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?

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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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.

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    $\begingroup$ 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

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