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.

  • 1
    $\begingroup$ Hint: it is a convexity argument, the cord joining $(0,0)$ and $(1,1)$ is above the curve, i.e. $x\ge f(x)$... $\endgroup$
    – zwim
    Commented Nov 30, 2022 at 16:11
  • 2
    $\begingroup$ $\lim_{x\to0^{+}}\frac{f(x)-f(0)}{x}\leq1$ does not imply $f(x)\leq x.$ $\endgroup$ Commented Nov 30, 2022 at 16:40
  • $\begingroup$ @AnneBauval yep, I was pretty sure that was wrong, thanks! $\endgroup$
    – itailitai
    Commented Nov 30, 2022 at 16:54

3 Answers 3


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. $f(x)\le x.$

  • $\begingroup$ Note that $f$ didn't need to be twice differentiable on $(0,1)$ but only to have a non-decreasing derivative on $(0,1).$ $\endgroup$ Commented Nov 30, 2022 at 16:43
  • $\begingroup$ Maybe they mentioned that $f$ is twice differentiable because in the next section I need to prove that if $f(0.5)\geq 0.5$ then $f$ is linear? Do I need to use it here? $\endgroup$
    – itailitai
    Commented Nov 30, 2022 at 17:42
  • $\begingroup$ No, twice differentiability is again useless for that question. $\endgroup$ Commented Nov 30, 2022 at 17:46

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 $x\in[0,1]$,

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


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 $[0,1]$.


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