Convexity is an important notion in mathematics with many applications to real-world problems. In this post, I would like to prove a few basic results for convex functions on Euclidean space. Recall that a subset of is said to be convex if, for all ,

Suppose is a real-valued function defined on a convex subset . We say that is a convex function if, for all ,

If the preceding inequality holds for the special case , then we say that is midpoint convex. There are number of interesting things we can say about convex functions. One is that every local minimum is a global minimum. Indeed, suppose is a local minimum of a convex function . Then there exists an open ball , such that for all . For ,

and therefore, since ,

Dividing both sides by completes the proof. Furthermore, any increasing convex function of a convex function is also convex. Suppose is increasing, and is convex. By convexity, for ,

and since is increasing on , we conclude that

We now restrict ourselves to convex functions defined on a convex subset of the real line. Any convex function defined on an interval is continuous. Indeed,

If we have a convex function , then we can define a function by

Observe that is symmetric in and . It turns out that is convex if and only if is monotonically nondecreasing for or fixed. Fix . First, I claim that we can write , where . Indeed,

By convexity, . Rearranging and dividing both sides by , we see that

Subtracting from both sides and dividing by in our first convexity inequality, we also obtain the inequality

Retracing our steps, we see that the preceding inequalities give a sufficient condition for to be convex, which completes the proof. We note for the interested reader.

We now prove some more interesting results about convex functions related to continuity, midpoint convexity, and Lebesgue measurability.

Lemma 1.Let be midpoint convex; ; and with . Then

*Proof.* First, I claim that

If , then the assertion is immediate from the hypothesis of midpoint convexity. Suppose the claim holds for . Then by midpoint convexity,

where we apply the induction hypothesis in the penultimate inequality. For arbitrary , there exists such that . Hence,

Let be the least common denominator of . Set , and . We have that

Hence,

Obviously every convex function is also midpoint convex, since we just take , but if we impose the additional hypothesis that is continuous, then turns out to be convex. We note for the reader’s benefit that this is Exercise 24, Chapter 4 of Baby Rudin.

Lemma 2.If is continuous and midpoint convex, then is convex.

*Proof. *Let and . Then there exists a sequence of positive rational numbers such that . Applying the preceding lemma to and using the continuity of , we have that

Lemma 3.If is midpoint convex and has a point of discontinuity , then .

*Proof. *By hypothesis that is discontinuous at , there exists such that , there exists such that . Set . Then by midpoint convexity,

If , then by choice of , . So either in any neighborhood of there exists a point such that . Set . By midpoint convexity,

By induction, we construct a sequence such that and as .

Sierpinski showed that if is midpoint and is also Lebesgue measurable, then is continuous. This result was not known to me, until I stumbled across it in a textbook.

Lemma 4.(Sierpinski) If is Lebesgue measurable and midpoint-convex, then is continuous.

*Proof.* Suppose that there exists a point of discontinuity . The previous lemma shows that there exists an interval and a sequence , such that . Set

Since is measurable, and are measurable, and it is evident that have the same Lebesgue measure. I claim that . Indeed, if , then either . Otherwise, , and setting , we obtain by midpoint convexity that

so . We then have that

which is a contradiction since is finite-valued (or finite-valued a.e.) and therefore as .

Combining the preceding lemmas, we conclude the following:

Proposition 5(Blumberg-Sierpinski) Let be a measurable, midpoint-convex function. Then is continuous, and a fortiori convex.

Lemma 6.Let be a measure space. If for some , then and

*Proof. *The first assertion is just interpolation. For the second assertion, let be given. If is a simple function, then

Since can be approximated by a simple functions from below, we have that

We now need to show that . Let . Then is measurable, and I claim that has positive measure by definition of essential supremum. Indeed,

Taking the ,

Since was arbitrary, we obtain the desired inequality.

Lemma 7.Let be a measure space such that , and be a convex function on . If , then

*Proof. *Since is convex, there exists such that

for all . Integrating both sides yields and using the fact that ,

Note that if is concave, then the inequality is just reversed. We conclude this post with the well-known inequality attributed to Johan Jensen.

Proposition 8.(Jensen’s Inequality) Suppose that . Then, for all ,If and for some , then

with the interpretation .

*Proof. *Let . If , then the stated inequality is trivially true, so assume otherwise. By Jensen’s inequality for a convex function,

which implies that

and therefore

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Really Nice, Helped a lot.

Hi,

how do you know that $ x_n \rightarrow x_0 $ as $ n \rightarrow \infty $ in lemma 3?

The proof appears to be incorrect. I intend to post the correction later today. Thank you for catching it!

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