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Let $X = C([0,1],\mathbb{R})$ be the set of continuous functions on $[0,1]$ equipped with the sup-norm $d(f,g) := \sup\limits_{x\in [0,1]} \{|f(x)-g(x)|\}$ for each $f,g \in X$.

Define a function $F: X \rightarrow X$ by $F(g)(x) := \int_{0}^x \cos(\frac{g(t)}{2}) dt$.

Prove that for each $g \in X$ we have that $F(g) \in X$, i.e. $F(g)(x)$ is continuous.

Can I just do this: $F(g)'(x) = \cos(\frac{g(t)}{2})$, and $|\cos(\frac{g(t)}{2})| \leq 1$ then since the derivative is bounded, from the Mean Value Theorem, $F(g)(x)$ is Lipschitz continuous hence uniformly continuous?

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    $\begingroup$ Your function is actually differentiable, by the fundamental theorem of Calculus. For continuity, all you need is the fact that $\cos(g(t)/2)$ is a (Riemann) integrable function on $[0,1]$ (being continuous there). $\endgroup$
    – GReyes
    Feb 27, 2019 at 9:26
  • $\begingroup$ @GReyes: $x\mapsto F(g)(x)$ is indead differentiable, but we are looking here for the continuity of $g\mapsto F(g)$. So differentiability of $x\mapsto F(g)(x)$ doesn't help... $\endgroup$
    – Surb
    Feb 27, 2019 at 9:38

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Strange way to prove continuity for such a function... If you can prove that $$|F(f)-F(g)|\leq K\|f-g\|_X,$$ for some constant $K$, then you are done.

Hint

There is $C$ s.t. for all $a,b\in\mathbb R$, $$|\cos(a)-\cos(b)|\leq C|a-b|.$$

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  • $\begingroup$ Since $\cos(x)$ is differentiable, there is a $c \in [\frac{f(t)}{2}, \frac{g(t)}{2}]$ s.t. $\cos'(c) = \frac{\cos(\frac{f(t)}{2})-\cos(\frac{g(t)}{2})}{\frac{f(t)}{2} - \frac{g(t)}{2}}$ by MVT. But then $|\frac{\cos(\frac{f(t)}{2})-\cos(\frac{g(t)}{2})}{\frac{f(t)}{2} - \frac{g(t)}{2}}| \leq |\cos'(c)| \iff | \cos(\frac{f(t)}{2}) - \cos(\frac{g(t)}{2})| \leq 1 \cdot | \frac{f(t)}{2} - \frac{g(t)}{2}| = \frac{1}{2} | f(t) - g(t) |$\\ $\endgroup$
    – goblinb
    Feb 27, 2019 at 11:20
  • $\begingroup$ So if we set $\delta = \frac{\epsilon}{\int\limits_0^x \frac{1}{2} dt}$ we have that $\int\limits_0^x | \cos(\frac{f(t)}{2}) - \cos(\frac{g(t)}{2})| dt \leq \int\limits_0^x \frac{1}{2} |f(t)-g(t)| dt < \int\limits_0^x \frac{1}{2} dt \frac{\epsilon}{\int_0^x \frac{1}{2} dt} = \epsilon$. Does that work? $\endgroup$
    – goblinb
    Feb 27, 2019 at 11:21
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    $\begingroup$ @goblinb: The idea is definitely here, but it's not well written. Espacially when you say : there is $c\in [f(t)/2,g(t)/2]$. After you complexify a bit, but it's correct. $\endgroup$
    – Surb
    Feb 27, 2019 at 12:50

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