Can we prove the weak convergence of sequence bounded by another weakly convergence sequence?

Question

A series $a_n$ in some Hilbert space $H$ weakly converges to zero (meaning $f(a_n) \to 0$ $\forall f \in H^{'}$, the dual space of $H$). It is also given that for another sequence $b_n$, norm of each element is bounded like $\|b_n\| \leq C \|a_n\|$ for some $C>0$. Now can we conclude that $b_n$ converges weakly to zero as well ?

Answer 1

No, take any sequence $\left\{a_n\right\}$ such that $a_n\to 0$ weakly and $\|a_n\|=1$ for all $n$. Then consider the sequence $\left\{b_n\right\}$ with $b_n=b$ for all $n$, where $b\neq 0$. Then $$\|b_n\|=\|b\|=\|b\|\|a_n\| $$ for all $n$, but clearly $\left\{b_n\right\}$ does not converge weakly to zero.

Answer 2

No. Suppose $(a_k)_k$ is an orthonormal system, then for any $x\in H$ we have $\sum_k \vert \langle x , a_k \rangle \vert^2 \le \Vert x \Vert^2 $ (Bessel's inequality), hence $\lim_{k\rightarrow \infty}\langle x , a_k \rangle =0$. By the Riesz representation theorem every functional is of the form $f = \langle x, \cdot \rangle$, hence $(a_k)_k$ automatically weakly converges to $0$.

On the other hand, the inequality $\Vert b_n \Vert\le 1$ does not imply that $(b_k)_k$ converges.

Answer 3

Consider the $\mathbb R$ as a Hilbert space. Does convergence of $\{b_n\}$ together with $|a_n| \leq C|b_n|$ imply convegence of $\{a_n\}$? Hint: $a_n=(-1)^{n}$.