# Sequence is norm convergent implies it's strongly convergent

1. Nov 8, 2015

### Chain

If a sequence of operators $\{T_n\}$ converges in the norm operator topology then:
$$\forall \epsilon>0$$ $$\exists N_1 : \forall n>N_1$$ $$\implies \parallel T - T_n \parallel \le \epsilon$$
If the sequence converges in the strong operator topology then:
$$\forall \psi \in H$$ $$\forall \epsilon>0$$ $$\exists N_2 : \forall n>N_2$$ $$\implies \parallel T\psi - T_n\psi \parallel \le \epsilon$$
Where H is the Hilbert space that the operators act on. I believe that norm convergence implies strong convergence since for $n>N_1$:
$$\parallel T\psi - T_n\psi \parallel = \parallel(T-T_n)\psi \parallel \le \parallel T - T_n \parallel \parallel \psi \parallel \le \epsilon \parallel \psi \parallel$$
Since the magnitude of any vector in the Hilbert space must be finite we can scale $\epsilon$ so that the RHS of the inequality is arbitrarily close to zero hence we have convergence in the strong operator topology.

However in my functional analysis book (methods of mathematical physics by Simon and Reed) it says that the map taking an operator to its adjoint is continuous in the norm topology but not the strong topology. This means if a sequence converges to an operator T then the sequence obtained by taking the adjoint of every operator in the original sequence also converges to some operator. Hence we have a sequence converging in the norm topology but not in the strong topology.

I would be very grateful if someone could point out my mistake (or perhaps a mistake in Simon and Reed however this is not the only example of norm convergence does not imply strong convergence that I have seen).

2. Nov 8, 2015

### Krylov

"some operator" is $T^{\ast}$.
No, this is not true.
Continuity of $B(H) \ni T \mapsto T^{\ast} \in B(H)$ in the norm topology means: If $T_n \to T$ in $B(H)$ in the operator norm, then $T_n^{\ast} \to T^{\ast}$ in the operator norm. From this we can of course conclude that $T_n^{\ast} \to T^{\ast}$ strongly as well, but that's all. In particular, in this context there is no single sequence in $B(H)$ that converges in operator norm but not strongly. So, there is no contradiction to the implication "convergence in operator norm implies strong convergence".

3. Nov 8, 2015

### Chain

But then if what you say is true then surely the map $T \mapsto T^*$ is also continuous in the strong operator topology which contradicts what is says in Simon and Reed.

Thank you for the quick response!

EDIT: In Simon and Reed it says the map $T \mapsto T^*$ is continuous in the weak and norm topologies but is only continuous in the strong topology if the Hilbert space is finite dimensional.

4. Nov 8, 2015

### Krylov

I don't see the contradiction. I said that $B(H) \ni T \mapsto T^{\ast} \in B(H)$ is continuous with respect to the operator norm. I also agree with S&R that this map fails to be continuous with respect to the strong operator topology if $H$ is infinite dimensional.

Convergence of a sequence in $B(H)$ in the operator norm implies strong convergence (as you already demonstrated yourself) of that sequence to the same limit, but continuity of a map on $B(H)$ with respect to the operator norm does not imply continuity of that map with respect to the strong operator topology.

I get the feeling that you may be mixing up sequences and maps?
My pleasure. I will be off for a bit now, though, but am happy to continue later. (However, others will probably pick it up before me.)

5. Nov 8, 2015

### Chain

Continuous maps always map convergent sequences to convergent sequences. So if the map $T \mapsto T^*$ is strongly continuous then it should map any strongly convergent sequence to another strongly convergent sequence.

Now I see the issue, we know the map will take a norm convergent sequence in to a norm (and hence strongly) convergent sequence but this doesn't necessarily mean it takes every strongly convergent sequence into a strongly convergent sequence. It might only map the strongly convergent sequences which are also norm convergent in to strongly convergent sequences.

Thank you Krylov for clarifying this.