# If $f(\frac{t_1+t_2}{2}) \leq \frac{f(t_1)+f(t_2)}{2}$, show that $f(\frac{t_1+t_2+ \cdots +t_n}{n})\leq \frac{f(t_1)+f(t_2)+\cdots f(t_n)}{n}$ [duplicate]

If $$f$$ is a continuous function on $$[0,1]$$ such that for all $$t_1, t_2 \in [0,1]$$,

$$f\left(\frac{t_1+t_2}{2}\right) \leq \frac{f(t_1)+f(t_2)}{2}$$

Show that $$f\left(\frac{t_1+t_2+ \cdots +t_n}{n}\right)\leq \frac{f(t_1)+f(t_2)+\cdots f(t_n)}{n}$$

I tried proving this statement by induction. I could do this for even $$n$$.

Let $$k=n/2$$, then, \begin{align} f\left(\frac{t_1+t_2+ \cdots +t_n}{n}\right) &= f\left(\frac{t_1 + \cdots +t_k}{n} + \frac{t_{k+1} + \cdots +t_n}{n}\right) \\ &=f\left( \frac{\frac{2}{n}(t_1 + \cdots +t_k)}{2}+\frac{\frac{2}{n}(t_{k+1} + \cdots +t_n)}{2}\right)\\ &\leq \frac{f\left(\frac{2}{n}(t_1 + \cdots +t_k)\right)+f\left(\frac{2}{n}(t_{k+1} + \cdots +t_n)\right)}{2}\\ &=\frac{f(\frac{t_1+\cdots +t_k}{k})+f(\frac{t_{k+1}+\cdots +t_n}{k})}{2}\\ &\leq\frac{f(t_1)+f(t_2)+\cdots f(t_n)}{2k}\\ &=\frac{f(t_1)+f(t_2)+\cdots f(t_n)}{n} \end{align}

However, I don't know how to split should $$n$$ be odd. One reasonable thing I thought of was to split the middle term. For instance, if $$n=3$$, We could split it as such $$t_1 + 0.5 t_2$$ and $$0.5 t_2 + t_3$$. However, I couldn't find a neat way to carry out the induction if I do it like this. Can someone give a hint?

• See the answer of Martin Sleziak here : math.stackexchange.com/questions/83383/… Commented Apr 30, 2020 at 10:21
• @ChocoSavour: Since $f$ is continuous we can indeed conclude that it is convex (as in the Q&A that you linked to). But actually the conclusion holds without the continuity condition and can be proven by induction only (as in the answer that I pointed to). Commented Apr 30, 2020 at 11:27
• @MartinR In the answer of Martin Sleziak (see the link below), the proof of "mid-point convex implies the property the author talk about" is also given without the assumption of $f$ continuous, see the part where he proved it by induction only (which is similar to yours) in the mid-part of his answer. Commented Apr 30, 2020 at 11:51
• Thanks for the responses, however, I'm not too familiar with convex functions or Jensen's inequality. Rozenberg's answer is the most intuitive for me. Commented Apr 30, 2020 at 12:02

Now, by your work you can use induction and the following reasoning as the proof for $$n=3$$: $$\frac{f(a)+f(b)+f(c)+f\left(\frac{a+b+c}{3}\right)}{4}\geq f\left(\frac{a+b+c+\frac{a+b+c}{3}}{4}\right)$$ or $$\frac{f(a)+f(b)+f(c)}{3}\geq f\left(\frac{a+b+c}{3}\right).$$ Can you end it now?