Limit of the scalar product with a sequence of linear operators

Question

Let $\ell^2$ over complex plane $\mathbb{C}$

Let $\{e_m\}_{m \in \mathbb{N}}$ be the canonical basis of $\ell^2$

Let $\{a_m\}_{m \in \mathbb{N}}$ be a sequence of positive real number such that $a_m \to +\infty$

Let $\{P_m\}_{m \in \mathbb{N}}$ be a sequence of linear operators $P_m : \ell^2 \to \ell^2$ such that $P_m$ converges strongly to $I$ ($I$ is the identity operator)

Let $v \in \ell^2$ such that $v \neq 0$ and $$ \lim_{m \to \infty} \langle a_m \cdot e_m , v \rangle = +\infty $$ My question is if $$ \lim_{m \to \infty} \langle a_m \cdot e_m , P_m(v) \rangle = +\infty $$ and if it is true, I would like a proof. Thanks.

Answer 1

Consider the operators $P_m : \ell^2 \to \ell^2$ defined by setting for $x = (x_n)_{n \in \mathbb{N}}$, $$(P_m x)_{n} = \begin{cases} x_n, \qquad n < m \\ 0, \qquad n \ge m \end{cases}$$

It is easy to check that $P_m$ converges strongly to the identity as $m \to \infty$ since $$\|P_m x - x\|_{\ell^2}^2 = \sum_{n \ge m} |x_n|^2.$$

However, $\langle e_m, P_m v \rangle = 0$ for every $m$.