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Let $M$ be a bounded subset of $C([0,1])$. Show that the set $\left\lbrace F(x) = \displaystyle\int_0^x f(t)dt | f \in M \right\rbrace$ is sequentially compact (meaning that any sequence in the set has a convergent subsequence).

My idea is that: Let $\lbrace F_n \rbrace$ be the sequence in the set and fix $x \in [0,1]$, then $\lbrace F_n(x) \rbrace$ is bounded, so by Bolzano-Weierstrass, there exists a subsequent $\lbrace F_{n_k}(x)\rbrace$ converges. So I guess if I repeat the same procedure, I may have a convergent subsequence? But for different $x$, I may have different subsequence, so I don't know what to handle this. Can somebody please help me with this?

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Use Arzelà-Ascoli: Let $f \in M$ be arbitrary and $F(x) = \int^x_0 f(t)~\mathrm{d}t$. Then $$ \lvert F'(x) \rvert = \lvert f(x) \rvert \leq \sup_{f \in M} \lVert f \rVert_{C([0, 1])} < \infty $$ since $M$ is bounded. So all functions $F$ in your set share the same Lipschitz-constant.

Furthermore note: $$ \lvert F(x) \rvert \leq \int^x_0 \lvert f(t) \rvert~\mathrm{d}t \leq x \sup_{f \in M} \lVert f \rVert_{C([0, 1])} \leq \sup_{f \in M} \lVert f \rVert_{C([0, 1])} $$ So all functions in your set are bounded by the same constant. So using Arzelà-Ascoli you can conclude that your space is compact.

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