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We got a real valued continuous function $g$ which is defined on $[0,1]$, $g(0)>0$, $g(1)=1$ and $g'(x) > 0$, $g''(x) > 0$ on $(0,1)$.

We need to prove that if $\lim_{x \to 1}g'(x) > 1$, then there are two solutions to $g(x)=x$ on $[0,1]$, if $\lim_{x \to 1}g'(x) \le 1$, then there must only one solution on $[0,1]$ for $g(x)=x$.

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    $\begingroup$ Consider the idea that having a fixed point is the same as crossing the diagonal, y=x. So subtract that function and look for 0's of the new function. That's my usual approach to problems like this $\endgroup$
    – Alan
    Feb 11, 2015 at 19:45

1 Answer 1

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Let $\lim_{x \to 1}g'(x) \le 1$ and there are more than $1$ solution, we peak a solution $q \ne 1$, obviously $q > 0$ and $\frac{g(1)-g(q)}{1-q}=1$, by the mean value theorem there must be a point $c$ such that $c \in (q,1)$ and $g'(c)=1$, it means that $\forall x > c$ $g'(x)>1$, moreover $g'(x)$ goes further away from 1 as $x$ grows, but then $\lim_{x \to 1}g'(x) > 1$ (contradiction).

Let $\lim_{x \to 1}g'(x) > 1$ and there is only one solution, obviously it has value $1$, it means that $\forall x \in (0,1)$ $f(x)=g(x)-x>0$ and $f(1)=g(1)-1=0$, also $f''(x)>0$ for $x \in (0,1)$, thus $f'(x)$ is strictly increasing and thus must be strictly negative $\forall x \in (0,1)$, so $f'(x)=g'(x)-1<0$ and $g'(x)<1$ $\forall x \in (0,1)$ thus $\lim_{x \to 1}g'(x) \le 1$ (contradiction).

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