Understanding fully the proof of the uniqueness of a fixed-point

Question

I was reading the proof of the uniqueness of the fixed-point.

The uniqueness is stated as follows:

...

If, in addition (to the fact that we know that $\exists$ a fixed-point in a range $[a, b]$), the derivative $g'(x)$ exists and there is a constant $p < 1$, such that the derivative satisfies $$|g'(x)| \leq p < 1 , \forall x \in (a, b)$$ then the fixed point $x^{*}$ is unique in this interval.

Proof

Suppose there are two fixed-points $x^*$ and $y^*$, such that $g(x^*) = x^*$ and $g(y^*) = y^*$, then we want to show that $y^* = x^*$.

We know that $g'(x) \leq p, \forall x \in (a, b)$

$$|x^* - y^*| = |g(x^*) - g(y^*)| = |g'(\alpha)| * |x^* - y^*| \leq p |x^* - y^*|$$ where $\alpha$ is an intermediate value between $x^*$ and $y^*$.

Obviously, the inequality can only hold with $p < 1 \iff x^* = y^*$.

I understood most of the things in this proof, except the most important (probably).

Questions

1. Why the inequality holds only if $p < 1$? Maybe it has something to do with the nature of $|g'(\alpha)|$? Or maybe to the nature of fixed points?

2. I understood that we are using the mean value theorem in the last equality, but why are we using $p$ there? (This is somehow a stupid question, but I like to hear your experienced opinions :)

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Edit

Part of the source of my confusion is due to the fact that I interpreted wrongly the statement

Obviously, the inequality can only hold with $p < 1 \iff x^* = y^*$

which is saying that the inequality holds when $p < 1$ if $x^* = y^* $ (and thus viceversa).

But I'm still confused why.

Answer 1

For Question 1, note that the inequality is

$$|x^*-y^*|\le p|x^*-y^*| \tag 1$$

If $x^*\ne y^*$, then upon dividing both sides of $(1)$ by $|x^*-y^*|$ reveals

$$1\le p$$

But $p< 1$, and so the contradiction implies that $x^*=y^*$.

Answer 2

We don't really need $p$. In fact we have the slightly stronger result

If $I\subseteq \Bbb R$ is a (possibly unbounded) interval and $g\colon I\to I$ is continuous on $I$ and differentiable on the interior of $I$ and $g'(x)\ne 1$ for all $x$ in the interior of $I$, then $g$ has at most one fixed point.

The proof is exactly the same: Suppose $g$ has two distinct fixpoints $x^*$ and $y^*$, where wlog. $x^*<y^*$. Then by the the intermediate value theorem $$y^*-x^*=g(y^*)-g(x^*)=g'(\xi)\cdot(y^*-x^*)$$ for some $\xi\in(x^*,y^*)$. Dividing by the non-zero number $y^*-x^*$ produces $g'(\xi)=1$, contradiction.