Show that if $f(0)=0,f(1)=1$ and $f''\ge0$ then $f(x)\leq x$ for every $x\in[0,1].$

Question

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).

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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.

Answer 1

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. $f(x)\le x.$

Answer 2

Since $f''(x)\geq0$ for all $x\in(0,1)$, we know that $f$ is a convex function on $[0,1]$ (see, for example, this question for details). Using convexity we thus have that, for any $x\in[0,1]$,

$$f(x)=f(1x+0(1-x))\leq f(1)x+f(0)(1-x)=x.$$

Answer 3

A more direct approach is given by @Lorago. However, if you are familiar with the maximum principles one gets a very quick and slick proof which carries over with little effort to higher dimensions.

Observe that $\Delta(f(x)-x) \geq 0$ on the set $[0,1]$ (here $\Delta$ is just the second derivative in the $x$ variable). The maximum principle now entails that the maximum of $f(x)-x$ lies on the boundary of the domain $[0,1]$. However, $f(0)-0=0=f(1)-1$ and so one concludes that $f(x) - x \leq 0$ on $[0,1]$.