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'm inclined to make this claim because the functions epigraph is $\{(x,t) : t \ge f(x)\}$. But to be a convex cone, it must be closed under the usual

$$\theta_1 (x_1,t_1) + \theta_2 (x_2,t_2)$$

for $\sum \theta = 1$ and $\theta_i \in [0,1]$. But that then implies:

$$\theta_1 t_1 + \theta_2 t_2 \ge f(\theta_1 x_1 + \theta_2 x_2) $$

which is the same as

$$f(\theta_1 x_1 + \theta_2 x_2) \le \theta_1 f(x_1) + \theta_2 f(x_2)$$

implying $f$ is convex.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

Your reasoning is almost correct. However, you switch from $t_i$ to $f(x_i)$ without justification – you should start out with $f(x_i)$ instead of $t_i$ from the beginning; this is a special case of the general property you're using.

share|improve this answer
    
Ah yes, I see what you mean, thanks! –  Palace Chan Sep 27 '12 at 2:48

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.