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
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
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!