Let $\varphi:[a,b]\to \mathbb R$ be continuous differentiable and $|\varphi'(x)|<\delta < 1\ \forall x\in[a,b]$. Let $x_1\in [a,b]$ with this define $(x_n)_{n\in \mathbb N}$ recursively with $x_n =\varphi(x_{n-1}), n>1,$ and let $x_n\in[a,b]$ for all $n\in \mathbb N$
Show with the mean value theorem that $$|x_{n+1}-x_{n}|<\delta|x_{n}-x_{n-1}|\ \ \ \ (1)$$ its easy to show that $x_n$ converges and with that it follows that $c=\lim_{x\to\infty} x_x$ sloves the fixed-point equation $$\varphi(c)=c \ \ \ (2)$$ show that c is unique with the mean value theorem
What I tried: 1) proof: If $\varphi$ is continous differentiable, then $\varphi'$ is bounded on the closed intervall $[a,b]$, by a constant $\delta$ therefor
$$\lvert \varphi(a) - \varphi(b)\rvert \le \delta \lvert a-b\rvert$$ this completes the proof.
2)
preconditions: $\varphi$ is a contraction, an arbitrary Cauchy series converges. proof: let $x_1,x_2\in[a,b]$ be fixpoints of $\varphi$. then must be true $$|x_1-x_2|=|\varphi(x_1)-\varphi(x_2)|<\delta|x_1-x_2|$$ that yields a contradiction which implies $\varphi(x_1)-\varphi(x_2)=0$ therefore $x_1=x_2$ which completes the proof
Things unclear: I don't understand completeness why do I need it as a precondition or do I not for the real-valued contraction theorem? But this one professor from whom I read the script and proof of the fixpoint theorem keeps stressing that you can't prove that $\mathbb R$ is uncountable, the mean value theorem, and so on without completeness. So completeness is basically handier than the supremum axiom in proofs from how I understood it but I still don't understand why we need it here or in the Banach fixpoint theorem. And am I correct in the assumption that 1) boils down to continuous differentiable implies Lipschitz?