# Real symmetric operators in $\ell^p$, for $p\neq 2$.

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 $p\neq 2$?

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