Strong convergence + convergence of norms -> norm convergence?

Question

Let $\{A_n\}$ and $A$ be boundend operators on a Hilbert space $H$, such that

(i) $A_n f \to A f$ for any $f\in H$;

(ii) $||A_n|| \to ||A||$.

Is that true that $A_n \to A$ in $\mathcal{L}(H)$?

Answer 1

No. Say $H=L^2(\Bbb R)$. Define $A_nf=\chi_{E_n}f$, where $E_n=[1/n,1]$ and $Af=\chi_Ef$ where $E=[0,1]$. Then $A_nf\to Af$ in norm for every $f$, $||A_n||=||A||=1$, but $||A-A_n||=1$.