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Fixed up convention in convex functions.

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This commit is contained in:
Bokuan Li
2026-06-24 14:18:45 -04:00
parent cddd7a4d55
commit b20ae09a0c
1 changed files with 27 additions and 8 deletions
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35
src/fa/convex/def.tex
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@@ -3,16 +3,16 @@
\begin{definition}[Epigraph]
\label{definition:epigraph}
Let $E$ be a vector space over $\real$, $A \subset E$ be convex, and $f: A \to (-\infty, \infty]$, then the \textbf{epigraph} of $f$ is the set
Let $E$ be a vector space over $\real$ and $f: E \to (-\infty, \infty]$, then the \textbf{epigraph} of $f$ is the set
\[
\text{epi}(f) = \bracs{(x, y) \in A \times \real| y \ge f(x)}
\text{epi}(f) = \bracs{(x, y) \in E \times \real| y \ge f(x)}
\]
\end{definition}
\begin{definition}[Convex Function]
\label{definition:convex-function}
Let $E$ be a vector space over $\real$, $A \subset E$ be convex, and $f: A \to (-\infty, \infty]$, then the following are equivalent:
Let $E$ be a vector space over $\real$ and $f: E \to (-\infty, \infty]$, then the following are equivalent:
\begin{enumerate}
\item For every $x, y \in E$ and $t \in [0, 1]$,
\[
@@ -32,7 +32,7 @@
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
(2) $\Rightarrow$ (1): Let $x, y \in E$ and $t \in [0, 1]$, then
\[
((1 - t)x + ty, (1 - t)f(x) + tf(y)) \in \text{epi}(f)
\]
@@ -43,7 +43,7 @@
\begin{lemma}
\label{lemma:convex-reverse}
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 $(1 - t)x + ty \in A$,
Let $E$ be a vector space over $\real$ and $f: E \to (-\infty, \infty]$ be convex, then for any $x, y \in E$ and $t \in \real \setminus [0, 1]$,
\[
f((1 - t)x + ty) \ge (1 - t)f(x) + tf(y)
\]
@@ -52,12 +52,31 @@
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 $tf(y) \le f(ty)$.
\end{proof}
\begin{proposition}
\label{proposition:convex-differential}
Let $E$ be a vector space over $\real$, $f: E \to (-\infty, \infty]$ be convex, and $x \in \bracs{f < \infty}$, then for any $h \in E$,
\[
\lim_{t \downto 0} \frac{f(x + th) - f(x)}{t} = \inf_{t > 0} \frac{f(x + th) - f(x)}{t}
\]
exists in $[-\infty, \infty]$.
\end{proposition}
\begin{proof}
Let $0 < s \le t$, then since $f$ is convex,
\begin{align*}
f(x + sh) &\le \paren{1 - \frac{s}{t}}f(x) + \frac{s}{t}f(x + th) \\
f(x + sh) - f(x) &\le \frac{s}{t}[f(x + th) - f(x)] \\
\frac{f(x + sh) - f(x)}{s} &\le \frac{f(x + th) - f(x)}{t}
\end{align*}
\end{proof}
\begin{proposition}
\label{proposition:convex-extension}
Let $E$ be a vector space over $\real$ and $A \subset E$ be convex, then:
Let $E$ be a vector space over $\real$, then:
\begin{enumerate}
\item For any $f, g: A \to (-\infty, \infty]$ convex and $\lambda \ge 0$, $\lambda f + g$ is convex.
\item For any convex functions $\cf \subset (-\infty, \infty]^A$, $\sup_{f \in \cf}f$ is convex.
\item For any $f, g: E \to (-\infty, \infty]$ convex and $\lambda \ge 0$, $\lambda f + g$ is convex.
\item For any convex functions $\cf \subset (-\infty, \infty]^E$, $\sup_{f \in \cf}f$ is convex.
\end{enumerate}
\end{proposition}