# How to prove that $F(g)(x) := \int_{0}^x \cos(\frac{g(t)}{2}) dt$ is continuous?

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?

• 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). Feb 27, 2019 at 9:26
• @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...
– Surb
Feb 27, 2019 at 9:38

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.
There is $$C$$ s.t. for all $$a,b\in\mathbb R$$, $$|\cos(a)-\cos(b)|\leq C|a-b|.$$
• 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 $|\frac{\cos(\frac{f(t)}{2})-\cos(\frac{g(t)}{2})}{\frac{f(t)}{2} - \frac{g(t)}{2}}| \leq |\cos'(c)| \iff | \cos(\frac{f(t)}{2}) - \cos(\frac{g(t)}{2})| \leq 1 \cdot | \frac{f(t)}{2} - \frac{g(t)}{2}| = \frac{1}{2} | f(t) - g(t) |$\\ Feb 27, 2019 at 11:20
• 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? Feb 27, 2019 at 11:21
• @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.