1

This is problem 3.32 in book Hilbert Space Operators in Quantum Physics.

Let $H$ be Hilbert space, $B_n$ converges strongly to $B$ in $\mathcal{B}(H)$ (linear bounded operator on $H$), this means $||B_n u -Bu||\to 0$ as $n\to\infty$ for every $u\in H$. Let $C$ be a compact operator, prove that $||B_n C - BC||\to 0$ as $n\to \infty$.

I am not sure if what I think is true: I consider an element $u\in H$ such that $||u||\leq 1$, then $Cu$ is bounded in $H$ because $C$ is compact, this implies that $||B_n Cu -BCu||\to 0$ as $n\to\infty$ for all $u\in H$, $||u||\leq 1$. Thus $$||B_n C -BC||=\sup_{u\in H, ||u||=1}||B_n Cu -BCu||\to 0.$$

Could anyone help me to check if this is true or it's wrong somewhere, please?

Davide Giraudo
  • 181,608
Muniain
  • 1,543

1 Answers1

1

If we manage to show that $\sup_n\left\lVert B_n\right\rVert$ is finite, then we can follow the lines of this thread. To this aim, define $$ F_N:=\bigcap_{n\geqslant 1}\left\{u\in H\mid \left\lVert B_nu\right\rVert\leqslant N\right\} $$ and use Baire's theorem.

Davide Giraudo
  • 181,608