Sorry, my initial problem statement was weird. I am correcting the problem statement. The following is the precise statement of the problem:
Let $D$ denote the set of functions (continuous) in $C([0, 1]; \mathbb{R})$ such that $|f(x)| ≤ 1$ for all $x \in [0, 1].$
For any $f\in D,$ we define $I(f)$ on $[0, 1]$ by the rule $I(f)(x) = \int_{0}^{x} f(t)dt.$ I want to show that for any sequence $f_n \in D,$ there is a subsequence ${f_n}_{k}$ such that $\{I({f_n}_{k}) \}$ converges uniformly on [0, 1].
Also, in my original post, I mistakenly wrote Banach Alaouglu's theorem instead of Arzela-Ascoli's theorem. Sorry for all sorts of confusion. Thanks again.
Thank you, Jonathan and Saptak. Another
I think you meant that for every sequence $f_{n}$ such that $||f_{n}||\leq 1$,there exists a subsequence $f_{n_{k}}$ such that $I(f_{n_{k}})$ converges uniformly on $[0,1]$.To do this,consider the closed unit ball $D$ in $C[0,1]$, and show that $I(D)$ is pre-compact.In order to proceed this way,observe that for every continuous $f$, $I(f)$ is of class $C^1$,hence,the collection $\{I(f):f\in D\}$ is uniformly Lipschitz, with Lipschitz constant $1$.Hence,what we have is an equicontinuous family of maps in $C[0,1]$.Now,by Arzela-Ascoli theorem, we are done.