# Prove that alpha convexity implies convexity

I encountered recently the concept of alpha convexity for functionals defined on normed spaces. The definition is as follow

A function $$f : K\subset E\to \mathbb{R}$$ (where $$K$$ is a convex subset of $$E$$) is said to be $$\alpha$$ convex if there exists $$\alpha>0$$ such that

$$f(\frac{x+y}{2})\leq \frac{f(x)+f(y)}{2} -\frac{\alpha}{8}\lVert x-y\rVert^{2}$$

My idea was to use the convexity of $$K$$ and consider $$w=\theta x + (1-\theta)y$$ in order to make appear some convex combination. However what I found did not help me as I wanted :

$$f(\frac{x+\theta x + (1-\theta)y}{2})\leq\frac{f(x)+f(\theta x + (1-\theta)y)}{2} -\frac{\alpha(1-\theta)^2}{8}\lVert x-y\rVert^{2}$$

I have also noticed that we have the convexity for $$\theta=\frac{1}{2}$$.

I would like to have not a already made proof but some helps or hints for this result please in order to work on this a little more.

Thanks in advance!

• I suspect that you need continuity. Commented Sep 19, 2023 at 1:10

## 1 Answer

If your function is additionally continuous, then this implies convexity. This is because $$\alpha$$-convexity implies midpoint-convexity, i.e., for every $$x,y\in K$$, $$f\left(\frac{x+y}{2}\right)\leq\frac{f(x)+f(y)}{2}.$$ However, it is a standard result that a continuous function which is midpoint-convex is also convex.