Fix point of function over disk Prove that there is a fix point of function $f:B(0,1) \to R^2$, where $B(0,1)$ is circle of radius 1, and $f(x,y)=\frac {1}{4}(ye^{x}-y,cosy)$. I tried to prove that f is contractive mapping (there is $0 < q < 1$ such that $d(f(a),f(b)) \leq qd(a,b)$), and than use Banach theorem.
 A: Using Brouwer theorem :
You know that $f$ is continous and $f(B(0,1))\subseteq B'\triangleq\overline{B} (0,1)$ where $\overline{B} (0,1)$ denotes the closed disk.
So $f: B' \to B'$ is continous, from a closed disk to itself.
By Brouwer theorem it has a fixed point.
Exercise to end the proof : verify it isn't on the edge of the disk.  (the factor $\dfrac{1}{4}$ is sufficient)
A: Maybe just without Banach's fixed-point theorem. At first we know that
\begin{align}
0 - \frac{1}{4} \cos(0) = - \frac{1}{4} < 0 \,\,\, \text{ and } \,\,\,\frac{\pi}{4} - \frac{1}{4} \cos\left(\frac{\pi}{4}\right) = \frac{\pi}{4} - \frac{1}{4 \sqrt{2}} > 0.
\end{align}
The intermediate value theorem then yields existence of some element $y_0 \in (0, \frac{\pi}{4})$ such that
\begin{align}
y_0 = \frac{1}{4}\cos(y_0).
\end{align}
Simply choosing $x_0 = 0$ gives
\begin{align}
\frac{1}{4}\left( y_0 \text{e}^{x_0} - y_0 \right) = x_0.
\end{align}
Since $(x_0, y_0) \in B(0,1)$ for obvious reasons we have found a fixed point.
