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There might be an obvious answer to the question, but it doesn't come to mind.

Suppose we have an infinite matrix $A=(a_{ij})$, which defines a bounded linear operator on $\ell^p$, i.e. for all sequences $(x_i)\in \ell^p, p>1$ $$ \sum_{i=1}^\infty\big|\sum_{j=1}^\infty a_{ij}x_j\Big|^p \leq C \Vert x \Vert_{p}^p.$$ For some positive constant $C$. Furthermore, assume that $a_{ij}=\overline{a_{ji}}$. Is it true that the spectrum of $A$ on $\ell^p $ is real ?

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    $\begingroup$ Do you know whether the point spectrum is always real? $\endgroup$
    – Jochen Glueck
    Commented Apr 1, 2020 at 22:49
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    $\begingroup$ Yes, Jochen, because the hypothesis implies that $A$ also is bounded on $\ell_q$, $1/p+1/q=1$, hence all eigenvectors are in $\ell_2$. Basically the same argument gives a positive answer to the OP's question (I think). $\endgroup$
    – Bill Johnson
    Commented Apr 1, 2020 at 23:36
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    $\begingroup$ @BillJohnson Your argument doesn't seem to use the fact $p>1$; but there are known examples of "self-adjoint" (i.e. conjugate symmetric) convolution operators on $l^1$(free group) whose spectrum has non-empty interior $\endgroup$
    – Yemon Choi
    Commented Apr 2, 2020 at 3:38
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    $\begingroup$ There are self-adjoint operators in $L^2$, generating (analytic) semigroups in $L^p$ for all $p$, such that the spectrum is not real for $p \neq 2$. One can take $L=r^2D_{rr}+2rD_r$ in the half-line. The spectrum is a parabola which degenerates into the negative half-line when $p=2$. The resolvents and the semigroups have similar bad properties, by the spectral mapping theorem. Maybe a discrete version can be contructed using them. $\endgroup$
    – Giorgio Metafune
    Commented Apr 2, 2020 at 7:57
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    $\begingroup$ I guess the only correct part of my comment above is that the point spectrum is real when $p<2$ (because the operator is bounded on $\ell_2$ and $\ell_p \subset \ell_2$). $\endgroup$
    – Bill Johnson
    Commented Apr 2, 2020 at 13:36

1 Answer 1

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It seems that I have found a counter example myself.

For the Hilbert matrix $$ H_\lambda:= \big( \frac{1}{1-\lambda+k+n} \big)_{k,n\geq 0}, \lambda < 1 $$ Rosenblum in "On the Hilbert Matrix I, Proceedings of the AMS" proves that the pointspectrum considered as an operator on $\ell^p, p>2$ contains the set $$ \{ \pi \sec(\pi u ) : | \Re ( u )| < 1/2-1/p \}. $$

If one could provide a more elementary counterexample I would be interested in looking at.

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  • $\begingroup$ However, I think it is true for $1 < p \le 2$ because, for $q \ge p$, if we denote $\sigma^q(A)$ the spectrum of the unbounded operator $A$ on $\ell^q$ with domain $\{x \in \ell^q \mid Ax \in \ell^q\}$ then we can show that the spectrum is increasing with $q$ meaning that $\sigma^q(A) \subset \sigma^r(A)$ for $p \le q \le r$. If I am not mistaken, for $1 < p \le 2$, the operator $A$ induce a self-adjoint operator on $\ell^2$ so $\sigma^p(A) \subset \sigma^2(A) \subset \mathbb{R}$. $\endgroup$
    – Zoïs Moitier
    Commented Apr 6, 2020 at 15:53
  • $\begingroup$ If you can provide the missing details please go ahead and post it as an answer, I will accept it. $\endgroup$
    – an_ordinary_mathematician
    Commented Apr 6, 2020 at 16:29

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