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Theorem

(a). If $g$ is continuous on $[a,b]$, $g(x) \in [a,b]$. Then $g$ has a fixed point in $[a,b]$.

(b). In addition, if there exists $k<1$ such that $|g'(x)| \leq k $. Then $g$ has unique fixed point in [a,b].

Why we can not modify the condition in part (b) to " $|g'(x)| < 1$ for all $x \in (a,b)$ ? "

Suppose that $p$ and $q$ are distinct fixed point in [a,b]. Then $|p-q| \leq |g'(x)||p-q| < |p-q|$ by MVT. Contradiction.

The proof in part (b) is still work. Or, is there some mistake?

sinoky
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    Indeed, if $g'$ is continuous, condition (b) is equivalent to say that $\max |g'| <1$. Maybe, there are some problems when $g'$ is not continuous. – Crostul Apr 25 '17 at 08:19
  • recall that if $g$ is continuous, then nothing can be said on $g'$... there can be points where $g'$ does not exist! Are you sure that this theorem is complete? Maybe some other hypotheses are missing. – the_candyman Apr 25 '17 at 09:17
  • add that g' exist, thanks – sinoky Apr 25 '17 at 09:47

1 Answers1

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Your proof works fine. The misunderstanding I think is that parts (a) and (b) are actually two different theorems. I don't know why there were put together like that. Part (a) is a simplified version of Brouwer's fixed-point theorem which says that if a continuous map maps a closed convex set into itself then it has a fixed point. Part (b) is a simplified version of Banach's fixed point theorem, which says that any contraction has a unique fixed point. If you have $|f'(x)|\le k<1$ you do not need to assume part (a). You get existence from $|f'(x)|\le k<1$, since this implies that $f$ is a contraction in $[a,b]$. Banach's fixed point theorem does not hold if the Lipschitz constant is one. However, if you have both (a) and (b) then you are perfectly right. To prove uniqueness you can relax the hypothesis in part (b).

Gio67
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