diff --git a/src/fa/convex/del.tex b/src/fa/convex/del.tex new file mode 100644 index 0000000..8e8958e --- /dev/null +++ b/src/fa/convex/del.tex @@ -0,0 +1,59 @@ +\section{Subgradients} +\label{section:subgradient} + +\begin{definition}[Subdifferential] +\label{definition:subgradient} + Let $E$ be a locally convex space over $\real$, $f: E \to (-\infty, \infty]$, $x \in \bracs{f < \infty}$, and $\phi \in E^*$, then $\phi$ is a \textbf{subgradient} of $f$ if for any $h \in E$, + \[ + f(x + h) \ge f(x) + \dpn{h, \phi}{E} + \] + + The set $\partial f(x)$ of all subgradients of $f$ at $x$ is the \textbf{subdifferential of $f$ at $x$}, and the mapping $\partial f$ is the \textbf{subdifferential} of $f$. +\end{definition} + +\begin{proposition} +\label{proposition:subgradient-gymnastics} + Let $E$ be a locally convex space over $\real$, $f, g: E \to (-\infty, \infty]$, and $x \in \bracs{f, g < \infty}$, then: + \begin{enumerate} + \item $\partial f(x)$ is a weak*-closed convex set. + \item $\partial(f + g)(x) \supset \partial f(x) + \partial g(x)$. + \end{enumerate} +\end{proposition} +% Proof omitted. + +\begin{proposition} +\label{proposition:subgradient-existence} + Let $E$ be a locally convex space over $\real$, $f: E \to (-\infty, \infty]$ be convex, and $x \in \bracs{f < \infty}$, then: + \begin{enumerate} + \item For any $\phi \in \partial f(x)$ and $h \in E$, + \[ + \dpn{h, \phi}{E} \le |f(x + h) - f(x)| + \] + \item If $f$ is continuous at $x$, then $\partial f(x)$ is non-empty, equicontinuous, and weak*-compact. + \end{enumerate} +\end{proposition} +\begin{proof}[Proof, {{\cite[Proposition 4.6]{Clarke}}}. ] + (1): By the subgradient inequality, + \[ + \dpn{h, \phi}{E} \le f(x + h) - f(x) \le |f(x + h) - f(x)| + \] + + (2): Since $f$ is continuous at $x$, $\text{epi}(f)^o \ne \emptyset$. Since $(x, f(x)) \not\in \text{epi}(f)^o$, by the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-1}, there exists $\phi \in E^*$ and $\lambda \in \real$ such that for any $(y, \alpha) \in \text{epi}(f)^o$, + \[ + \dpn{y, \phi}{E} + \lambda \alpha < \dpn{x, \phi}{E} + \lambda f(x) + \] + + By continuity of $f$ at $x$, there exists $\alpha_0 \in \real$ such that $\bracs{x} \times [\alpha_0, \infty) \subset \text{epi}(f)^o$. Thus $\lambda < 0$. + + By \autoref{proposition:convex-interior-closure}, $\text{epi}(f) \subset \ol{\text{epi}(f)^o}$. Therefore for any $(y, \alpha) \in \text{epi}(f)$, + \[ + \lambda^{-1}\dpn{y, \phi}{E} - \alpha \le \dpn{x, \phi}{E} - f(x) + \] + + so $\phi \in \partial f(x) \ne \emptyset$. + + (3): By \autoref{proposition:subgradient-gymnastics} and the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}. +\end{proof} + + + diff --git a/src/fa/convex/index.tex b/src/fa/convex/index.tex index 56ad9e8..60a43c3 100644 --- a/src/fa/convex/index.tex +++ b/src/fa/convex/index.tex @@ -2,3 +2,5 @@ \label{chap:convex-functions} \input{./def.tex} +\input{./del.tex} +\input{./legendre.tex} \ No newline at end of file