Self-Adjointness for Banach Spaces

Question

Good evening. Is there a reasonable notion of being self-adjoint for the adjoint operator on Banach Spaces? Kind regards, Alex

Answer 1

Among the first hits in Google are

The theory of self-adjoint operators in banach spaces with a Hermitian form, V. A. Shtraus, Siberian Mathematical Journal, 1978, Volume 19, Issue 3, pp 483-489

and

Self-adjoint operators on real Banach spaces, Paweł Wójcik, Nonlinear Analysis: Theory, Methods & Applications, 2013, Volume 81, pp 54–61.

The former uses continuous and Hermitian bilinear forms $Q:X\times X\to\mathbb{R}$ and calls an operator $A$ on $X$ $Q$-self-adjoint if $Q(Ax,y) = Q(x,Ay)$.

In a similar spirit I could imagine that the following construction could make sense:If you have a uniformly smooth Banach space $X$ you could use the duality mapping $J:X\to X^*$ defined by $J(x) = \{x^*\in X^*\ :\ \langle x^*,x\rangle_{X^*\times X}=\|x\|^2=\|x^*\|^2\}$ (which is single valued and uniformly continuous in this case) to define $Q(x,y) = \langle J(x),y\rangle_{X^*\times X}$ (however, not bilinear). Then it seems natural to require that $Q(Ax,y) = Q(x,Ay)$ (however, I haven't seen this construction used anywhere).

Answer 2

There are ways to define hermitian operators in complex Banach spaces without the latter having additional structure and these were studied in some detail in the 70's. One way is to demand that the numerical range be real, another that the semi-group of operators generated by $iT$ consist of isometries. A chapter of Dowson's book on spectral operators is devoted to them and this is probably the best entrance point to this topic and its relevance in spectral theory.