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Recently four mathematicians claimed to have proven the invariant subspace problem, which is the problem that states

Does every bounded operator on a separable Hilbert space have a non-trivial invariant subspace?

They claimed to have proven the existence of a non-zero weak limit that is orthogonal to the entire space and that gave rise to a contradiction.

For those interested this is the link to the paper:

Roshdi Khalil, Yousef Abdelrahman, Alshanti Waseem Ghazi, and Abu Hammad Ma’mon, "The Invariant Subspace Problem for Separable Hilbert Spaces" Axioms 13, no. 9: 598 (2024) DOI:10.3390/axioms13090598.

So my question is, since the paper was published in a journal does that mean the problem is closed?

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    $\begingroup$ Without looking at the paper, I'd say the proof is probably wrong. Otherwise the authors would have chosen a first rate journal. The journal Axioms is not even on the MR reference list. $\endgroup$
    – Antonius
    Commented Sep 4 at 6:03
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    $\begingroup$ MDPI, and Axioms in particular, is a predatory publisher. The referee requests they occasionally send me ask for reviews due in two weeks, which is completely ridiculous; this kind of review process is obviously just a fig leaf. And, of course, they ask authors for exorbitant article processing charges (a recent spam from them I found in my inbox mentions 2400 CHF). Also, I was under the impression that Axioms was supposed to be a logic journal; what that has to do with functional analysis? $\endgroup$
    – Emil Jeřábek
    Commented Sep 4 at 7:01
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    $\begingroup$ The club of ISP-solvers has quite a number of members, it might be chaired by Enflo (arxiv.org/abs/2305.15442) and de Branges (math.purdue.edu/~branges/invariantsubspaces.pdf). $\endgroup$
    – Jochen Wengenroth
    Commented Sep 4 at 7:41
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    $\begingroup$ @JochenWengenroth: In defense of the new club members whose paper is linked in the question, I have to say that their paper is at least sufficiently clearly written that I could find the mistake within 20 minutes or so. In contrast, when I tried to read Enflo's recent paper, I just gave up after the same period of time. $\endgroup$
    – Jochen Glueck
    Commented Sep 4 at 9:04
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    $\begingroup$ I find it beyond ridiculous that this question should be closed. It is a perfectly reasonable question, since it is about an important unsolved problem in research mathematics. There is nothing gained by closing such a question. In fact, it led to Jochen Glueck's very useful debunking of the paper in his answer below. $\endgroup$
    – Daniel Asimov
    Commented Sep 5 at 14:08

1 Answer 1

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No, the proof is wrong.

Conceptual reason why it can't be correct: The argument in the paper, were it correct, would actually show much more, namely that for every bounded linear operator $T$ on a separable Hilbert space $H$ and every non-zero $x \in H$, it cannot happen that both the span of the orbit of $x$ and the span of the orbit of $Tx$ are dense. This is wrong, as one can see be considering a hypercyclic operator on $\ell^2$.

(Note: In the first version of the answer I said that a stronger claim would follow from the proof in the paper, but this is not correct, as pointed out by the OP in a comment under this answer.)

What's wrong in the proof? The functional $f$ in formula (20) is not well-defined, in general, since the $y_n$ need not be linearly independent (and if they are, it's still not clear why it should be possible to extend $f$ continuously to all of $H$).

I don't know whether I have overlooked further errors in the proof.

Minor remark. There are a few further odd things about Theorem 1: The authors don't use that the underlying scalar field is complex. But over the real field the claim is wrong even in two dimensions. By the way, the statement of Theorem 1 is also wrong over the complex field if $H$ is one- or zero-dimensional.

Edit 2024-12-02. Today, a preprint appeared on arXiv which also discusses what is wrong with the proof. I have not read it in detail, but it also seems to focus on the functional $f$ not being well-defined.

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    $\begingroup$ Now, the article names Palle Jorgensen as its editor. What might be going on there? $\endgroup$
    – Jochen Glueck
    Commented Sep 4 at 6:54
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    $\begingroup$ I agree with you that $f$ may not be well-defined, but in terms of a conceptual problem, I don't think that the paper makes the claim you say, since if $T$ is an operator with no invariant subspaces that is enough to conclude that every orbit is dense, since if there was a non-dense orbit it would be an invariant subspace. $\endgroup$
    – euleroid
    Commented Sep 4 at 9:39
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    $\begingroup$ @euleroid: Thanks, you're right that proof in the paper, if it were correct, would actually show a slightly weaker claim than I initially wrote. I've edited my answer accordingly. Please note, though, that I'm not saying that the paper explicitly make this claim. I'm just saying that the proof would imply this claim if the proof were correct. Note that the proof does not really use that $T$ has no invariant subspace; it just uses that (1) is true for at least one vector $x$ and for the vector $Tx$. $\endgroup$
    – Jochen Glueck
    Commented Sep 4 at 10:13

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