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

Please can you check my proof:

$\implies$: Assume there exists $c \in (0,1)$ such that $g(c) = c$. If $g'(1) \le 1$ and $g(1) = 1$ then since $g'' > 0$ it follows that $g(x) > x$ for all $x \in [0,1)$ which would contradict $g(c) = c$ where $c \in (0,1)$.

$\Longleftarrow$: Assume $g'(1) > 1$ and $g(1) = 1$. Then there exists $\varepsilon > 0$ such that $x \in (1-\varepsilon,1) $ implies $g(x) < x$. On the other hand, $g(0) > 0$. Hence by the intermediate value theorem there exists $c \in (0,1)$ such that $g(c) = c$.

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the proof is fine. the only point: you could better justify that statement: $\exists\epsilon>0:\forall x\in\left(1-\epsilon,1\right):g\left(x\right)<x$ (should work by taylor or something, dunno... but it cannot be hard) –  Max Mar 8 '14 at 6:37
here, some spoiler for you: math.stackexchange.com/questions/9728/… –  Max Mar 8 '14 at 6:42

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