Why is the topology of the norm of dual of Banach space $X$ identical to that of uniform convergence on the bounded subset of $X$?

Question

Let $X$ be a Banach space; the associated dual space is denoted by $X^∗$.

Why is the topology of the norm of $X^*$ identical to that of uniform convergence on the bounded subset of $X$?

An idea please.

Answer 1

Since the topology on $X^{\ast}$ is induced by a norm, the topology is entirely determined by the convergent sequences. So you want to show that for any sequence $\{f_n\}_{n \in \mathbb{N}} \subseteq X^{\ast}$ and $f \in X^{\ast}$ we have $f_n \rightarrow f$ in the norm topology if and only if $f_n \rightarrow f$ uniformly on every bounded subset $A$.

"$\Rightarrow$" Assume $f_n \rightarrow f$ in the norm topology, then by definition $$\sup_{x \in X} \frac{\vert f_n(x) - f(x) \vert}{\Vert x \Vert} \rightarrow 0 ~~.$$

So in particular for any fixed $n \in \mathbb{N}$ we have

$$\sup_{x \in X, \Vert x \Vert \leq n} \frac{\vert f_n(x) - f(x) \vert}{n} \leq \sup_{x \in X, \Vert x \Vert \leq n} \frac{\vert f_n(x) - f(x) \vert}{\Vert x \Vert} \leq \sup_{x \in X} \frac{\vert f_n(x) - f(x) \vert}{\Vert x \Vert} \rightarrow 0 ~~.$$

Hence, on every unit ball of radius $n \in \mathbb{N}$ around $0$ the sequence converges uniformly. In other words, $\forall \varepsilon > 0$ there is an $n_0 \in \mathbb{N}$ s.t. for all $x \in B_n(0)$ and $n \geq n_0$ we have $\vert f_n(x) - f(x) \vert < \varepsilon$.

Now consider any $A$ bounded. Then by definition there is some $n \in \mathbb{N}$ s.t. $A \subseteq B_n(0)$ and the claim follows.

"$\Leftarrow$" Assume $f_n \rightarrow f$ uniformly on any bounded set. Then it does so in particular on $B_1(0)$ and we have

$$\sup_{x \in X, \Vert x \Vert \leq 1} \vert f_n(x) - f(x) \vert \rightarrow 0 ~~,$$

which is equivalent to convergence in the operator norm.