(Proof Correction): Suppose $f: [0,1] \rightarrow [0,1]$ which is cont. diff. and $|f'(x)| \leq 1/2$, then it has unique fixed point.

Question

My proof: (assume existence is proved) Suppose their exist two fixed points $y$ and $z$, then $\mid f(y)-f(z) \mid = \mid y-z \mid .\mid f'(x) \mid$ for some $y \leq x \leq z$ by mean value property. Since $y, z$ are fixed points, we have $\mid y- z \mid = \mid y-z \mid .\mid f'(x) \mid$ or $\mid f'(x) \mid = 1$ which is a contradiction. Hence the result. What I don't understand is where did I made use of the assumption that $f$ is continuosly differenciable.

Answer 1

The existence of a fixed point follows from just continuity and examining $$ g(x)=f(x)-x $$ Note that $g(0)\geq 0$ and $g(1)\leq 0$. If either one is actually $0$, we have our fixed point. If neither are, we may conclude the function has a zero on the interval by the intermediate value theorem.

Your proof looks perfectly fine to me and it is ok to relax the restriction to just differentiable on $[0,1]$ with your bound on the derivative.