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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.

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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.

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

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