Definition 13.1.2 (Convex Function).label Let $E$ be a vector space over $\real$, $A \subset E$ be convex, and $f: A \to (-\infty, \infty]$, then the following are equivalent:
- (1)
For every $x, y \in E$ and $t \in [0, 1]$,
\[f((1 - t)x + ty) \le (1 - t)f(x) + tf(y)\] - (2)
$\text{epi}(f)$ is convex.
If the above holds, then $f$ is convex.
Proof. (1) $\Rightarrow$ (2): Let $(x, \alpha), (y, \beta) \in \text{epi}(f)$ and $t \in [0, 1]$, then
\begin{align*}(1 - t)\alpha + t\beta&\ge (1 - t)f(x) + tf(y) \\&\ge f((1 - t)x + ty)\end{align*}
so $((1 - t)x + ty, (1 - t)\alpha + t\beta) \in \text{epi}(f)$.
(2) $\Rightarrow$ (1): Let $x, y \in A$ and $t \in [0, 1]$, then
\[((1 - t)x + ty, (1 - t)f(x) + tf(y)) \in \text{epi}(f)\]
so $f((1 - t)x + ty) \le (1 - t)f(x) + tf(y)$.$\square$
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