The Minkowski (Gauge) Functional

# The Minkowski (Gauge) Functional

Recall from the Absorbent Sets page that if $X$ is a linear space and $E \subseteq X$ then $E$ is said to be absorbent if for every $x \in X$ there exists a $\lambda > 0$ such that $\lambda x \in E$.

Given an absorbent subset $U$ of a linear space $X$ we can define a special type of functional.

 Definition: Let $X$ be a linear space and let $U \subseteq X$ be an absorbent set. The Minkowski Functional (or Gauge Functional) of $U$ is the functional $p_U : X \to [0, \infty)$ defined for all $x \in X$ by $p_U(x) = \inf \{ \lambda > 0 : x \in \lambda U \}$.

Observe that we require that $U$ is an absorbent set so that $p_U(x) < \infty$ for every $x \in X$.

 Theorem 1: Let $X$ be an algebra and let $U \subseteq X$. a) If $U$ is an absolutely convex, absorbent, sub-semigroup, then the Minkowski functional $p_U$ is an algebra seminorm on $X$. b) If $U$ is an absolutely convex, absorbent, sub-semigroup, and radially bounded, then the Minkowski functional $p_U$ is an algebra norm on $X$.
• Proof of a): We will show that $p_U$ satisfies properties 2, 3, and 4 of the definition of an algebra seminorm.
• First, let $x \in X$ and let $\alpha \in \mathbf{F}$. If $\alpha = 0$ then $\alpha x = 0$. Since $U$ is absorbent $0 \in U$ and $0 \in \lambda U$ for every $\lambda > 0$. Therefore $p(\alpha x) = 0$. Also, $|\alpha|p(x) = 0p(x) = 0$ too.
• If $\alpha \neq 0$ then we have that:
(1)
\begin{align} \quad \{ \lambda > 0 : \alpha x \in \lambda U \} = \left \{ \lambda > 0 : x \in \frac{1}{|\alpha|} \lambda U \right \} \end{align}
• Therefore:
(2)
\begin{align} \quad p_U(\alpha x) = |\alpha|p_U(X) \end{align}
• Now let $x, y \in X$. By the definition of infimum, for all $\epsilon > 0$ there exists a $\lambda_1, \lambda_2 > 0$ such that $x \in \lambda_1 U$, $y \in \lambda_2 U$, and:
(3)
\begin{align} \quad \lambda_1 & \leq p_U(x) + \epsilon \\ \quad \lambda_2 & \leq p_U(y) + \epsilon \end{align}
• Let:
(4)
\begin{align} \quad a = \frac{\lambda_1}{\lambda_1 + \lambda_2}, \: b = \frac{\lambda_2}{\lambda_1 + \lambda_2} > 0 \end{align}
• Then $a, b \in \mathbf{F}$ are such that $|a| + |b| = 1$. Since $U$ is absolutely convex, we have that for all $z, w \in U$ that:
(5)
\begin{align} az + bw = \left ( \frac{\lambda_1}{\lambda_1 + \lambda_2} \right ) z + \left ( \frac{\lambda_2}{\lambda_1 + \lambda_2} \right ) w \in U \end{align}
• Since the above inclusion holds for every $z, w \in U$ we see that:
(6)
\begin{align} \quad \left ( \frac{\lambda_1}{\lambda_1 + \lambda_2} \right ) U + \left ( \frac{\lambda_2}{\lambda_1 + \lambda_2} \right ) U \subseteq U \end{align}
• Or equivalently:
(7)
\begin{align} \quad \lambda_1 U + \lambda_2 U \subseteq (\lambda_1 + \lambda_2)U \end{align}
• Since $x \in \lambda_1 U$ and $y \in \lambda_2 U$, we see that $(x + y) \in (\lambda_1 + \lambda_2)U$. Thus:
(8)
\begin{align} \quad p_U(x + y) = \inf \{ \lambda > 0 : (x + y) \in \lambda U \} \leq \lambda_1 + \lambda_2 \leq p_U(x) + p_U(y) + 2\epsilon \end{align}
• The above inequality holds for every $\epsilon > 0$, and so, $p_U(x + y) \leq p_U(x) + p_U(y)$ for all $x, y \in X$.
• Lastly, let $x, y \in X$. Since $U$ is a sub-semigroup we have that $x, y \in U$ implies that $x \cdot y \in U$. Then:
(9)
\begin{align} \quad \{ \lambda > 0 : x, y \in \lambda U \} \subseteq \{ \lambda > 0 : x \cdot y \in \lambda U \} \end{align}
• Recall that if $A$ and $B$ are bounded sets of real numbers then $A \subseteq B$ implies $\inf A \geq \inf B$. From above we get
(10)
\begin{align} \quad \quad \inf \{ \lambda > 0 : x \cdot y \in \lambda U \} \leq \inf \{ \lambda > 0 : x, y \in U \} \leq \inf \{ \lambda > 0 : x \in U \} \inf \{ \lambda > 0 : y \in U \} \end{align}
• That is, $p_U(x \cdot y) \leq p_U(x) p_U(y)$ for all $x, y \in X$. Thus, $p_U$ is an algebra seminorm on $X$. $\blacksquare$
 Corollary 2: Let $X$ be a algebra and let $U \subseteq X$. If $U$ is absolutely convex, absorbent, and is a sub-semigroup then $\{ x \in X : p_U(x) < 1 \} \subseteq U \subseteq \{ x \in X : p_U(x) \leq 1 \}$.

Corollary 1 says that if $U$ satisfies the properties of being absolutely convex, absorbent, and is a sub-semigroup, then $U$ is contained between the open and closed unit balls with respect to the $p_U$ seminorm.

• Proof: Let $z \in \{ x \in X : p_U(x) < 1 \}$. Then $p_U(z) = \inf \{ \lambda > 0 : z \in \lambda U \} < 1$. So there exists a $\lambda^*$ with $0 < \lambda^* < 1$ for which $z \in \lambda^* U$, or equivalently:
(11)
\begin{align} \quad \frac{1}{\lambda^{*-1}} z \in U \end{align}
• Since $U$ is absolutely convex, $U$ is convex. Consider the line segment joining the points $0$ and $\frac{1}{\lambda^{*-1}} z$, both of which are in $U$, and so the line segment is in $U$ by convexity. Points on this line are of the form:
(12)
\begin{align} \quad t \frac{1}{\lambda^{*-1}} z \end{align}
• Where $t \in [0, 1]$. When $t = \lambda^* \in [0, 1]$, we see that $z \in U$. Thus:
(13)
\begin{align} \quad \{ x \in X : p_U(x) < 1 \} \subseteq U \quad (*) \end{align}
• Now let $z \in U$. Then $z \in 1U$ and so $p_U(z) = \{ \lambda > 0 : z \in \lambda U \} \leq 1$, thus:
(14)
\begin{align} \quad U \subseteq \{ x \in X : p_U(x) \leq 1 \} \quad (**) \end{align}
• Combining $(*)$ and $(**)$ gives us that $\{ x \in X : p_U(x) < 1 \} \subseteq U \subseteq \{ x \in X : p_U(x) \leq 1 \}$. $\blacksquare$