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 $g'(1)\gt 1$.
Proof.
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.