Another Definition of Convex function

Question

Let $f$ be a continuous convex function, which means that $f(\alpha x+(1-\alpha) y)\le \alpha f(x)+(1-\alpha) f(y)$ for any $x,y$ in the domain and $\alpha\in[0,1]$.

I read a statement saying that if $f$ is continous, then in order to show $f$ is convex, it suffices to show the inequality holds for $\alpha=\frac{1}{2}$ or any number in $(0,1)$. I don't know how to prove this statement. Can someone give me a hint? Thanks!

Answer 1

Given a continuous function $f$ in $I$ satisfying $ f\big(\frac{x+y}{2}\big)\le \frac{f(x)+f(y)}{2},$ $\forall x,y\in I$ you want to prove that $f$ is a convex function.

Let $n\in {\mathbb{Z}}_{\ge 1}$ and $p\in\{1,2,..., 2^n\}$. One can show by induction that:

$$f\bigg(\frac{p}{2^n}x+\left(1-\frac{p}{2^n}\right)y\bigg)\le \frac{p}{2^n}f(x)+\left(1-\frac{p}{2^n}\right)f(y) \tag{**}$$.

Since the set $\{\frac{p}{2^n}\}$ is dense in $[0,1]$, each $\lambda$ in $[0,1]$ is limit of a subsequence $(\frac{p}{2^n})_n$. Take the limit of both side in relation $(**)$

Answer 2

Use induction to prove that inequality holds when $\alpha = \frac i {2^{n}}$ and then use the fact that numbers of the form $\frac i {2^{n}}$ form a dense set in $[0,1]$.