Strong convexity on sets?

Consider the definition of convex functions: $$f(tx+(1-t)y) \le t f(x)+(1-t)f(y)$$ It is easy to show the definition of the convexity on sets with respect to the above definition (Specifically for a convex hull of some points $x_1, ...., x_n$, it is the points $\sum_i \alpha_i x_i$ and $\sum_i \alpha_i = 1$).

Consider the definition of strong convex functions: $$f(tx+(1-t)y) \le t f(x)+(1-t)f(y) - \frac{1}{2} m t(1-t) \|x-y\|_2^2$$ for some $m > 0$. How can we reinterpret this definition for sets (or discrete points)?

• I think @gerw's answer is interesting and I voted it up. But frankly I have to ask why you're asking. Oh, sure, as a mental exercise I get it. But what would the purpose of the definition be? What could we do with these strongly convex sets that we could not do with non-strongly convex sets? My guess is that if you have a specific purpose in mind, a definition is likely to follow readily. – Michael Grant Nov 2 '14 at 19:26
• Thanks for asking. I am curious if we can model a family of submodular functions (en.wikipedia.org/wiki/Submodular_set_function) such that their continuous extension (e.g. something like Lovasz extension) is strongly convex. Any comments on than? – Daniel Nov 2 '14 at 20:04
• I do not have an idea at the moment, no, but I will ruminate on it! – Michael Grant Nov 3 '14 at 4:05

A set $A$ is strongly convex, if there is some $m > 0$ with $$x,y \in A \;\Rightarrow\; \lambda \, x + (1-\lambda) \, y + m \, \lambda \, (1-\lambda) \, \mathbb{B} \subset A,$$ where $\mathbb{B}$ is the unit ball. I must admit that I do not know if this definition is used anywhere.
• I think that is not possible. You could, however, define a $m$-strongly-convex hull by intersecting all $m$-strongly-convex sets containing your initial set. – gerw Nov 2 '14 at 19:18