# Distance decreasing transformation and fixed point theorem

Let T be a distance decreasing transformation of the plane into itself. Prove that T leaves exactly one point of the plane fixed.

I clearly need to use the fixed point theorem here but couldn't think of a way to apply it to the transformation case. And what should I do to prove that no other point will be fixed? What are your opinions?

Uniquenes of the fixed point is clear. Suppose that $P_1$ and $P_2$ are fixed points. Then $$0\le\|P_1-P_2\|=\|T(P_1)-T(P_2)\|<\|P_1-P_2\|\implies\|P_1-P_2\|=0.$$
But existence may fail. Let $$f(t)=t+\frac\pi2-\arctan t,\quad f'(t)=\frac{t^2}{1+t^2}.$$ By the Mean Value Theorem, $|f(t)-f(s)|<|t-s|$, but $f$ has no fixed points. Then $$T(x,y)=(f(x),f(y))$$ is a distance decreasing transformation without fixed points.