19.1 Convexity
Definition 19.1.1 (Epigraph).label Let $E$ be a vector space over $\real$ and $f: E \to (-\infty, \infty]$, then the epigraph of $f$ is the set
Definition 19.1.2 (Convex Function).label Let $E$ be a vector space over $\real$ and $f: E \to (-\infty, \infty]$, 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 $x, y \in E$ and $t \in [0, 1]$, then
so $f((1 - t)x + ty) \le (1 - t)f(x) + tf(y)$.$\square$
Lemma 19.1.3.label Let $E$ be a vector space over $\real$ and $f: E \to (-\infty, \infty]$, then $f$ is convex if and only if $\bracs{f < \infty}$ is convex and $f|_{\bracs{f < \infty}}$ is a convex function.
Proof. If $f$ is convex, then for any $x, y \in \bracs{f < \infty}$ and $t \in [0, 1]$,
so $\bracs{f < \infty}$ is convex, and the restriction $f|_{\bracs{f < \infty}}$ is a convex function.
On the other hand, for any $x, y \in E$ and $t \in [0, 1]$, if $f(x) = \infty$ or $f(y) = \infty$, then
Otherwise, $x, y \in \bracs{f < \infty}$, and
by convexity of $f|_{\bracs{f < \infty}}$.$\square$
Lemma 19.1.4.label Let $E$ be a vector space over $\real$ and $f: E \to (-\infty, \infty]$ be convex, then for any $x, y \in E$ and $t \in \real \setminus [0, 1]$,
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 19.1.5.label Let $E$ be a vector space over $\real$, $f: E \to (-\infty, \infty]$ be convex, and $x \in \bracs{f < \infty}$, then for any $h \in E$,
exists in $[-\infty, \infty]$.
Proof. Let $0 < s \le t$, then since $f$ is convex,
$\square$
Proposition 19.1.6.label Let $E$ be a vector space over $\real$, then:
- (1)
For any $f, g: E \to (-\infty, \infty]$ convex and $\lambda \ge 0$, $\lambda f + g$ is convex.
- (2)
For any convex functions $\cf \subset (-\infty, \infty]^{E}$, $\sup_{f \in \cf}f$ is convex.
Post a Comment