# Strong topology and uniform convergence

Let $$X$$ be a normed space and $$\{f_n\}\in X^{\ast}$$. I want to prove the following: $$\{f_n\}$$ converges (under strong topology) in $$X^{\ast}$$ if and only if $$\{f_n\}$$ converges uniformly on the unit ball in $$X.$$

My attempt: $$f_n$$ converges to $$f$$ in $$X^{\ast}$$ when $$\begin{equation*} \lim_{n\rightarrow\infty}\| f_n-f\|=0. \end{equation*}$$ $$f_{n}$$ converges uniformly when for all $$\epsilon>0$$, there exists natural number $$N$$ such that $$n\geq N$$ implies $$\begin{equation*} |f_n(x)-f(x)|<\epsilon \end{equation*}$$ It seems convergence under strong topology is more strong than uniform convergence. Can anyone help me to prove the statement?

Let $$B_1(0)$$ denote the unit ball in $$X$$ and $$f_n,f,X,X^{\ast}$$ as in the question.

So what you want to show is that $$\Vert f_n - f \Vert \rightarrow 0$$ if and only if $$\sup_{x \in B_1(0)} \Vert (f_n - f)(x) \Vert \rightarrow 0$$.

$$\Rightarrow$$: Let $$\Vert f_n - f \Vert \rightarrow 0$$, then we have

$$\sup_{x \in B_1(0)} \Vert (f_n - f)(x) \Vert \leq \sup_{x \in B_1(0)} \Vert f_n - f \Vert \Vert x \Vert \leq \Vert f_n - f \Vert \rightarrow 0$$

Here the first inequality is given by the standard bound and the second, since the supremum is over such $$x \in X$$ with $$\Vert x \Vert \leq 1$$.

$$\Leftarrow$$: Let $$\sup_{x \in B_1(0)} \Vert (f_n - f)(x) \Vert$$, then by definition of the operator norm we have

$$\Vert f_n - f \Vert = \sup_{x \in B_1(0)} \Vert (f_n - f)(x) \Vert \rightarrow 0$$