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!

  • 3
    $\begingroup$ I suspect that you need continuity. $\endgroup$
    – copper.hat
    Sep 19 at 1:10

1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .