I'm reading about Gelfand triple from Wikipedia page and from Brezis' Functional Analysis. Could you confirm if my below understanding is fine?
Let $(H, \langle \cdot , \cdot \rangle_H)$ be a 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 (linear) 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 $$ i^* (\varphi ) = \varphi \restriction V \quad \forall \varphi \in H^*. $$
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 $\operatorname{im } i^*$ is dense in $(V^*, [\![ \cdot ]\!])$. By Riesz representation theorem, there is a canonical isometric isomorphism $j:H \to H^*$. Then we write $$ V \overset{i}{\hookrightarrow} H \overset{j}{\cong} H^* \overset{i^*}{\hookrightarrow} V^*. $$
The inner product $\langle \cdot , \cdot \rangle_H$ is compatible with the dual pairing $\langle \cdot, \cdot \rangle_{V^*, V}$, i.e., $$ \langle u, v \rangle_H = \langle i^* \circ j (u), v \rangle_{V^*, V} \quad \forall u \in H, v \in V. $$
Assume now that the norm $[\cdot]$ of $V$ is induced by an inner product $\langle \cdot, \cdot \rangle_V$. By Riesz representation theorem, there is a canonical isometric isomorphism $\ell:V \to V^*$. It is generally not true that $$ \ell = i^* \circ j \circ i. $$