Showing that $f$ is convex

Question

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.

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

Answer 1

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

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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.

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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.

Answer 2

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.