$\textbf{The question is as follows:}$
Suppose the sequence of operators $A_n:\ell_2 \to \ell_2$ given by $$A_n x = (x_1, x_2, \ldots , x_n, 0 ,0, \ldots)$$ Prove that $A_n$ converges strongly to $I$ but not by norm.
Sorry I saw some where a solution for this question as follows:
Define the projection $P_n:X \to X$ for $X = \ell_2(\mathbb N)$ by $P_n(x_1, \ldots , x_n, x_{n+1}, x_{n+2}, \ldots )= (x_1, \ldots , x_n, 0, 0,\ldots )$. Then we have $\|P_n - P_m\| = 1$ for $m\neq n,$ so $(P_n)$ does not converges uniformly.
Nevertheless, if $x \in \ell_2(\mathbb N)$ is any fixed vector, we have $P_n x \to x$ as $n$ goes to $\infty$. Thus $P_n \to I$ strongly.
Everything is Okay with this proof else than $\|P_n - P_m\|=1$! I cannot understand how can it work?
Can someone please let me know how can it be?
Thanks!
Let $n>m$. Then if $x$ is the vector which is 1 everywhere except the $m+1$ place, $P_m x=0$ and $P_n x=x$ (can you see that?). Thus, $(P_n-P_m)(x)=x$ and by the definition of the norm operator $$\lVert P_n-P_m \rVert =\sup \frac{(P_n-P_m)(y)}{\lVert y\rVert}\geq \frac{(P_n-P_m)(x)}{\lVert x\rVert} =1$$