Let $f$ be twice differentiable in $(a,b)$ and $f''>0$ in the same interval. If $a<x_1<x_2<b,$ show that $$f\left(\frac{x_1+x_2}{2}\right)\leq\frac{f(x_1)+f(x_2)}{2}.$$

I'm not sure how to start. I know at least that $f'$ is strictly increasing in $(a,b).$ Should I check separately for the cases $x_1\neq x_2$ and $x=x_2$? I'm supposed to use the mean value theorem, but I don't know how. Any insight I need to make in order to start?

  • 4
    $\begingroup$ Hint: first suppose $f(x_1) = f(x_2) = 0$, and suppose to the contrary that $f(\frac{x_1+x_2}{2}) > 0$. Then what does the mean value theorem applied to the interval $[x_1, \frac{x_1+x_2}{2}]$ tell you? What about if you apply it to the interval $[\frac{x_1+x_2}{2}, x_2]$? $\endgroup$ – Daniel Schepler Dec 19 '17 at 22:20

Rewrite this as $f(\frac{x_1+x_2}{2}) - f(x_{1}) \leq f(x_2) - f(\frac{x_1+x_2}{2}) $ and then apply the Mean Value Theorem, noting that $f'$ is increasing.

  • $\begingroup$ I don't understand how I should apply the MVT on this. $\endgroup$ – Parseval Dec 20 '17 at 8:36
  • $\begingroup$ Naki: So following your advice, according to the Mean Value Theorem, $\exists \ c,k\in \mathbb{R} : $ $$f'(c)\left(\frac{x_1+x_2}{2}-x_1\right)\leq f'(k)\left(x_2-\frac{x_1+x_2}{2}\right)$$ which is equivalent to $$f'(c)(x_1-x_2)\leq f'(k)(x_2-x_1)\Leftrightarrow f'(c)(x_1-x_2) \leq -f'(k)(x_1-x_2).$$ Which finally leaves me with $f'(c)\leq -f'(k).$ Not sure what to do with this information. It also is clearly wrong. $\endgroup$ – Parseval Dec 20 '17 at 9:08
  • $\begingroup$ No, you get $x_{2}-x_{1}$ for both terms. Think about it as the middle of two points $a$ and $b$ being equidistant of both $a$ and $b$. $\endgroup$ – Elie Louis Dec 20 '17 at 9:27
  • $\begingroup$ Ok so the conclusion is $f'(c)\leq f'(k)$ which is clearly true since $c\leq k?$ $\endgroup$ – Parseval Dec 20 '17 at 9:30
  • $\begingroup$ Yes, because $f'$ is increasing. $\endgroup$ – Elie Louis Dec 20 '17 at 9:30

What you want here is the convexity of $f$. Since you know that $f'$ is monotonically increasing, I will outline a proof for you of how to show $f$ is convex.

Proof outline. Let $a< x < y < b$ and consider the slope $m$ of the line $L$ joining the points $(x,fx)$ and $(y,fy)$. By the mean value theorem, there exists a point $c\in (x,y)$ such that $f'(c) = m$. Suppose that there is a point $\theta\in(x,y)$ such that $f(\theta)>L(\theta)$. Consider separately the two cases where $\theta\in(x,c)$ and $\theta\in(c,y)$ and derive contradictions. (Draw pictures.)

  • $\begingroup$ I learned back in highschool that if $f''>0$, then the function is convex. Why do I need to show that it's convex when I know that it is convex? $\endgroup$ – Parseval Dec 19 '17 at 22:34
  • $\begingroup$ @Parseval if you know it’s convex, then you know it’s convex, and you’re done. I’m just showing you how to get from $f’’>0$ to $f$ is convex. $\endgroup$ – Alex Ortiz Dec 19 '17 at 22:36
  • $\begingroup$ @Parseval, the definition of convexity is not that the second derivative is positive since a general convex function is not even differentiable everywhere, so it requires justification to show that the positivity of the second derivative implies convexity. $\endgroup$ – Alex Ortiz Dec 19 '17 at 22:39
  • $\begingroup$ @Parseval DId you learn how to prove it's convex? (Did you even learn the definition of "convex"?) $\endgroup$ – David C. Ullrich Dec 19 '17 at 23:59
  • $\begingroup$ @DavidC.Ullrich Hi David, no we never learned the definition of convex. Is there a way around this without this knowledge? $\endgroup$ – Parseval Dec 20 '17 at 7:43

