For a linear map $T$, we denote by $R(T)$ its range and by $N(T)$ its kernel. I'm reading about Gelfand triple at page 136 in Brezis' Functional Analysis.
Let $(H, \langle \cdot , \cdot \rangle_H)$ be a real Hilbert space and $|\cdot|$ its induced norm. Let $V$ be a dense linear subspace of $(H, \langle \cdot , \cdot \rangle_H)$. Assume that the vector space $V$ has its own norm $[ \cdot ]$ such that $(V, [\cdot])$ is a Banach space. We assume that the inclusion map $i: (V, [\cdot]) \to (H, |\cdot|), v \mapsto v$ is continuous. Let $(V^*, [\![ \cdot ]\!])$ be the dual space of $(V, [\cdot])$. Let $(H^*, \|\cdot\|)$ be the dual space of $(H, |\cdot|)$. Let $i^*:(H^*, \|\cdot\|) \to (V^*, [\![ \cdot ]\!])$ be the adjoint operator of $i$. Then $$ i^* (\varphi ) = \varphi |_V \quad \forall \varphi \in H^*. $$
- It follows from $V$ is dense in $(H, \langle \cdot , \cdot \rangle_H)$ that $i^*$ is injective.
- It follows from $i$ is continuous that $i^*$ is continuous.
- If $(V, [\cdot])$ is reflexive, then $R(i^*)$ is dense in $(V^*, [\![ \cdot ]\!])$.
Could you explain how the reflexivity of $V$ implies the denseness of $R(i^*)$?
It is not as complicated as you may think.
Let $T \in V^{**}$ such that $T=0$ on $R(i^*)$. It suffices to prove that $T \equiv 0$. Let $J:V \to V^{**}$ be the canonical evaluation map. Because $V$ is reflexive, there is $v \in V$ such that $J(v)=T$. This means $\langle T, \varphi \rangle = \langle \varphi, v \rangle$ for all $\varphi \in V^*$. We have $\langle T, \varphi \rangle=0$ for all $\varphi \in R(i^*)$, so $$ \langle \varphi, v \rangle=0 \quad \forall \varphi \in H^*. $$
It follows that $v=0$. This completes the proof.