I was reading book by John Matthews on Numerical Analysis wherein I was going through the proof of Fixed Point Theorem. In its hypothesis it assumes that f, f' belong to C[a,b], K is a positive constant, f(x) belongs to [a,b] for all x in [a,b] and if |f'(x)|<=K<1 for all x in [a,b] then the iteration function will converge to a fixed point p in [a,b].
Before the proof begins there is a remark which says "Because f is continuous on an interval containing p, it is permissible to use the simpler criterion |f'(x)|<=K<1."
What I don't understand is how can we say that the derivative is bounded and that too specifically less than 1.
All I could think of was Lipschitz criterion which says |f(y)-f(x)| < M|y-x|, which implies |f'(x)|< M. Since the domain of f is closed and bounded, f is uniformly continuous. But that does not imply Lipschitz continuity, right?
I am stumped by the assumption that the derivative is bounded. Any help will be of great use. Thanks in advance.
We have the hypothesis that $|f'(x)| \le K <1$ for all $x \in [a,b].$ By the mean value theorem we get
$$|f(x)-f(y)| \le K|x-y|$$
for all $x \in [a,b].$