Existence of a function that fulfills an equation I am just revising for my exams and came across this question: Show that in the Banach-space of functions that are continuous in the interval $[-1,1]$, together with the supremum-norm, there is exactly one function that fullfills the equation $f(x) = \frac{1}{2}\ln\left(1 + f(x)^2\right) + \sin(\ln(x + 2))$. I think that this is probably a Banach Fixpoint problem but I am not sure how to approach it. Any ideas?
 A: You should indeed use the Banach fixed-point theorem.
The "obvious" map to consider is $T : C[-1,1] \to C[-1,1]$ $$(Tf)(x) = \frac{1}{2}\ln(1+f(x)^2) + \sin(\ln(x+2))$$
Check that $T$ is a contraction mapping.
The inequality
$$\vert \ln(1+x^2)-\ln(1+y^2) \vert \leq \vert x - y \vert$$
might prove useful.
(The latter inequality follows e.g. from the mean-value theorem and the observation $$\frac{d}{dx}\ln(1+x^2)=2x/(1+x^2)\in [-1,1]$$ for all $x\in\mathbb{R}$).
A: For each $x\in I=[-1,1]$, define $T_{x}(y):=\frac{1}{2}\ln(1+y^{2})+\sin(\ln(x+2))$. Observe that $T_{x}'(y)=\frac{y^{2}}{1+y^{2}}\leq \frac{1}{2}$ on $I$, so we can show $|T_{x}(y')-T_{x}(y)|\leq \frac{1}{2}|y'-y|$ and $T_{x}$ has a unique fixed point for each $x$ by Banach fixed point theorem. Define $f(x)$ as fixed point of $T_{x}$ for each $x\in I$. 
To show that $f(x)$ is continuous, define $g(y)=y-\frac{1}{2}\ln(1+y^{2})$ then $g'(y)>0$ on $I$, so $g$ has an inverse which is also continuous. Note that $g(f(x))=\sin(\ln(x+2))$.
