Limit, supremum and boundedness

Question

Let $a_n$ be a sequence that converges in norm to $a$ and let $S\neq \emptyset$ be such that for all $n\ge 0$ we have $\sup_{s\in S} \langle s, a_n \rangle $ is finite. I am wondering if we can conclude that $\sup_{s\in S} \langle s, a \rangle $ is finite.

Answer 1

D - I think that's not the true.

Let H = $l^2(\mathbb{N})$ (Hilbert) and $a_n := \Sigma_{i=1}^n \frac{1}{i} \delta_i$ (where $\delta_i$ is the standard base), $a_n \to a$ with $a = \Sigma_{i=1}^{\infty} \frac{1}{i}\delta_i \in l^2(\mathbb{N})$. Let $S = \{x \in l^2: x = n^3 \delta_n\}$. We have, because $(a_n)_j = 0$ for $j>n$ $$ sup_{s \in S}\langle s,a_n \rangle < \infty $$

and $$ sup_{s \in S}\langle s,a \rangle = \infty $$