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Let $g:[0,1] \rightarrow \mathbb{R}$ be twice differentiable with $g''(x)\gt 0$ for all $x\in[0,1]$. If $g(0)>0$ and $g(1)=1$, show that $g(d)=d$ for some point $d\in(0,1)$ if and only if $g'(1)\gt 1$.


Suppose there exits $d\in(0,1)$ such that $g(d)=d$. Then by the MVT applied on $[d,1]$, $f'(c)(d-1)=g(d)-g(1) $for some $c\in(d,1)$. But $g(d)=d$ and $g(1)=1$ so $f'(c)=1$. Now, since $g''(x)\gt 0$ for $x\in[0,1]$, $g'$ is increasing on this interval, therefore $1\gt d$ implies $g'(1)-g'(d)\gt 0$. Assume $g'(1)\gt 1$ and let $f(x)=g(x)-x$. Then $f(0)=g(0)-0\gt 0$ by hypothesis. I want to show $f(x)\lt 0$ for some $x$ in a neighborhood of $1$. $g'(1)= \lim\limits_{x\to1}\frac{g(x)-g(1)}{x-1}$ so there exists $\delta\gt 0$ such that given $\epsilon\gt 0$ with $x\in(1-\delta,1)$ implies $|g'(1)-\frac{g(x)-1}{x-1}|\lt \epsilon$. If $\epsilon=g'(1)-1\gt 0$, then $g'(1)-\epsilon\lt \frac{g(x)-1}{x-1}$ which implies $g(x)\lt x$ for $x\in(1-\delta,1)$. The result follows by the IVT.

Any comments, corrections or different solutions are welcome.

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Consider $h(x) = g(x) - x$ and a similar argument like what you have made. The argument will look easier if you deal with $h$ instead of directly with $g$ –  user17762 Nov 10 '10 at 16:46
Sometimes it is easier to prove the contrapositive of a statement than to prove the statement directly. –  bobobinks Jan 28 '11 at 19:21
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2 Answers

up vote 1 down vote accepted

To put it more concisely: If there is a $\xi\in]0,1[$ with $g(\xi)=\xi$ then by the mean-value theorem there is a $\xi'\in ]\xi,1[$ with $g'(\xi')=1$, whence $g'(1)>1$, as $g'$ is strictly increasing. Conversely, if $g'(1)>1$ then $g(a)<a$ for some $a$ immediately to the left of $1$. As $g(0)>0$ there must be a $\xi\in]0,a[$ with $g(\xi)=\xi$.

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Suppose $g'(1)<1$. Let $f(x)=g(x)-x$. Then if $x\in(0,1), f'(x)=g'(x)-1< g'(1)-1<1-1=0$, since g'(x) is increasing.

Hence $g(x)-x=f(x)>f(1)=g(1)-1>0 \;\;\;\forall x\in(0,1)$, since by the preceding f is strictly decreasing.

Now assume $f(x)=g(x)-x\neq0 \;\;\;\forall x\in (0,1)$.

Then $g(x)>x \;\;\;\forall x\in(0,1)$ (otherwise $\exists x_0$ such that $f(x_0)<0$ and since $f(0)>0$ we'd have a contradiction to the IVT).

Hence, $\forall x\in(0,1), \frac{g(1)-g(x)}{1-x}<\frac{g(1)-x}{1-x}=1$. It follows $g'(1)\leq1$.

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