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.

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)
