There have been several questions asked on various aspects of Gelfand triples. However, I have not yet found an answer to the following question:
Let $V$ be a Banach space, $H$ be a Hilbert space such that we have a dense embedding $$ i :V \hookrightarrow H. $$ We then consider the "adjoint map" $$ i^{*} : H^{*} \rightarrow V^{*} $$ which is defined via $$ _{V^{*}}\langle i^{*}(\phi), v \rangle_{V} :=~ _{H^{*}} \langle \phi, i(v) \rangle_{H} $$ (intuitively, if $V \subset H$ as sets, one can think of the action of $i^{*}$ as being the restriction of a functional on $H$ to a functional on $V$).
We would like this map $i^{*}$ to have three properties:
- continuous
- injective
- with dense image, i.e. $i^{*}(H^{*}) \subset V^{*}$ is a dense subset w.r.t. the norm on $V^{*}$.
If we have established this, we can proceed to find the Gelfand triple $$ V \hookrightarrow H \equiv H^{*} \hookrightarrow V^{*} $$ with dense embeddings.
The first property is easy to check, the second follows from the density of $i$. However, for the third property, one needs (or rather: seems to need) something more. Brezis, in his functional analysis book, for example, assumes reflexivity (Rem. 5.2.3 on p. 136 in the 2011 edition) and so do Liu/Röckner (SPDEs: An Introduction, Ch. 4.1, p. 69), Zeidler (Linear Monotone Operators, Ch. 23.4, p. 416) and all other books I could get my hands on.
As I will show below, reflexivity is sufficient indeed, but here is my question: Is reflexivity also necessary for property 3?
That it works if $V$ is assumed to be reflexive can be seen as follows: Let $c$ denote the canonical map $$ c : V \rightarrow V^{**}, ~ v \mapsto \psi_{v}, ~ \psi_{v}(v^{*}) := v^{*}(v), $$ i.e. $c(v)(v^{*}) := v^{*}(v)$. If $V$ is reflexive, $c$ is surjective. Now to prove the density of $i^{*}(H^{*}) \subset V^{*}$ we need to check that if a continuous functional $\psi$ on $V^{*}$, i.e. $\psi \in V^{**}$, is zero on the image, then it must already be the zero functional, i.e. $\psi = 0$. So the condition we need to check is $$ _{V^{**}} \langle \psi, i^{*}(\phi) \rangle_{V^{*}} = 0 \quad \forall \phi \in H^{*} \quad \Rightarrow \quad \psi = 0. $$ By reflexivity, $\psi = c(v)$ for a $v \in V$. So let $\phi \in H^{*} $ be arbitrary. By the definition of $c$ and $i^{*}$ we get \begin{align*} 0 &= ~_{V^{**}} \langle \psi, i^{*}(\phi) \rangle_{V^{*}} = ~_{V^{**}} \langle c(v), i^{*}(\phi) \rangle_{V^{*}} = ~_{V^{*}} \langle i^{*}(\phi), v \rangle_{V} \\ &= ~_{H^{*}} \langle \phi, i(v) \rangle_{H}. \end{align*} Since this holds true for every $\phi \in H^{*}$, by Hahn-Banach (or rather, since $H^{*}$ separates points on $H$), this implies $i(v) = 0$ which by the injectivity of $i$ yields $v = 0$ and hence by linearity of $c$ that $\psi = c(v) = 0$.
