Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given that $$f\left(\frac{x+y}{2}\right)\leqslant \frac{f(x)+f(y)}{2}~,$$ how can I show that $f$ is convex.
Thanks.


Edit: I'm sorry for all the confusion. $f$ is assumed to be continuous on an interval $(a,b)$.

share|improve this question
1  
You can't. At least, not without assuming that $f$ is continuous. –  Willie Wong Nov 18 '11 at 15:06
1  
Lebesgue measurability is enough @WillieWong. This is a theorem by Sierpinski. (If I recall correctly). –  Jonas Teuwen Nov 18 '11 at 15:11
4  
The trick in the continuous case is to set $z = \frac{p}{2^{n + 1}} x + \frac{q}{2^{n + 1}} y$ with $p + q = 2^{n + 1}$ and proceed by induction. –  Jonas Teuwen Nov 18 '11 at 15:15
    
Am I wrong here? Assuming continuity, isn't that the definition of convexity? (coincidentally though it is also called Jensen's Inequality) –  picakhu Nov 18 '11 at 15:35
    
@picakhu: Convexity is that $f(t_1x+t_2y)\leq t_1f(x) + t_2f(y)$ when $t_1+t_2=1$ - that is, the whole line segment betwen $(x,f(x))$ and $(y,f(y))$ is "above" the graph of $f$. –  Thomas Andrews Nov 18 '11 at 15:39

3 Answers 3

up vote 8 down vote accepted

Bellow is the proof of the fact that every midpoint convex is rationally convex which I copied from my older post on a different forum.

If you add the condition that $f$ is continuous, then from rational convexity you will get convexity. (Note that if you are interested only in continuous functions, then it suffices to show the validity of $f(t x + (1-t)y)\le t f(x) + (1-t)f(y)$ for $t=\frac k{2^n}$ as suggested in Jonas' comment. The proof of this fact is a little easier. I've given a little more involved proof, since the relation between midpoint convexity and rational convexity seems to be interesting on its own.)

Maybe I should also mention that midpoint convex functions are called Jensen convex by some authors.

Note that without some additional conditions on $f$, midpoint convexity does not imply convexity, see this question: A counterexample for Big Rudin's Chapter 3 Exercise 4


Let $f: R\to R$ be a midpoint convex function, i.e. $$f\left(\frac{x+y}2\right) \le \frac{f(x)+f(y)}2$$ for any $x,y \in R$.

We will show that then this function fulfills $$f(t x + (1-t)y)\le t f(x) + (1-t)f(y)$$ for any $x,y\in R$ and any rational number $t\in\langle0,1\rangle$.

Hint: Cauchy induction: see wikipedia or AoPS.

Proof. It is relatively easy to see, that it suffices to show $f([x_1+\dots+x_k]/k)\le [f(x_1)+\dots+f(x_k)]/k$ for any integer $k$ (and any choice of $x_1,\dots,x_k\in R$).

The case $k=2^n$ is a straightforward induction.

Now, if $2^{n-1}<k\le 2^n$ then we denote $\overline x=\frac{x_1+\dots+x_k}k$. Now from $$f(\overline x)=f\left(\frac{x_1+\dots+x_k+\overline x+\dots+\overline x}{2^n}\right) \le\frac{f(x_1)+\dots+f(x_k)+(2^n-k)f(\overline x)}{2^n}$$ we get $kf(\overline x) \le f(x_1+\dots+f(x_k)$ by a simple algebraic manipulation.


The fact that measurability of $f$ is enough for the implication midpoint convex $\Rightarrow$ convex to hold was mentioned in some of the comments above and in answers to the question I linked. Some references for this fact:

Constantin Niculescu, Lars Erik Persson: Convex functions and their applications, p.60:

H. Blumberg [31] and W. Sierpinski [226] have noted independently that if $f : (a, b) \to \mathbb R$ is measurable and midpoint convex, then $f$ is also continuous (and thus convex). See [212, pp. 220.221] for related results.

[31] H. Blumberg, On convex functions, Trans. Amer. Math. Soc. 20 (1919), 40–44.

[212] A. W. Roberts and D. E. Varberg, Convex Functions, Academic Press, New York and London, 1973.

[226] W. Sierpinski, Sur les fonctions convexes mesurables, Fund. Math. 1 (1920), 125–129.

Marek Kuczma: An introduction to the theory of functional equations and inequalities, p.241. He mentions the book T. Bonnesen and W. Fenchel, Theorie der konvexen Körper, Berlin, 1934 as an additional reference.

share|improve this answer
    
Not to be too much of a whiner, but that "Spoiler" trick is annoying when the point of asking a question here is to get an answer. It's not like accidentally glancing at your text is going to reveal the answer, so if I want to avoid reading it, I avoid reading it. In addition, there is no way to show your text on displays that do not have "mouse-over" - it stays blank on my iPad no matter what I do, for example. There might be fora and questions where it is appropriate, but this doesn't sem to be one of them. –  Thomas Andrews Nov 18 '11 at 15:51
    
Sorry @Thomas, I've removed it. –  Martin Sleziak Nov 18 '11 at 16:01

Can you provide any more information about $f$? For example, the property holds for continuous functions $f: I \rightarrow \mathbb{R}$, $I$ being an interval of real numbers. I think this result is due to Jensen [1].

Theorem (Jensen). Let $f: I\rightarrow\mathbb{R}$ be a continuous function. Then $f$ is convex if and only if it is midpoint convex, i.e. for $x,y$ in $I$ we have

$$ f\left(\frac{x+y}{2}\right) \leq \frac{f(x)+f(y)}{2} $$

[1] J. L. W. V. Jensen, Sur les fonctions convexes et les inégalités entre les valeurs moyennes, Acta Math., 30 (1906), 175-193.

share|improve this answer

Given a weight lambda, take its binary expansion. Think of what midpoint convexity implies to the inequalities involving the partial sums of this binary expansion. Then the result follows by the continuity of f.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.