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.

I feel like this should be straightforward, but does anyone have a proof of the following?

Let $f: \mathbb{R}^n \to \mathbb{R}$ satisfy the following. For each coordinate $i$, for an arbitrary vector $x_{-i}$, define $f_i(y)$ to be $f$ restricted to the $i$th parameter, fixing the others $x_{-i}$; then $f_i(y)$ is convex. Then $f$ is a convex function.

Also, if there is a more standard math notation I should use to describe this problem, what would that be?

(Edit: or a counterexample...)

Thanks!


I should add what I've tried.

$g(\lambda \vec{x} + (1-\lambda)\vec{y}) = g(\lambda x_1 + (1-\lambda)y_1,\dots,\lambda x_n + (1-\lambda)y_n)$

$\leq \lambda g(x_1,\lambda x_2 + (1-\lambda)y_2,\dots) + (1-\lambda) g(y_1,\lambda x_2 + (1-\lambda)y_2,\dots)$

I don't see this going anywhere good, we end up with something nasty like

$\lambda^n g(\vec{x}) + \lambda^{n-1}(1-\lambda) \sum_i g(y_i,\vec{x}_{-i}) + \dots + (1-\lambda)^n g(\vec{y})$

I think. Maybe we can get it from there?

share|improve this question

1 Answer 1

up vote 0 down vote accepted

Simple counterexample (assuming I understood your setup): $xy$ is nonconvex while trivially convex (linear) when fixing one of the variables.

share|improve this answer
    
Ha, yep. Thanks! –  mathison Oct 12 '12 at 8:49

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.