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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.

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.

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 $x \in H$, the span of orbit $\{T^nx : n \in \mathbb{N}_0\}$ is not dense (since the choice of $x$ is arbitrary in the proof). This is obviously wrong (consider the right shift on $\ell^2$ and $x$ the first canonical unit vector).

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.

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.

No, it's wrong.

Conceptual reason why it can't be true: 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 $x \in H$, the span of orbit $\{T^nx : n \in \mathbb{N}_0\}$ is not dense (since the choice of $x$ is arbitrary in the proof). This is obviously wrong (consider the right shift on $\ell^2$ and $x$ the first canonical unit vector).

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, they might still not form a Schauder basis, so it's 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.

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 $x \in H$, the span of orbit $\{T^nx : n \in \mathbb{N}_0\}$ is not dense (since the choice of $x$ is arbitrary in the proof). This is obviously wrong (consider the right shift on $\ell^2$ and $x$ the first canonical unit vector).

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.