Need help with a twice differentiable and bounded function.

Question

Let $f : \mathbb{R} \to \mathbb{R} $ a bounded and twice diffentiable function so that $\begin{equation} \forall x \in \mathbb{R}, f"(x) \geq 0 \end{equation}. $

My point is to prove that $f$ is constant.

So if I can prove that $\forall x \in \mathbb{R}, f'(x) = 0$ then my job is done.

So far, I only know that $f'$ is increasing but I don't know how I can use that to end the proof.

Can anyone help me please? Thanks

Thanks for including your progress on the question; that always helps. Can a function with an increasing derivative be bounded? — Matthew Leingang, Sep 01 '19 at 19:21
@amsmath You are correct, of course! I was reacting to an earlier version of the problem that had $f'' \leq 0$ rather than $f'' \geq 0$ as it now shows. I deleted the obsolete comment. — Matthew Leingang, Sep 01 '19 at 19:23
Possible duplicate of Show bounded and convex function on $\mathbb R$ is constant — Martin R, Sep 01 '19 at 19:26
Hint: Pick two points $a < b$ and suppose $f(a) < f(b)$. Since $f'' \geq 0$, prove that all $x \geq b$ lie above the line through $(a,f(a))$ and $(b,f(b))$. — Dzoooks, Sep 01 '19 at 19:28
I can reach the point where $f(c) \geq f'(a)(c-a)+f(a)$ with $c \geq b$ but what can I do with $f'(a)$. I feel like I can somehow use the mean-value theorem but I'm lost. Can you help me with that please? — Pablito, Sep 01 '19 at 19:59

Andreas Blass · Answer 1 · 2019-09-02T11:47:22.240

If there is an $a\in\mathbb R$ with $f'(a)>0$, then for all $x>a$ we have $f(x)\geq f(a)+(x-a)f'(a)$. [Proof: By the mean value theorem, we have $f(x)= f(a)+(x-a)f'(y)$ for some $y$ between $a$ and $x$. And $f'(y)\geq f'(a)$ because, as you noticed, $f'$ is increasing.] This contradicts the assumption that $f$ is bounded; specifically, $f(x)$ won't be bounded above as $x\to+\infty$.

If there is an $a$ with $f'(a)<0$, then a similar argument shows that $f(x)$ cannot be bounded above as $x\to-\infty$.

So $f'(a)=0$ for all $a$, as required.

score 0 · Accepted Answer · answered Sep 01 '19 at 20:22

This is what I did, if anything's wrong just hit me up.

Let $a < b $ so that $f(a) < f(b)$.

By the mean-value theoreme, $c \in ]a,b[ $ exists so that $f'(c)= \frac{f(b)-f(a)}{b-a}<0$.

Let $x \geq b$.

Since $f$ is convex, we have $f(x) \geq f'(b)(x-b)+ f(b).$

But $f'$ is increasing and $b \geq c$ thus $f'(b) \geq f'(c)$.

Finally, $f(x) \geq \frac{f(b)-f(a)}{b-a}(x-b)+f(b).$

So $\forall x \geq b, f(x) \geq \frac{f(b)-f(a)}{b-a}(x-b)+f(b).$

As $x \to +\infty$, we can see that $f$ is not bounded hence the contradiction.

As a conclusion, $f$ is constant.

Need help with a twice differentiable and bounded function.

2 Answers2