Fixed point and extrema Let $\varphi_{a,b}:\mathbb{R}\ni x \mapsto \cos(ax+b)\in \mathbb{R}$. Show that for every $(a,b)\in (-1,1)\times\mathbb{R}$ there exist exactly one fixed point $s(a,b)$ of $\varphi_{a,b}$.
If it is true, we can define function $\psi:(a,b)\mapsto s(a,b)$.  
What is the biggest r such that $\psi$ is $C^r$ ?
What are local and global extrema of $\psi$ ?
Any hints ? 
 A: Here are hints for the first question.
Consider $f(x)=\cos(ax+b)$ and $g(x)=x$.
$f(x)$ is bound between $-1$ and $1$, while $g(x)$ goes from $-\infty$ to $\infty$. Since both $f(x)$ and $g(x)$ are continuous, there must be at least one $c$ where $f(c)=g(c)$.
Find $f'(x)$ and show that $|f'(x)|<1$. So $(f-g)'(x)$ is nonzero. Then the mean value theorem prohibits $f(x)-g(x)$ from having two roots. (Or you could show this using the integral of $f'(x)$.)
A: For the first part you can simply show that $x \to \cos(ax+b)$ is a contraction mapping (and so has a unique fixed point) provided $|a| < 1$. This requires the trig identity of the sum of cosines: 
\begin{equation}
|\cos(ax_1 + b) - \cos(ax_2 + b)| = \left|-2\sin\left(\frac{a(x_1+x_2)+b}{2}\right)\sin\left(\frac{a(x_1-x_2)+b}{2}\right)\right|\leq2\left|\sin\left(\frac{a(x_1+x_2)+b}{2}\right)\sin\left(\frac{a(x_1-x_2)+b}{2}\right)\right|
\end{equation}
And then using a few more basic facts about trig functions to get a bound like $|a||x_1-x_2|$. It might be helpful to write $y_1 = ax_1 + b$, $y_2 = ax_2 +b$ and then work with $y_1, y_2$ as then the annoying $b$ terms resolve themselves. 
For part 2) you can set $\psi(a,b) = \cos(a\psi (a,b) + b)$ and then start taking partial derivatives with respect to $a$ and $b$ to examine smoothness/critical points, but the calculations are a bit involved. Nothing intrinsically difficult, just a bit fiddly. 
