# How does the reflexivity of $V$ imply the denseness of $R(i^*)$?

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 $$i^* (\varphi ) = \varphi |_V \quad \forall \varphi \in H^*.$$

1. It follows from $$V$$ is dense in $$(H, \langle \cdot , \cdot \rangle_H)$$ that $$i^*$$ is injective.
2. It follows from $$i$$ is continuous that $$i^*$$ is continuous.
3. 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 $$R(i^*)$$?

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.