Let $f:[0,1]\to\mathbb{R}$ be continuous on $[0,1]$ and twice differentiable on $(0,1)$ such that $\forall x\in(0,1),\ f''(x)\geq0$ .
Suppose also that $f(0)=0,f(1)=1$, show that $\forall x\in[0,1], f(x)\leq x$.
My attempt:
Since the function satisfies the conditions of Lagrange's mean value theorem, then there exists a point $c\in(0,1)$
such that $f'(c)=\frac{f(1)-f(0)}{1-0}=1$.
Now, since the function is twice differentiable then its derivative is also differentiable and therefore continuous on (0,1).
This is where I'm stuck, I'm not sure if that's correct but maybe if I could show that the derivative is continuous on $[0,1]$
I could then use Lagrange's mean value theorem once more and
conclude that there exists a point $c_{2}\in(0,c)$ s.t. $f''(c_{2})=\frac{f'(c)-f'(0)}{c-0}=\frac{1-f'(0)}{c}\overset{f''(x)\geq0}{\overbrace{\geq}}0\Rightarrow f'(0)\leq1.$
Now, using the definition $\lim_{x\to0^{+}}\frac{f(x)-f(0)}{x}\leq1\Rightarrow f(x)\leq x$.
But i'm not sure if that's true because it feels lacking, and even if it's true I don't know how to show that the derivative is continuous on $[0,1]$.
I would appreciate your help, many thanks.
Best Answer
Let $g(x):=f(x)-x,$ so that $g(0)=g(1)=0$ and $g'$ is non-decreasing on $(0,1).$
By contradiction, if $\exists c\in(0,1)\quad g(c)>0$ then, by Lagrange's mean value theorem, $\exists a\in(0,c)\quad g'(a)>0$ and $\exists b\in(c,1)\quad g'(b)<0.$ But this is incompatible with $g'$'s monotonicity.
We thus proved that $\forall x\in[0,1]\quad g(x)\le0,$ i.e. $f(x)\le x.$