Proving continuity of $f$ Here is a problem I encountered some time back: Suppose $f$ is bounded for $a\leq x\leq b$ and for every pair of values $x_1$ and $x_2$ with $a\leq x_1\leq x_2 \leq b$,
$f(\frac{1}{2}(x_1+x_2))\leq \frac{1}{2}(f(x_1)+f(x_2))$. Prove that $f$ is continuous for $a<x<b$.
I tried to solve it but I could not really come with anything...
Here's an attempt.The idea is not due to me but to a friend;
By the condition given, $f(\frac{2x+2\delta}{2})\leq \frac{1}{2}(f(x+2\delta)+f(x))$,i.e. $f(x+\delta)-f(x)\leq \frac{1}{2}f(x+2\delta)-f(x)$ and in this manner, $f(x+\delta)-f(x)\leq \frac{1}{2}f(x+2\delta)-f(x)\leq \frac{1}{2^2}(f(x+4\delta)-f(x))\leq$
$ \dots \dots \dots \dots $
$\frac{1}{2^n}(f(x+2^n\delta)-f(x))$ where $a<x+2^n\delta<b$
As $\delta\to 0$,$f(x+2^k\delta)\to f(x)$ for $k=1,2,\dots n$ i.e. $f$ is continuous in the interval $(a,b)$.
I tried posting this flawed attempt on Aops , but no one has suggested how to finish off the proof using what I used.I will be happy if someone could suggest something.Thanks!
 A: Functions satisfying the inequality
$$
f\Bigl(\frac{x_1+x_2}2\Bigr)\le\frac{f(x_1)+f(x_2)}2
$$
are called midpoint convex. This is a slightly weaker condition than convexity. Continuous midpoint convex functions are convex. A beautiful result due to Sierpinski is that Lebesgue measurable midpoint convex functions are convex.
To prove that a bounded midpoint convex is continuous, argue by contradiction. Supose $f$ is discontinuous at $x_0\in(a,b)$. Without loss of generality we may assume $x_0=0$, $f(x_0)=0$.
First step.  There exists a sequence $\{x_n\}\subset(a,b)$, such that $\lim_{n\to\infty}x_n=0$ and $\lim_{n\to\infty}f(x_n)=m\ne0$. We may assume that $m>0$.
Second step. The sequence $\{2\,x_n\}$ also converges to $0$ and
$$
\begin{align}
f(x_n)=f\Bigl(\frac{0+2\,x_n}2\Bigr)\le\frac{f(0)+f(2\,x_n)}2 &\implies f(2\,x_n)\ge2\,f(x_n)\\&\implies\liminf f(2\,x_n)\ge2\,m.
\end{align}
$$
Iteration shows that
$$
\liminf f(2^k\,x_n)\ge2^k\,m,
$$
which is impossible since $f$ is bounded.
You can find this and much more in this book.
A: More generally I want to show that If $f$ is a convex function on an open interval $I$ then $f$ is continous: $f$ is said to be convex on $I$ if $\forall a,b\in I$ and every     $t\in(0,1)$, if $f$ satisfies 
$$f(tb+(1-t)a)\le tf(b)+(1-t)f(a).$$ 
Now for $a=b$, the definition says nothing. If $a<b$,then $$a<tb+(1-t)a<b;$$ and conversely, for each  $x\in (a,b)$ we can write $x=tb+(1-t)a$ by taking $t=(x-a)/(b-a)$, so $1-t=(b-x)/(b-a)$. Thus the following  is a simple restatement of of the definition: 
Proposition ONE. 

Suppose that $f:I\rightarrow \mathbb{R}$, then $f$ is convex iff $\forall a,b$ and $a<x<b$ we have 
  $$f(x)\le \frac{x-a}{b-a}f(b)+\frac{b-x}{b-a}f(a).$$ 
  And now try to prove this one by using the previous discussion: 

Proposition TWO. 

If f is a convex function on an interval $I$ and $a<b<c\in I,$ then 
  $$f(x)\ge\frac{x-a}{b-a}f(b)+\frac{b-x}{b-a}f(a) \;\; \forall b<x<c$$ 
  and 
  $$f(x)\ge\frac{c-x}{c-b}f(b)+\frac{x-b}{c-b}f(c) \;\; \forall a<x<b.$$ 

Now the Final Claim 

A convex function on an open interval is continous. 

Proof 

If $b\in I$ , then there exist $a$ and $c\in I$ with $a<b<c$. Then the last two proposition give the inequalities 
  $$\frac{x-a}{b-a}f(b)+\frac{b-x}{b-a}f(a)\le f(x)\le \frac{x-b}{c-b}f(c)+\frac{c-x}{c-b}f(b)\;\; \forall b<x<c$$
  from which it follows that 
  $$f(b+)=f(b).$$ 
  Similarly $$\frac{c-x}{c-b}f(b)\le f(x) \le \frac{x-a}{b-a}f(b)+\frac{b-x}{b-a}f(a) \;\; \forall a<x<b$$ 
  which gives 
  $$f(b-)=f(b).$$ 
  Thus $f$ is continuous at $b$,$b$ was arbitrary so $f$ is continuous on $I$,in your case $t=1/2.$

