Convergence of Operators

Definition 1.18: Let $T_n,T \in \mathcal{B}(\mathcal{H})$. We say that:

• $T_n \rightarrow T$ in the operator norm, (or uniformly) if $\|T_n - T\| \rightarrow 0$ as $n \rightarrow \infty$;
• $T_n \rightarrow T$ strongly, if for any $x \in \mathcal{H}$ one has $\|T_nx - Tx\| \rightarrow 0$ as $n \rightarrow \infty$;
• $T_n \rightarrow T$ weakly, if for any $x,y \in \mathcal{H}$ one has $(T_nx,y) \rightarrow (Tx,y)$ as $n \rightarrow \infty$.

Theorem 1.19: Uniform convergence implies strong convergence, and strong convergence implies weak convergence.

Proof: Exercise. $\blacksquare$

The following examples show the implication does not go in the other direction.

Example 1.20: Consider operators on $\ell^2$.

• Let $T_n = \frac{1}{n}I$. Then $T_n \rightarrow 0$ uniformly.
• Consider the powers of the backwards shift operator on $\ell^2$,
  $$(S^*)^n (x_0,x_1,...) = (x_n,x_{n+1},...).$$
  Then $(S^*)^n \rightarrow 0$ strongly, but not uniformly.
• Consider the powers of the shift operator in $\ell^2$, i.e.:
  $$S^n(x_0,x_1,...)=(\overbrace{0,0,...,0}^n,x_0,x_1,...).$$
  Then $S^n \rightarrow 0$ weakly but not strongly.