Let us consider the space $L^2:=L^2(\mathbb{R}^n,\mathbb{C})$ and the associated scalar product $S(f,g):=\int f \overline g$. In distribution theory, we have a situation where we have to deal with two different identifications (which makes things a little bit tricky) : 


* if we identify $L^2$ with its antidual, an element $f\in L^2$ is an antilinear form $g\mapsto S(f,g)$
* if we identify $L^2$ as a distribution space, an element $f\in L^2$ is a linear form $g\mapsto \int f g $.


I am aware of the following facts : 


* we can anti-identify (through a anti-isomorphism) $L^2$ to its dual (as an Hilbert space can always be), but this just shifts the problem, since the first identification becomes the map $f\mapsto (g\mapsto \int \overline f g)$.
* We can **define** distributions as antilinear maps (and not linear), and define the identification of a function as a distribution by : $f\mapsto \int f \overline g$. This way, the two identifications coincide, and there is no need to keep track of the two different identifications made.


These observations made me think that maybe the discrepancy can be solved in another way, where instead of redefining distributions, we redefine (or more precisely refine) the abstract definition of Hilbert spaces.


More precisely, let us say that a (complex) **Hilbert space with real structure** is a triple $(H,b,C)$ where $C$ is a antilinear involution on $H$, and $b$ is a bilinear map such that 
$\langle \cdot,\cdot\rangle = b(\cdot,C(\cdot))$ is a scalar product on $H$ giving $H$ the structure of a Hilbert space. 
With this definition it seems that a Hilbert space with real structure is naturally isomorphic (through a bijective **linear** isometry, namely $f\mapsto b(f,\cdot)$) to its **dual** (and not its antidual), which is indeed the natural situation we have in the case of $L^2$ (where $C:f\mapsto \overline f$ and $b(f,g):=\int f g$). 


**My question** is then : why do we study (complex) Hilbert spaces in general and not (complex) Hilbert spaces with real structure (as previously defined) ? Are there Hilbert spaces that appear naturally *without* an obvious real structure ? 


Or again, formulated differently : if we formulated the whole theory of Hilbert space with the definition given before (with a real structure), would we gain anything in applications to abstract it further to the case of Hilbert space without real structure ?



On the definition of Hilbert spaces and real structures on Hilbert spaces
Good question. It does seem like most naturally occurring complex Hilbert spaces come equipped with a natural real structure. Incidentally, I pointed out in [this paper](https://arxiv.org/abs/math/9908119) that the same is true of Hilbert modules over C\*-algebras, and in this case you can use the natural involution to turn a left action into a right action and vice versa. So many naturally occurring Hilbert modules are automatically Hilbert bimodules.


The first example that comes to mind of a Hilbert space without an obviously natural real structure is the [Bargmann-Segal space](https://en.wikipedia.org/wiki/Segal%E2%80%93Bargmann_space) of entire analytic functions on $\mathbb{C}^n$ whose integral against $e^{-|z|^2}$ is finite. I guess there is a natural isomorphism of this space with $L^2(\mathbb{R}^n)$, and you can then import the real structure of the latter, so maybe it's not a completely convincing example.


Edit: better example, the [Bergman space](https://en.wikipedia.org/wiki/Bergman_space) of holomorphic functions on a domain in $\mathbb{C}$ which are square integrable against Lebesgue measure. I don't see any obvious real structure for a general domain.


