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For a function to be convex it should have a convex domain besides the first-order, second-order conditions. Why do we need that the domain of the function should also be convex?

Edit:

What I am trying to say is that its possible for a function to be convex on a non-convex domain. Here is a link: link

Then why add the convexity condition specifically???

Shirin
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1 Answers1

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You need the condition on the domain, because the definition of the convex function implies calculating $f(\lambda x+(1-\lambda)y)$ for every $\lambda \in (0,1)$ and for every $x,y \in D$.

For the function to be well defined you need that $$ x,y \in D,\lambda \in (0,1) \Rightarrow \lambda x+(1-\lambda)y \in D$$ and that is exactly the property which characterizes convex domains.

Beni Bogosel
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  • You are right that what the definition says. But there must be some reason for this definition. I mean can you give me a geometrical example where if I remove this condition and check if the function is still convex it fails. – Shirin Feb 16 '13 at 10:38
  • @Shaun: If you remove the condition on the domain, no function defined on it can be convex... It makes no sense to talk about convex functions on non convex domains... – Beni Bogosel Feb 16 '13 at 10:41