This is a follow-up of a popular exercise found in Rudin's Real and Complex analysis.

It is known that if a continuous function $f:\left]a,b\right[\to \bf R$ satisfies the inequality $f((x+y)/2)\le 1/2(f(x)+f(y))$ for every $a<x<y<b$, then $f$ is convex on $\left]a,b\right[$.

In general, if one does not assume $f$ continuous, the conclusion is incorrect, e.g. because one can prove (assuming the axiom of choice) the existence of pathological functions $f:\bf R\to R$ that are $\bf Q$-linear but not $\bf R$-linear.

My question: does the conclusion still hold if one only assumes that $f$ is measurable?

Another question for the sake of curiosity: does there exist a function $f$ satisfying the above inequality, that are neither continuous nor additive?

  • 4
    $\begingroup$ As noted at en.wikipedia.org/wiki/Convex_function#cite_note-3 , "a real-valued Lebesgue measurable function that is midpoint-convex will be convex by the Sierpinski theorem". $\endgroup$ – Iosif Pinelis Nov 18 '18 at 4:19
  • 1
    $\begingroup$ If $f: \mathbb{R} \to \mathbb{R}$ satisfies $f((x+y)2) = (1/2)(f(x) + f(y))$, then $g(x) = \exp(f(x))$ satisfies $g((x+y)/2)^2 = g(x)g(y) \leq ((g(x) + g(y))/2)^2$, forcing midpoint convexity but also discontinuity and non-additivity. $\endgroup$ – Todd Trimble Nov 18 '18 at 4:25
  • 1
    $\begingroup$ For the second questipn: alternatively, you may jusr take the sum of a convex and an additive functions. $\endgroup$ – Ilya Bogdanov Nov 18 '18 at 6:13

Since second question was addressed in comments I will answer only the first one.

I will prove that $f$ is continuous on $(a, b)$ from which by exercise from Rudin everything follows.

Consider sets $L_n = \{x\in (a, b) : f(x) < n\}$. They are measurable and their union is the whole $(a, b)$ so for some $n$ we have $|L_n| > 0$. Note that by midpoint-convexity we have $\frac{1}{2}(L_n + L_n) \subset L_n$. By the classical result $\frac{1}{2}(L_n + L_n)$ contains some interval. Thus, $f < n$ on some $I \subset (a, b)$. For simplicity let us assume that $n = 0$ (this can be done by considering $g = f -n$). Now for some time we will only work on $I$ and not on the whole $(a, b)$.

Let $x\in I$. Let us prove that $f$ is continuous in $x$. Let $n$ be some natural number. It is easy to prove by induction that $\frac{1}{2^n}f(w) + \frac{2^n - 1}{2^n}f(z) \ge f(\frac{1}{2^n}w + \frac{2^n - 1}{2^n}z)$ for all $w, z\in I$. If we assume that $z = x$ and $\frac{1}{2^n}w + \frac{2^n - 1}{2^n}z = y$ so close to $x$ that $w$ is also in $I$ then we have

$$ \frac{2^n - 1}{2^n} f(x) > \frac{1}{2^n}f(w) + \frac{2^n - 1}{2^n}f(z) \ge f(y).$$

On the other hand if we put $z = y$ and $\frac{1}{2^n}w + \frac{2^n - 1}{2^n}z = y$ (again we assume that $y$ is so close to $x$ that $w\in I$) we get $\frac{2^n - 1}{2^n}f(y) > f(x)$. From these two facts we get that $f$ is continuous in $x$ by letting $n$ go to infinity. Since $x$ is arbitrary we have that $f$ is continuous on $I$.

Let $(c, d)$ be the biggest interval containing $I$ on which $f$ is continuous and assume that $(c, d) \subsetneq (a, b)$, for example $d < b$. Let $[r, s]\subset (c, d)$ be any interval (so that $f$ is bounded on $[r, s]$ by some constant $M$). It is easy to see that we can get any point $w \in [s, \frac{b+d}{2}]$ as linear combination of some point from $[r, s]$ and $\frac{b+d}{2}$ with weights which are dyadic rationals. From this it is easy to see that $f(w) \le \max(M, f(\frac{d+b}{2}))$. So $f$ is bounded on $[r, s]$ and on $[s, \frac{b+d}{2}]$ and so it is bounded on $[r, \frac{b+d}{2}]$. repeating argument from the beginning of this post we get that $f$ is continuous on $(r, \frac{b+d}{2})$ and this contradicts maximality of $(c, d)$. Thus $f$ is continuous on $(a, b)$ and we are done.

  • 1
    $\begingroup$ Ouch, I somehow was able to miss comment of Iosif Pinelis stating that the answer to the first question is known even though I read comment of Ilya Bogdanov about second one. $\endgroup$ – Aleksei Kulikov Nov 18 '18 at 12:35
  • $\begingroup$ Many thanks for these comments, which completely clarify my questions. $\endgroup$ – jacaboul Nov 18 '18 at 16:59
  • $\begingroup$ In fact, once proven that $f$ is bounded on an nonempty open subinterval, one is done, as it is also true (and not very difficult to show) that if a mid-convex function is discontinuous at a point, then it is unbounded in any open subinterval of $]a,b[$. $\endgroup$ – jacaboul Nov 18 '18 at 18:30

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

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