If $H$ is an Hilbert space with inner product denoted by $(\cdot , \cdot )$, then $$||u||=\sup \{(u,v):\:||v|| \leq 1 \}$$
Question: In the case that $H$ is separable, is there any countable family $\{ w_n\} \subset H$ such that $||u||=\sup_{n \in \mathbb N} \, (u,w_n)$ ?
I know that in this case $H$ admists an orthonormal basis and the Parseval identity holds, but I don't know how to use it
The hints received in comments were useful (thank you)
The separable Hilbert space $H$ admits a countable orthonormal basis $\{e_i\}$ and a dense subspace is given by the set $E$ of finite linear combinations with rational coefficients of the element of the basis. Denote by $ E_1=\{w_k:k \in \mathbb N\}$ the set of elements of $E$ with norm at most $1$.
Fix $u \in H$. I want to prove that
$$\sup\{(u,v):\,v \in H, ||v||\leq 1 \}=\sup_{k \in \mathbb N}\,(u,w_k)$$
and since one inequality is trivial I only need to prove this "$\le$"
For any $v \in H$ there exists a sequence $(v_n)$ in $E$ such that $v_n \to v$ in $H$. If $||v|| \leq 1$, then $v_n \in E_1$ for $n >>1$. Given $\epsilon >0$ arbitrary, choose $m >>1$ such that $||v-v_m||\leq \epsilon/||u||$. One has
$$(u,v_m)=(u,v)-(u,v-v_m)\geq (u,v)-\epsilon$$
thus $\sup_{k \in \mathbb N}\,(u,w_k) \geq (u,v)- \epsilon$. Taking the supremum over $v \in H$ with $||v||\leq 1$ and letting $ \epsilon \to 0^+$, the claim follows.