See difference from $f(x_1)$ to $f({{x_1 + x_2}\over 2})$, which is at exactly a middle point need to be smaller as a difference between $f({{x_1 + x_2}\over 2})$ and $f(x_2)$ because a second derivative is > 0. So middle value of this f(x) need to be higher as this : enter image description here


Let $x_1$ be fixed. Let $g(x_2) = f\left(\frac{1}{2}(x_1 + x_2)\right)$ and $h(x_2) = \frac{1}{2}(f(x_1) + f(x_2))$ be defined as functions of $x_2$ on the interval $[x_1, b]$. We wish to show $g(x_2) \leq h(x_2)$ for all $x_2 \in [x_1, b]$. Since $g(x_1) = h(x_1)$, it suffices to show that $g'(x_2) \leq h'(x_2)$ on $[x_1, b]$.

Differentiate $g$ with respect to $x_2$ giving $\frac{1}{2}f'\left(\frac{1}{2}(x_1+x_2)\right)$. Do the same with $h$ yielding $\frac{1}{2}f'(x_2)$.

Since $f''>0$, it follows that $f'$ is increasing. Since $x_2 = \frac{1}{2}(x_2 + x_2) \geq \frac{1}{2}(x_1+x_2)$, it follows that $g'(x_2) = \frac{1}{2}f'\left(\frac{1}{2}(x_1+x_2)\right) \leq \frac{1}{2}f'(x_2) = h'(x_2)$ as desired.


If $f''(x)>0$ you can show that the strictly inequality holds:


For the proof one way is to show that:

$$f''(x)>0 \implies f'(x) \ \text{strictly increasing} \iff f(x) \ \text{is strictly convex}$$

To proof that $f'(x) \ \text{strictly increasing} \iff f(x) \ \text{is strictly convex}$ let's consider $a<c<b$, thus for the MVT:

$$\frac{f(c)-f(a)}{c-a}=f'(d)<f'(e)=\frac{f(b)-f(c)}{b-c}\quad d \in(a,c), \quad e\in(c,b)$$

Now consider $a<x_1<z<x_2<b$ and $\lambda\in[0,1]$, we want to show that:

$$f(z)=f(\lambda x_1+(1-\lambda) x_2)<\lambda f(x_1)+(1-\lambda) f(x_2)$$

For the previous result by MVT we have that:





$${f(z)}<f(x_1)\frac{(x_2-z)}{(x_2-x_1)}+f(x_2)\frac{(z-x_1)}{(x_2-x_1)}=\lambda f(x_1)+(1-\lambda) f(x_2) \quad \square$$

For the special case $\lambda=\frac12$ thus:



The inequality is a particular case of "Jensen's inequality" for convex functions.



  • $\begingroup$ ??? The problem is to show $f$ is convex. Once we know it's convex the inequality is just a special case of the definition - we don't need Jensen. $\endgroup$ – David C. Ullrich Dec 19 '17 at 23:57
  • $\begingroup$ @DavidC.Ullrich since $f''>0$ it's convex, the first link is indeed on convex functions, I've given Jensen as a reference for a generalization; do you think it's not much clear? $\endgroup$ – gimusi Dec 20 '17 at 0:03
  • $\begingroup$ The problem is precisely to show that $f''>0$ implies that $f$ is convex. Saying "since $f''>0$ it's convex" is just proving this by saying we know it's true. $\endgroup$ – David C. Ullrich Dec 20 '17 at 0:09
  • $\begingroup$ ah ok, it's a basic property he can find the proof in any good book of calculus! Thanks for the clarification. $\endgroup$ – gimusi Dec 20 '17 at 0:14
  • $\begingroup$ @DavidC.Ullrich I've added the complete proof. $\endgroup$ – gimusi Dec 20 '17 at 2:04

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