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Is the invariant subspace problem known for $\ell^2(\mathbb{N})$ or for more general $L^2$ spaces, i.e. does every bounded linear operator $T \colon \ell^2(\mathbb{N}) \to \ell^2(\mathbb{N})$ have a non-trivial (closed) T-invariant subspace?

Zahlenteufel
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    Every separable Hilbert space is isometrically isomorphic to $\ell^2(\mathbb{N})$. Based on that, what would you be able to conclude? – Disintegrating By Parts Dec 26 '21 at 07:43
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    Thanks! I somehow forgot about this (so my question is equivalent to the invariant subspace problem (for non separable Hilbert spaces one can easily construct a non trivial invariant subspace)). How can I close the question (or do I have to delete the question or do you want to write this as an answer?) – Zahlenteufel Dec 26 '21 at 13:54
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    It is okay for you to write an answer, and then select it as an answer. In fact, it is encouraged because you are here to find answers. It will be a good thing to do. – Disintegrating By Parts Dec 26 '21 at 20:20

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The answer is, we don't know.

The invariant subspace problem is known for non separable Hilbert spaces and for finite dimensional ones, e.g. see this answer here. As noted by Disintegrating By Parts, every infinite dimensional and separable Hilbert space is isometrically isomorphic to $\ell^2(\mathbb{N})$.

Hence, the question is equivalent to asking whether every bounded linear operator $T \colon \mathcal{H} \to \mathcal{H}$ has an non-trivial invariant subspace, where $\mathcal{H}$ is a separable Hilbert space.

Zahlenteufel
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