Let $$ V = \big \{ v = (v_n)_{n \ge 1} : \sum_n n^2 |v_n|^2 < \infty \big \} $$ be equipped with an inner product $\langle u, v \rangle_V := \sum_n n^2 u_n v_n$. Then $(V, \langle \cdot, \cdot \rangle_V)$ is a Hilbert space. Consider the vector space $$ U = \big \{ f = (f_n)_{n \ge 1} : \sum_n \frac{1}{n^2} |f_n|^2 < \infty \big \} $$ with the dual pairing $\langle \cdot, \cdot \rangle_{U, V}$ given by $$ f (v) :=\langle f, v \rangle_{U, V} := \sum_n f_n v_n. $$
I would like to verify that
Theorem $U=V^*$.
Could you have a check on my below attempt?
Let $f \in U$ and $v \in V$. Let $|\cdot|$ be the induced norm of $\langle \cdot, \cdot \rangle_V$. By Cauchy–Schwarz inequality, $$ f(v) = \sum_n (f_n/n) (v_n/n) \le \sqrt{\sum_n \frac{1}{n^2} |f_n|^2} |v|. $$
Because $\sum_n \frac{1}{n^2} |f_n|^2 < \infty$, we get $U \subset V^*$. Let's prove the reverse, i.e., $V^* \subset U$. Let $i:V \to V^*$ be the canonical isometric isomorphism. Fix $u \in V$. Then $$ \langle i (u), v \rangle_{V^*, V} = \langle u, v \rangle_V = \sum_n (n^2u_n) v_n \quad \forall v \in V. $$
Notice that $u \in V$ implies $(n^2 u_n)_{n \ge 1} \in U$. It follows that $i(u) \in U$ and thus $V^* \subset U$.