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Consider the spaces $\ell^p$, for $1 \leq p \leq \infty$. Suppose we have a bounded operator $S: \ell^\infty \to \ell^\infty$ such that $S(\ell^p) \subseteq \ell^p$ for every $p \geq 1$ and such that $S$ restricted to $\ell^p$ is also a bounded operator into $\ell^p$.

Furthermore, assume that $S$ is symmetric in the sense that $$ \langle S f, g \rangle =\langle f, S g \rangle $$ for every $f \in \ell^p$ and every $g \in \ell^q$, for $1\leq p < \infty$, where $\langle \cdot, \cdot \rangle$ denotes the usual pairing between $\ell^p$ and $\ell^q$.

Lastly, assume $\overline{\big(Sf\big)}=S\big(\overline{f}\big)$, where $\overline{g}$ denotes the complex conjugate of the sequence $g$.

Question: have these type of operators been studied? In particular, what is known about their spectrum? For example: if $p=2$ is not hard to be convinced that $S$ is self-adjoint and hence has real spectrum. Is the same true for $p\neq 2$?

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  • $\begingroup$ Is $S$ assumed linear? $\endgroup$ – C.Ding Jul 4 '17 at 5:35
  • $\begingroup$ You'd better ask this question on mathoverflow. $\endgroup$ – Norbert Jul 4 '17 at 10:36
  • $\begingroup$ @C.Ding Yes. $S$ is linear. $\endgroup$ – Ruben Jul 5 '17 at 7:05
  • $\begingroup$ @Norbert Thanks. I'll do that. $\endgroup$ – Ruben Jul 5 '17 at 7:05
  • $\begingroup$ Is this all happening on $\mathbb{R}^n$? $\endgroup$ – mathworker21 Jul 9 '17 at 9:50

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