I'm reading Gelfand triple from Brezis' Functional Analysis.
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 $i^* (\varphi ) = \varphi \restriction V$ for all $\varphi \in H^*$. 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 $\ell = i^* \circ j \circ i$.
Then he gives an example to illustrate the last point, i.e.,
Let $$ H = \big \{ u = (u_n)_{n \ge 1} : \sum_n |u_n|^2 < \infty \big \}, $$ where $\langle u, v \rangle_H := \sum_n u_n v_n$. Let $$ V = \big \{ u = (u_n)_{n \ge 1} : \sum_n n^2 |u_n|^2 < \infty \big \}, $$ where $\langle u, v \rangle_V := \sum_n n^2 u_n v_n$. We identity $V^*$ with the space $$ V^* = \big \{ f = (f_n)_{n \ge 1} : \sum_n \frac{1}{n^2} |f_n|^2 < \infty \big \}, $$ which is bigger then $H$. The dual pairing $\langle \cdot, \cdot \rangle_{V^*, V}$ is given by $$ \langle f, v \rangle_{V^*, V} = \sum_n f_n v_n, $$ and the canonical isometric isomorphism $\ell :V \to V^*$ is given by $$ (\ell (v))_n = n^2 v_n \quad \forall v \in V. $$
Could you confirm if my below understanding is fine?
Clearly, $H^* = H$ with the dual pairing $\langle \cdot, \cdot \rangle_{H^*, H}$ given by $$ \langle u, v \rangle_{H^*, H} = \langle u, v \rangle_H, $$ and the canonical isometric isomorphism $j :H \to H^*$ given by $$ j (v)=v \quad \forall v \in H. $$
Now fix $v, t \in V$. Let $u:= \ell (v)$. Then $u_n = n^2 v_n$ for all $n \in \mathbb N^*$. We have $$ \langle u, t \rangle_{V^*, V} = \sum_n n^2 v_n t_n. $$
Let's determine $u' := i^* \circ j \circ i (v)$. We have $i (v) = v$ and thus $j \circ i (v) = v$. Then $u' = v \restriction V$. Hence $$ \langle u', t \rangle_{V^*, V} = \langle v, t \rangle_H = \sum_n v_n t_n. $$
It follows that $u \neq u'$.