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



Self-Adjointness for Banach Spaces
Among the first hits in Google are 



> 
> [The theory of self-adjoint operators in banach spaces with a Hermitian form](http://link.springer.com/article/10.1007/BF01875301), V. A. Shtraus,
>  Siberian Mathematical Journal, 1978, Volume 19, Issue 3, pp 483-489
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and



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> [Self-adjoint operators on real Banach spaces](http://www.sciencedirect.com/science/article/pii/S0362546X13000035), 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).


