I was looking for a example for sequence of bounded operators which converges strongly but does not converge in Norm. I've found the following example somewhere on internet, but unable to prove this.

Let $H$ be a separable Hilbert space and $(e_n)$ be ONB for $H$. Define $T_n \in B(H)$ as $T_n(x)= \langle x,e_n\rangle e_1$ Show that $T_n$ converges strongly but not in Norm.

I think $T_n$ converges to zero strongly. Is this correct? Please help.


1 Answer 1


Yes, $T_n$ converges strongly to zero by the Riemann-Lebesgue lemma: For any $x\in H$, $$ \lim_{n\to \infty} \langle x,e_n\rangle = 0 $$ which itself follows from Bessel's inequality $$ \sum_{n=1}^{\infty} |\langle x,e_n\rangle | ^2 \leq \|x\|^2 $$

Furthermore, $T_n$ does not converge in norm, because if $n\neq m$ $$ \|T_n - T_m\| \geq \|T_n(e_n) - T_m(e_m)\| = \|e_1\| = 1 $$


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