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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?

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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

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

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