Does Convergence of Maps Evaluated at Points Imply Convergence in Operator Norm?

Question

Suppose that I have $T,T_n \in B_H$, for some Hilbert space $H$. Is the following implication true?

$$ \|(T-T_n)x\| \rightarrow 0 \ \forall x\in H \ \Rightarrow \ \|T-T_n\| \rightarrow 0, \ \text{ie} \ T_n \rightarrow T.$$

Is it valid if and only if the convergence to zero if uniform? I think I have a proof in that case, but I'm not sure:

For an operator $A$, $$\|A\| = \sup\lbrace {\|Ax\| \over \|x\|} \ | \ x \in H \rbrace = \sup\lbrace \|Ax\| \ | \ x \in B_H \rbrace \\ \Rightarrow \|T - T_n\| = \sup\lbrace \|(T-T_n)x\| \ | \ x \in B_H \rbrace \rightarrow 0,$$ since $\|(T-T_n)x\| \rightarrow 0$ for all $x \in H.$

Without uniform convergence, then surely we can't make the final claim, but with uniform convergence that it does seem legitimate to me.

Any help would be most appreciated - I've been stuck on this for ages!

Answer 1

One should always mention the set on which uniform convergence is considered. The convergence cannot be uniform on all of $H$ unless $T_n=T$ for all large $n$.

If your assumption is that convergence is uniform on the unit ball, the proof is correct.

Or you could adopt a point of view that makes the statement tautological: the operator norm is the $C(B_H)$ norm (uniform norm on the space of continuous functions) and convergence in the uniform norm is precisely the uniform convergence.

The rest has been answered by Jose27 already:

Pointwise convergence is called Strong (operator) convergence in this setting. To see that this doesn't imply norm convergence consider the shift operator $S$ in $\ell^2$ given by $$(x_1,x_2,\ldots )\mapsto (x_2, x_3\ldots).$$ And define $T_n=S^n$ (composition). -- Jose27