After reviewing the (locally convex) topological vector spaces that I know, the only examples I could find where there is an isomorphism from the space to its (anti)dual, are Hilbert spaces. So my question is :
Are there topological vector spaces $V$ such that the topology does not come from a Hilbert structure, and such that there exists an isomorphism $\chi : V \to V'$, where $V'$ denotes the antidual of $V$ (continuous antilinear forms on $V$) ?
An interesting family of examples comes from number theory (or algebraic geometry, depending on who you ask): If you have a field $k$, the Laurent power series field $k((t))$ has an ultrametric topology where $\{ t^n k[[t]] \}_{n \in \mathbb{Z}}$ form a neighborhood basis of zero. This space is isomorphic to its topological dual: by adjoining $(dt)^{1/2}$, one obtains the perfect residue pairing $$\langle f(t) (dt)^{1/2}, g(t) (dt)^{1/2} \rangle = \operatorname{Res} f(t)g(t) dt.$$ You can do a similar trick with finite dimensional vector spaces over $k((t))$.
If you want nontrivial antilinearity, you may choose $k$ to be a separable quadratic extension of some underfield $F$, and change the residue pairing to be sesquilinear: $$\langle f(t) (dt)^{1/2}, g(t) (dt)^{1/2} \rangle = \operatorname{Res} f(t) \bar{g}(t) dt.$$
`If $X$ is any reflexive space, then $X \oplus X^{*}$ is isomorphic to its dual $X^* \oplus X^{**}$.
Take a reflexive TVS $V$, and consider $V \times V^\ast$.
