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Let $g : [0, 1] \rightarrow \mathbb{R}$ be twice differentiable with $g^{\prime \prime}(x) > 0$ for all $x \in [0,1]$. Suppose that $g(0) > 0$ and $g(1) = 1$. Prove if $g$ has a fixed point in $(0,1)$, then $g^{\prime}(1) > 1$.

My attempt: Define a function $h(x)=g(x)-x$. Since $g$ has a fixed point, say $c \in (0,1)$, we have $h(c)=g(c)-c=0$.

Notice that we have $h(c)=h(1)=0$, by Rolle's Theorem, there exists $d \in (c,1)$ such that $h^{\prime}(d)=0$

Applying Mean Value Theorem on $h$ on $[d,1]$, there exists $e \in (d,1)$ such that $h^{\prime \prime}(e)=\frac{h^{\prime}(d)-h^{\prime}(1)}{c-1}$. Notice that we have $h^{\prime \prime}(x)=g^{\prime \prime}(x) >0 $ for all $x \in [0,1]$. Hence, $-h^{\prime}(1)<0 \implies g^{\prime}(1) > 1$

Can anyone explain to me why we need to use Rolle's theorem here?

Idonknow
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2 Answers2

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We don't use Rolle's theorem,since $g^{\prime\prime}>0$,then $g^{\prime}(x)$ is strictly increasing,also $g$ has a fixed point c in $(0,1)$ and $1$ is fixed point for $g$,we can use only Mvt theorem.($\frac{g(1)-g(c)}{1-c}=g^{\prime}(t)=1,t\in(c,1)$). Of course I mean $g^{\prime}(1)$ as $\lim_{x\to1^-}g^{\prime}(x)$ because $1$ is end of $[0,1]$.

R Salimi
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The Rolle's theorem enables us to get a root of $h'$ on some point lying left to $1$ which helps to conclude the result using the strict monotonicity of $h'$ (Since $h''>0$ on $[0,1]\implies h'$ is strictly increasing on $[0,1]$)

Sugata Adhya
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