Let $X = C([0,1],\mathbb{R})$ equipped with the sup norm $d(f,g) = \sup_{x \in [0,1]} \,\{|f(x)-g(x)|\}$ for each $f,g \in X$. Define $F: X \to X$ by $$ F(g)(x) = \int_{0}^x \cos\left(\frac{g(t)}{2}\right)\, \mathrm{d}t. $$
Show that $F$ has a unique fixed point in $X$.
I've been trying to show that $F(g)(x)$ is a contraction, because since I assume that $X$ is complete, $F(g)(x)$ would then have a unique fixed point by Banach's Fixed Point Theorem, but to no avail. Help much appreciated.
It is also given that for each $g \in X$, $F(g)(x)$ is continuous, and also that $$ g(x) = \cos\left(\frac{x}{2}\right) $$ is Lipschitz continuous, if that helps.
Your map $F$ is indeed a contraction because for all $0\le x\le 1$, $$\begin{align*} \left|F(g)(x)-F(h)(x)\right|&\le \int_0^x \left|\cos\left(\frac{g(t)}2\right)-\cos\left(\frac{h(t)}2\right)\right| dt\\&\le\int_0^1 \frac 1 2\left|g(t)-h(t)\right| dt \\&\le \int_0^1 \frac 12 d(g,h)\ dt=\frac 1 2 d(g,h). \end{align*}$$ (Note that $|\cos(x)-\cos(y)|=|x-y|\cdot|\sin(sx+(1-s)y)|\le |x-y|$ by mean value theorem.) Therefore $d(F(g),F(h))\le \frac 12 d(g,h)$ and you can apply Banach fixed point theorem.