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Recently I encountered two problems:

Notation: For $f:S\to \mathbb{R}$, where $S\subset \mathbb{R}$, if $f$ is continuous on $S$, then denote $f\in \mathcal{C}(S)$.

  1. Suppose $f:(a,b)\to \mathbb{R}$ is such that at every point in $(a,b)$ the single side limit of $f$ exists, and that $$f(\frac{x+y}{2})\leqslant \frac{f(x)+f(y)}{2},\; \forall x,y \in (a,b).$$ Show that $f\in \mathcal{C}(a,b)$.
  2. Suppose $f:(a,b)\to \mathbb{R}$ is such that $$f[\lambda x_1+(1-\lambda)x_2]\leqslant \lambda f(x_1)+(1-\lambda)f(x_2),\; \forall x_1,x_2\in (a,b),\forall \lambda \in (0,1).$$ Show that $f\in \mathcal{C}(a,b)$.

I found them a bit alike, and then I came up with the following question:

Suppose $f:I\to \mathbb{R}$, where $I$ is an interval, is such that $$f(\frac{x+y}{2})\leqslant \frac{f(x)+f(y)}{2},\; \forall x,y \in I.$$ Is it possible that $f\notin \mathcal{C}(I)$?

I've tried both proving and disproving the statement but failed. My guess is a counterexample exists. Please help.

Juggler
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1 Answers1

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It is really important that you have the one-sided limits $f(x+)$ and $f(x-)$. Let $x>x_0$. Then $$ f\left(\frac{x+x_0}2\right)\le\frac{f(x)+f(x_0)}2. $$ Letting $x\to x_0$ yields $$ f(x_0+)\le\frac{f(x_0+)+f(x_0)}2, $$ that is, $f(x_0+)\le f(x_0)$. Similarly, you prove that $f(x_0-)\le f(x_0)$. In particular, $f(x_0)\ge\frac{f(x_0+)+f(x_0-)}2$.

On the other hand, $$ f(x_0) = f\left(\frac{(x_0+\epsilon)+(x_0-\epsilon)}2\right)\le\frac{f(x_0+\epsilon)+f(x_0-\epsilon)}2 $$ and so, letting $\epsilon\to 0$ gives $f(x_0)\le\frac{f(x_0+)+f(x_0-)}2$. Hence, $$ f(x_0) = \frac{f(x_0+)+f(x_0-)}2 $$ But also (see above) $f(x_0+)\le f(x_0)$ and $f(x_0-)\le f(x_0)$. This yields $f(x_0) = f(x_0+) = f(x_0-)$.

As to 2. see for example here: Proof every convex function is continuous (Problem 10 Convex Functions Spivak)

amsmath
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