Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

in advance excuse my use of non technical language.

This function $f(\alpha)$ and the set $S$ have the following property ($\lambda \in (0,1)$):

$\alpha_1,\alpha_2 \in S \nRightarrow \lambda\alpha_1+(1-\lambda) \alpha_2 \in S$

$\forall \alpha_1, \alpha_2 \in S$ we have that $f(\lambda\alpha_1+(1-\lambda) \alpha_2) \leq \lambda f(\alpha_1)+(1-\lambda) f(\alpha_2)$

I'm asking whether such a function $f$ has a technical name.

share|improve this question
Yuval:> shouldn't convex function be defined on convex sets? –  user1963 Feb 8 '11 at 13:42
Yes, but you appear to have defined $f$ on some superset of $S$ that is itself convex. (Your second line assumes $f$ is defined on $\lambda\alpha_1 + \left(1 - \lambda\right)\alpha_2$, but your first line states that this may not lie in $S$.) –  Rawling Feb 8 '11 at 15:34
Rawling: thanks for the clarification. Can we refer to f as being convex on the non-convex set S ? The aim is to avoid causing misunderstanding and/or confusion among mathematicians. –  user1963 Feb 8 '11 at 15:49
It depends. If you want your second line to hold for all $\lambda$, you could say that $f$ is convex on the convex hull of $S$. If you want your second line to hold only for $\lambda$ s.t. the "midpoint" lies within $S$, you'd probably have to do some explaining. –  Rawling Feb 8 '11 at 16:11
show 1 more comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.