Lef $K$ be an operator defined on $L^{\infty}((0,1)\times(0,1))$ into itself. $$K: L^{\infty}((0,1)\times(0,1)) \rightarrow L^{\infty}((0,1)\times(0,1)) $$ And maps the ball $B(0,M)$ into itself.
My question if I prove that the map $K$ is a contraction: for all $f_1,f_2 \in L^{\infty}((0,1)\times (0,1))$ with $0<k<1$ $$||K(f_1)-K(f_2)||_{\infty} \leq k ||f_1-f_2||_{\infty}$$ Can I conclude that $K$ has a fixd point ?
This is map on a complete metric space, so any contraction on it has fixed point by the Contraction Mapping Theorem. See https://en.wikipedia.org/wiki/Contraction_mapping