Let $X = C([0,1])$ be the Banach space of continuous functions on $[0,1]$ (with the supremum norm) and define a map $F:X\rightarrow X$ by $$F(f)(x)=\int^{x}_{0} \cos(f(t)^{2})dt, \space x \in [0; 1].$$ Show that $FX=\{F(f):f \in X \} \subset X$ is relatively compact.
This comes down to using the Arzela-Ascoli Theorem. I have shown pointwise boundedness. I am stuck on showing equicontinuity. So far I have tried the following $$\begin{align} |(F(f)-F(g))(x)| &= \Big{|} \int^{x}_{0} \cos(f(t)^{2}-\cos(g(t)^{2}) dt \Big{|} \\ &\leq \int^{x}_{0}| \cos(f(t)^{2})-\cos(g(t)^{2})| dt\\ &= \int^{x}_{0} \Big|f(t)^{2}-g(t)^{2}| dt \end{align}.$$
I am stuck at this point, as I do not know how to bound the term inside the integral. Any hints on how to proceed from here or whether this approach is incorrect, and if so, which approach to take? Thank you in advance.
I think that using Arzela-Ascoli is the correct approach here. But let's think a bit more carefully about what we need in order to apply Arzela-Ascoli...
(1) Uniform boundedness. We must show that there exists an $M \geq 0$ such that $$ \left| \int_0^x \cos (f(t)^2) dt \right| \leq M$$ for any $f \in C([0,1])$ and any $x \in [0, 1]$.
Hopefully it should be clear that this works with $M = 1$. You might like to verify this using the observation that $| \cos(f(t)^2) | \leq 1$ for any $f \in C([0,1])$ and any $t \in [0,1]$.
(2) Equicontinuity. We must show that, for any $\epsilon > 0$, there exists a $\delta > 0$ such that $$ | x - y | < \delta \implies \left| \int_x^y \cos (f(t)^2) dt \right| < \epsilon $$ for any $f \in C([0,1])$ and any $x, y \in [0,1]$.
I believe this works with $\delta = \epsilon$. Again, you might like to prove this using the fact that $| \cos(f(t)^2) | \leq 1$ for any $f \in C([0,1])$ and any $t \in [0,1]$...