounded linear functionals on $X$. Let $\langle x,f\rangle$ be the value of $f\in X^*$ on $x\in X$. If $T$ is a bounded linear operator on $X$, then its adjoint is a bounded linear operator $T^*$ on $X^*$ which is defined by $\langle x,T^*f\rangle=\langle Tx,f\rangle$ for all $x\in X$ and $f\in X^*$. There exist Banach space $X$ such that not every bounded linear operator on $X^*$ is the adjoint of an operator on $X$. The question if every operator on $X^*$ which is an adjoint has a non-trivial closed invariant subspace is stil unsolved. The problem is open even for the infinite dimensional separable complex Hilbert space. It is known, for instance, that there exists a bounded linear operator $S$ on $l^1$ which does not have non-trivial closed invariant subspaces. Of course, $S$ is not an adjoint operator.

Janko Bracic
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  • Regarding the concrete example of $\ell^2(\mathbb C)$: What would be an invariant subspace for the right shift operator $Re_n = e_{n+1}$? –  Jan 03 '15 at 07:02
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    Let $k$ be a positive integer and $M_k$ the closed linear subspace spaned by $e_j$ $(j\geq k)$. Then $M_k$ is a closed invariant subspace for $R$. Closed invariant subspaces of the shift operator have been described by Beurling more than 60 years ago. – Janko Bracic Jan 03 '15 at 07:40