2
$\begingroup$

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.

2 Answers 2

1

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.

  • 0
    Ah yes, I see what you mean, thanks!2012-09-27
0

the idea you used shows $f$ is convex, for showing that $f$ is convex homogenous or equivalently $epi f$ is convex cone you need to show that

$f(\theta_1 x_1 + \theta_2 x_2) \le \theta_1 f(x_1) + \theta_2 f(x_2)$ for all $\theta_{i} \ge 0 $