# Show that the set $\lbrace F(x) = \displaystyle\int_0^x f(t)dt | f \in M\rbrace$ is sequentially compact.

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?

• Do you know about the Arzelà-Ascoli theorem? (or rather, can you use it yet?) Feb 23, 2023 at 7:39

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.