In what way is convergence in operator norm ‘better’ than other forms of convergence

Question

If we show a sequence of operators $T_n$ converges uniformly/in operator norm to some operator $T$ this implies the sequence converges strongly/pointwise, which in term means it converges weakly.

I know in some cases we can only show weak convergence so we have to be content with that. So what have we 'lost' when we can only show weak convergence? In contrast, what do we gain by having convergence in operator norm as opposed to the other forms of convergence? Actually what do we gain as we move up the hierarchy of convergences:

1. Weakly
2. Strongly
3. Uniformly

I am interested in what we gain in a mathematical sense, and also how how it relates to applications in physics or numerical analysis.

Answer 1

Here is one example. If a sequence of unitary operators converges in norm, the limit is a unitary operator. Same is true if convergence is strong. But if they converge weakly, all you can say about the limit is that it is a contraction; more specifically, the weak closure of the set of unitaries is the whole unit ball. The same is true for the set of projections.

On the other hand, weak and strong closures agree on convex sets, so they produce the same selfadjoint algebras (i.e., von Neumann algebras).