Subgradient

Subgradient

Das Subdifferential ist eine Verallgemeinerung des Gradienten. Für eine Funktion f:\mathbb{R}^n\rightarrow\mathbb{R} ist das Subdifferential im Punkt \bar x gegeben durch

\partial f(\bar x)=\{g\in\mathbb{R}^n:f(x)-f(\bar x)\geq g^{T}(x-\bar x) für alle x\in\mathbb{R}^n\}

Die Elemente g\in\partial f(\bar x) heißen Subgradienten.

Beispiel

Subgradienten einer konvexen Funktion

Das Subdifferential von der Funktion f:\mathbb{R}\rightarrow\mathbb{R}:x\mapsto|x| ist gegeben durch:

\partial f(\bar x)=\begin{cases}\{-1\} & \bar x<0\\
\left[-1,1\right] & \bar x=0\\ \{1\} & \bar x>0\end{cases}


Beschränktheit

Sei f:\mathbb{R}^n\rightarrow\mathbb{R} stetig und sei X\subset\mathbb{R}^n beschränkt. Dann ist die Menge \bigcup_{\bar x\in X}\partial f(\bar x) beschränkt.

Beweis

Sei f:\mathbb{R}^n\rightarrow\mathbb{R} stetig und sei X\subset\mathbb{R}^n beschränkt. Setzte \epsilon:=\sup |f(\overline{U_1(X)})| wobei \overline{U_1(X)}=\{x\in\mathbb{R}^n\,|\,{\rm dist}(x,X)\leq1\}. Angenommen \bigcup_{\bar x\in X}\partial f(\bar x) ist nicht beschränkt, dann gibt es für R: = 2ε ein \bar x\in X und ein g\in\partial f(\bar x) mit \|g\|_2>R=2\epsilon. Sei x:=\frac{1}{\|g\|_2} g+\bar x. Somit sind \bar x,x\in\overline{U_1(X)}. Wir erhalten die Abschätzung

g^T(x-\bar x)=\frac{1}{\|g\|_2}g^T g=\|g\|_2 > 2\epsilon\geq\left|f(x)-f(\bar x)\right|\geq f(x)-f(\bar x) .

g ist also kein Subgradient. Das ist ein Widerspruch.

\Box

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