110 lines
3.7 KiB
TeX
110 lines
3.7 KiB
TeX
\section{Convexity}
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\label{section:convex-functions}
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\begin{definition}[Epigraph]
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\label{definition:epigraph}
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Let $E$ be a vector space over $\real$ and $f: E \to (-\infty, \infty]$, then the \textbf{epigraph} of $f$ is the set
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\[
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\text{epi}(f) = \bracs{(x, y) \in E \times \real| y \ge f(x)}
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\]
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\end{definition}
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\begin{definition}[Convex Function]
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\label{definition:convex-function}
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Let $E$ be a vector space over $\real$ and $f: E \to (-\infty, \infty]$, then the following are equivalent:
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\begin{enumerate}
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\item For every $x, y \in E$ and $t \in [0, 1]$,
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\[
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f((1 - t)x + ty) \le (1 - t)f(x) + tf(y)
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\]
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\item $\text{epi}(f)$ is convex.
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\end{enumerate}
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If the above holds, then $f$ is \textbf{convex}.
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\end{definition}
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\begin{proof}
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(1) $\Rightarrow$ (2): Let $(x, \alpha), (y, \beta) \in \text{epi}(f)$ and $t \in [0, 1]$, then
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\begin{align*}
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(1 - t)\alpha + t\beta &\ge (1 - t)f(x) + tf(y) \\
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&\ge f((1 - t)x + ty)
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\end{align*}
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so $((1 - t)x + ty, (1 - t)\alpha + t\beta) \in \text{epi}(f)$.
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(2) $\Rightarrow$ (1): Let $x, y \in E$ and $t \in [0, 1]$, then
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\[
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((1 - t)x + ty, (1 - t)f(x) + tf(y)) \in \text{epi}(f)
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\]
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so $f((1 - t)x + ty) \le (1 - t)f(x) + tf(y)$.
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\end{proof}
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\begin{lemma}
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\label{lemma:convex-domain}
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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.
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\end{lemma}
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\begin{proof}
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If $f$ is convex, then for any $x, y \in \bracs{f < \infty}$ and $t \in [0, 1]$,
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\[
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f((1 - t)x + ty) \le (1 - t)f(x) + tf(y) < \infty
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\]
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so $\bracs{f < \infty}$ is convex, and the restriction $f|_{\bracs{f < \infty}}$ is a convex function.
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On the other hand, for any $x, y \in E$ and $t \in [0, 1]$, if $f(x) = \infty$ or $f(y) = \infty$, then
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\[
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f((1 - t)x + ty) \le = \infty (1 - t)f(x) + tf(y)
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\]
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Otherwise, $x, y \in \bracs{f < \infty}$, and
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\[
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f((1 - t)x + ty) \le (1 - t)f(x) + tf(y)
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\]
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by convexity of $f|_{\bracs{f < \infty}}$.
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\end{proof}
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\begin{lemma}
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\label{lemma:convex-reverse}
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Let $E$ be a vector space over $\real$ and $f: E \to (-\infty, \infty]$ be convex, then for any $x, y \in E$ and $t \in \real \setminus [0, 1]$,
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\[
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f((1 - t)x + ty) \ge (1 - t)f(x) + tf(y)
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\]
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\end{lemma}
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\begin{proof}
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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 $tf(y) \le f(ty)$.
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\end{proof}
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\begin{proposition}
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\label{proposition:convex-differential}
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Let $E$ be a vector space over $\real$, $f: E \to (-\infty, \infty]$ be convex, and $x \in \bracs{f < \infty}$, then for any $h \in E$,
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\[
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\lim_{t \downto 0} \frac{f(x + th) - f(x)}{t} = \inf_{t > 0} \frac{f(x + th) - f(x)}{t}
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\]
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exists in $[-\infty, \infty]$.
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\end{proposition}
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\begin{proof}
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Let $0 < s \le t$, then since $f$ is convex,
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\begin{align*}
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f(x + sh) &\le \paren{1 - \frac{s}{t}}f(x) + \frac{s}{t}f(x + th) \\
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f(x + sh) - f(x) &\le \frac{s}{t}[f(x + th) - f(x)] \\
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\frac{f(x + sh) - f(x)}{s} &\le \frac{f(x + th) - f(x)}{t}
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\end{align*}
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\end{proof}
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\begin{proposition}
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\label{proposition:convex-extension}
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Let $E$ be a vector space over $\real$, then:
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\begin{enumerate}
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\item For any $f, g: E \to (-\infty, \infty]$ convex and $\lambda \ge 0$, $\lambda f + g$ is convex.
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\item For any convex functions $\cf \subset (-\infty, \infty]^E$, $\sup_{f \in \cf}f$ is convex.
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\end{enumerate}
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\end{proposition}
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