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I see there is a post almost related to this, but I think the conditions are slightly different.

Suppose $f$ is differentiable on (a,b) and continuous on [a,b], with $a\leq f(x) \leq b, \forall x \in [a,b]$ and $\mid f'(x)\mid<1,\forall x \in (a,b)$. Prove that $f(x) = x$ has a unique solution in $[a,b]$

So, I let, $g(x) = f(x) - x \Rightarrow g(a) = f(a) - a \geq 0$ and $g(b) = f(b) - b \leq 0$. So, from the Intermediate value theorem, we know, $\exists c \in [a,b]$, s.t.$g(c ) = 0 \Rightarrow g(c ) = f(c ) - c = 0 \Rightarrow f(c ) = c$, but then again, I don't understand how is that going to help us prove uniqueness. Any help? Thanks.

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    $\begingroup$ if two solutions, MVT en.wikipedia.org/wiki/Mean_value_theorem $\endgroup$ – Will Jagy Jul 13 '15 at 21:09
  • $\begingroup$ If there were two, f(c) = c and f(d) = d the Mean Value Theorem would give us a point, P, between c and d with f'(P) = 1. $\endgroup$ – lulu Jul 13 '15 at 21:09
  • $\begingroup$ @WillJagy, makes much sense in terms of M.V.T. $\endgroup$ – Jellyfish Jul 13 '15 at 21:22
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Suppose that we have two solutions $a,b$, where $a\neq b$ then by MVT:

$$\frac{f(a)-f(b)}{a-b}=f'(\xi)$$

where $\xi \in (a,b)$. But $a,b$ are solutions, so:

$$\frac{f(a)-f(b)}{a-b}=\frac{a-b}{a-b}=1$$

Thus $f'(\xi)=1$. But $f'(\xi)<1$.

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Uniqueness comes from $$ g'(x)=f'(x)-1\le|f'(x)|-1<0,\quad x\in[a,b] $$ which means that $g$ is strictly decreasing function, hence, can cross the $x$-axis at most once.

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  • $\begingroup$ @ProbabilityGuy You are right, it must be the non-strict inequality (fixed). $\endgroup$ – A.Γ. Jul 13 '15 at 21:21
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Suppose, by contradiction, that there exists $2$ solutions, say $c_1,c_2$. WLOG, assume $c_1 < c_2 $. By applying mean value theorem, we can find some $\xi \in (c_1,c_2) \subseteq [a,b] $ so that

$$ |c_2 - c_1| = |f(c_2)-f(c_1) | \leq |f'( \xi ) ||c_2 - c_1| < |c_2- c_1| $$

which gives a non-sense.

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