Let $f: \mathbb R \to \mathbb R$ be a continuous function with a continuous derivative. In short $f \in C^1$.

We know that $0<c\leq f'(x) \leq d < \infty$. We want to prove that $\exists! x_0 \in \mathbb R: f(x_0)=0$.

Disclaimer: I am fully aware that there is a simple proof involving Rolle's theorem. This is not the proof I am after.

I am trying to somehow invoke Banach Fixed Point Theorem in order to show that this function has a single root.

Something I tried, to give you a general idea of what I'm looking for:

Define $g(x) = \frac{1}{d}(f(x)-cx)$. We have that $0 \leq g'(x) \leq 1-\frac{c}{d}<1$

Now, since $f$ is $C^1$, we also have that $g$ is $C^1$. And more specifically, $g$ is continuous and with continuous derivative on all $[x_1,x_2]$ segments of the real line.

So Lagrange mean value theorem applies. $\forall x_1,x_2 \in \mathbb R \exists c \in \mathbb R: g(x_1)-g(x_2)=g'(x)(x_1-x_2)$

If we add modules we get $|g(x_1)-g(x_2)| = |g'(c)||x_1-x_2| \leq |1-\frac{c}{d}||x_1-x_2|$

And so we have for all $x_1,x_2:$ $|g(x_1)-g(x_2)| \leq |1-\frac{c}{d}||x_1-x_2|$, which means $g$ is a contraction.

So from Banach Fixed Point Theorem, $g$ admits a unique fixed point $g(x_0)=x_0$.

The problem is: $g(x_0)=x_0$ does not imply that $f(x_0)=0$. Sure, it admits a fixed point but that does not help us prove that $f$ has a unique zero. if $g(x) = f(x)+x$ then that would have solved our problem, but no such luck.

How can I use Banach fixed point theorem in a similar way to prove $f$ has a unique zero?

  • $\begingroup$ I need to build some function $g$ such that $g$ is a contraction AND $g(x_0)=x_0$ implies $f(x_0)=0$. $\endgroup$ – Oria Gruber Dec 27 '15 at 15:00

Try $$g(x)=-\frac{f(x)}{2d}+x.$$

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  • $\begingroup$ Thanks, found it on my own too, but forgot to close the question :) Accepted regardless. $\endgroup$ – Oria Gruber Dec 27 '15 at 21:54

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