# Is the spectrum of a "self adjoint" operator real on $\ell^p$?

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 ?

• Do you know whether the point spectrum is always real? Apr 1, 2020 at 22:49
• 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). Apr 1, 2020 at 23:36
• @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 Apr 2, 2020 at 3:38
• 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. Apr 2, 2020 at 7:57
• 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 $\ell_p \subset \ell_2$). Apr 2, 2020 at 13:36

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 $$\{ \pi \sec(\pi u ) : | \Re ( u )| < 1/2-1/p \}.$$
• 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 $\sigma^p(A) \subset \sigma^2(A) \subset \mathbb{R}$. Apr 6, 2020 at 15:53