13.1 Convexity
Definition 13.1.1 (Epigraph).label Let $E$ be a vector space over $\real$, $A \subset E$ be convex, and $f: A \to (-\infty, \infty]$, then the epigraph of $f$ is the set
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
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
so $f((1 - t)x + ty) \le (1 - t)f(x) + tf(y)$.$\square$
Lemma 13.1.3.label Let $E$ be a vector space over $\real$, $A \subset E$ be convex, and $f: A \to (-\infty, \infty]$ be convex, then for any $x, y \in E$ and $t \in \real \setminus [0, 1]$ such that $(1 - t)x + ty \in A$,
Proof. Via an affine transformation, assume without loss of generality that $x = 0$ and $f(x) = 0$. By exchanging $x$ and $y$, assume without loss of generality that $t > 1$. In which case, since $f$ is convex and $y = t^{-1}ty$, $f(y) \le t^{-1}f(ty)$ and $tf(y) \le f(ty)$.$\square$
Proposition 13.1.4.label Let $E$ be a vector space over $\real$ and $A \subset E$ be convex, then:
- (1)
For any $f, g: A \to (-\infty, \infty]$ convex and $\lambda \ge 0$, $\lambda f + g$ is convex.
- (2)
For any convex functions $\cf \subset (-\infty, \infty]^{A}$, $\sup_{f \in \cf}f$ is convex.
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