Let $X,Y$ be infinite dimensional Hilbert spaces, $A_n:X\to Y$, a sequence of compact operators converging strongly to $0$, that is $A_nx\to 0,\forall x\in X$. If $x_n$ and $z_n$ are bounded sequences in $X$, can we say that $\langle A_nx_n,z_n\rangle \to 0$?
Does the following inner product converge?
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functional-analysis
operator-theory
1 Answers
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No. Consider consider $X=Y=\ell^2(\Bbb N)$ and let $A_n$ be the projection onto $e_n$. The $A_n$ are compact and converge strongly to zero, however with $x_n=z_n=e_n$ you've got $$\langle A_nx_n,z_n\rangle=\langle e_n,e_n\rangle=1$$ for all $n$.
Indeed, if $A_n\not\to0$ in norm topology then you will always find sequences $x_n,y_n$ so that $\langle A_nx_n,z_n\rangle$ does not converge to zero. For if $A_n\not\to0$ in norm you've always got a subsequence $A_{n_k}$ so that $\|A_{n_k}\|≥C$ for some $C>0$. Then for such $C$ you can find a non-zero $x_{n_k}\in \ell^2$ so that $\|A_{n_k}x_{n_k}\|≥C\|x_{n_k}\|$. Let $z_{n_k}=A_{n_k}x_{n_k}$ then $$\left\langle A_{n_k}\frac{x_{n_k}}{\|x_{n_k}\|},\frac{z_{n_k}}{\|z_{n_k}\|}\right\rangle=\left\|A_{n_k}\frac{x_{n_k}}{\|x_{n_k}\|}\right\|≥C.$$