Convergence in operator norm equivalent to uniform convergence on bounded sets

Question

Let $E$ be a Banach space and let $\{T_n\}_{n=1}^{\infty}$ be a sequence of bounded linear operators on $E$. I am trying to prove the following claim:

Claim: $T_n \rightarrow T$ in operator norm (i.e. $\lim\limits_{n \to \infty} \|T_n - T\| = 0)$ if and only if $T_n \rightarrow T$ uniformly on every bounded subset of $E$ (i.e. $\lim\limits_{n \to \infty} \sup\limits_{x \in S} \|T_n(x) - T(x) \| = 0$ for every bounded subset $S \subset E$).

I am having a little trouble with the "only if" part of the proof. Here is what I have so far:

Proof attempt:
($\implies$) Suppose that $T_n \rightarrow T$ in the operator norm. Let $S \subset E$ be a bounded set and let $x_0 \in E$. Pick $r > 0$ large enough so that $S$ is contained in the closed ball $B_r(x_0) := \{x \in E: \|x-x_0\| \leq r \}$. Since the map $x \mapsto \|x\|$ is continuous and $B_r(x_0)$ is compact, there exists an $M > 0$ such that $\|x\| \leq M$ for all $x \in B_{r}(x_0)$. We then have $$ \|T_n(x) - T(x)\| \leq \|T_n - T \| \|x\| \leq \|T_n - T\| M \quad \forall x \in B_{r}(x_0). $$

Thus, $$\sup_{x \in S} \|T_n(x) - T(x)\| \leq \sup_{x \in B_{r}(x_0)} \|T_n(x) - T(x)\| \leq \|T_n - T\| M. $$ And since $\|T_n - T\| M \rightarrow 0$ as $n \to \infty$, it follows that $\lim\limits_{n \to \infty} \sup\limits_{x \in S} \|T_n(x) - T(x)\| \to 0$.

So far so good (I think). Now here's the direction where I'm not so sure:

$(\impliedby)$ Suppose $T_n \rightarrow T$ on every bounded subset of $E$. Pick some $x_0 \in E$. Then for each $r \in \mathbb{N}$, we have $$ \lim_{n \to \infty} \sup_{x \in B_{r}(x_0)} \|T_n(x) - T(x)\| = 0. $$

Then taking the limit in $r$, we have $$ \lim_{r \to \infty} \lim_{n \to \infty} \sup_{x \in B_{r}(x_0)} \|T_n(x) - T(x)\| = 0. \hspace{1cm} (*)$$

Now the following would suffice to give the result:

1. The limits in $(*)$ can be interchanged.
2. $\lim\limits\limits_{r \to \infty} \sup\limits_{x \in B_{r}(x_0)} \|T_n(x) - T(x)\| = \sup\limits_{x \in E} \|T_n(x) - T(x)\|$

My main questions: Are 1) and 2) true? For 2) I am somewhat skeptical and am not sure how to rigorously justify interchanging the limits. For 2) I haven't quite proved it yet, but I am pretty confident it is true. (Or am I wrong?) Any insights would be appreciated.

Answer 1

Recall what it means for a sequence of operators $S_n$ to converge to $0$ in operator norm: $S_n\to 0$ in operator norm if $\lim_{n\to\infty}\sup_{\|x\|\le 1}\|S_n x\| = 0$. The set $\{x:\|x\| \le 1\}$ is bounded, so by assumption, we have the uniform convergence $\sup_{\|x\|\le 1}\|(T_n-T)x\|\to 0$. Hence $T_n-T$ converges to $0$ in operator norm.

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I think your confusion stems from the "false start" of letting $r$ be arbitrary, and trying to do something with a sequence of $r\to\infty$. The focus should be on some particularly relevant bounded subsets of $E$, such as the unit ball.