# A question about real convex functions

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 $$a, then $$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?

• 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". – Iosif Pinelis Nov 18 '18 at 4:19
• 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. – Todd Trimble Nov 18 '18 at 4:25
• For the second questipn: alternatively, you may jusr take the sum of a convex and an additive functions. – Ilya Bogdanov Nov 18 '18 at 6:13

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.

• 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. – Aleksei Kulikov Nov 18 '18 at 12:35
• Many thanks for these comments, which completely clarify my questions. – jacaboul Nov 18 '18 at 16:59
• 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 $]a,b[$. – jacaboul Nov 18 '18 at 18:30