Properties of Gauges of Absolutely Convex and Absorbent Sets

Proposition 1: Let $E$ be a vector space and let $A, B \subseteq E$ be absolutely convex and absorbent subsets of $E$, with gauges $p_A$ and $p_B$ respectively. Then: (1) For all $\alpha \in \mathbf{F}$, $\alpha \neq 0$, the gauge of $\alpha A$, denote it by $p_{\alpha A}$, is given by $p_{\alpha A} = |\alpha|^{-1}p_A$. (2) If $A \subseteq B$ then $p_B(x) \leq p_A(x)$ for all $x \in E$. (3) The gauge of $A \cap B$, denote it by $p_{A \cap B}$, is given by $p_{A \cap B} = \sup \{ p_A, p_B \}$. (4) If $\{ x : p_A(x) < 1 \} \subseteq B \subseteq \{ x : p_A(x) \leq 1 \}$ then $p_A = p_B$.

• Proof of (1): Let $\alpha \in \mathbf{F}$, $\alpha \neq 0$. Since $A$ is an absolutely convex and absorbent set, so is $\lambda A$. Thus the gauge $p_{\alpha A}$ of $\alpha A$ is well-defined, and for each $x \in E$:

(1)

\begin{align} p_{\alpha A}(x) = \inf \{ \lambda : \lambda > 0 \: \mathrm{and} \quad x \in \lambda (\alpha A) \} \end{align}

• Since $A$ is balanced, $\alpha A = |\alpha| A$. Thus $x \in \lambda (\alpha A)$ if and only if $x \in \lambda (|\alpha| A)$, if and only if $\displaystyle{|\alpha|^{-1} x \in \lambda A}$. So by setting $y = |\alpha|^{-1} x$, we see that:

(2)

\begin{align} \quad p_{\alpha A}(x) = \inf \{ \lambda : \lambda > 0 \: \mathrm{and} \quad |\alpha|^{-1} x \in \lambda A \} = |\alpha|^{-1} \inf \{ \lambda : \lambda > 0 \: \mathrm{and} \: y \in \lambda A \} = p_A(y) = p_A(|\alpha|^{-1}x) = |\alpha|^{-1} p_A(x) \end{align}

• Thus $p_{\alpha A} = |\alpha|^{-1} p_A$. $\blacksquare$

• Proof of (2): Suppose that $A \subseteq B$. Let $x \in E$. Let $\lambda > 0$ such that $x \in \lambda A$. Since $A \subseteq B$, we have that $\lambda A \subseteq \lambda B$, and so $x \in \lambda B$. So $p_B(x) \leq \lambda$. Taking the infimum over all $\lambda > 0$ with $x \in \lambda A$, and we conclude that for each $x \in E$:

(3)

\begin{align} \quad p_B(x) \leq p_A(x) \quad \blacksquare \end{align}

• Proof of (3): Since $A$ and $B$ are absolutely convex and absorbent, so is $A \cap B$, so that the gauge of $A \cap B$ makes sense. Observe that $A \cap B \subseteq A$ and $A \cap B \subseteq B$, and so from (2) we have that $p_A \leq p_{A \cap B}$ and $p_B \leq p_{A \cap B}$ on $E$, and thus:

(4)

\begin{align} \quad \sup \{ p_A, p_B \} \leq p_{A \cap B} \quad (\star) \end{align}

• For the reverse inequality, let $x \in E$ and let $\lambda, \mu > 0$ be such that $x \in \lambda A$ and $x \in \mu B$.

• Since $A$ and $B$ are balanced, we have that $\lambda A \subseteq \sup \{ \lambda, \mu \} A$ and $\mu B \subseteq \sup \{ \lambda, \mu \} B$ (see one of the propositions on the Properties of Balanced Sets of Vectors page). Thus $x \in \lambda A$ and $x \in \mu B$ implies that:

(5)

\begin{align} \quad x \in (\sup \{ \lambda, \mu \} A) \cap (\sup \{ \lambda, \mu \} B) = \sup \{ \lambda, \mu \} (A \cap B) \end{align}

• (Indeed, the equality above holds since in general, for all $t > 0$ and for all subsets $A$ and $B$ of a vector space $E$, $(tA) \cap (tB) = t(A \cap B)$, since if $x \in (tA) \cap (tB)$ then $x = ta$ and $x = tb$ for some $a \in A$ and $b \in B$. Then $ta = tb$ implies that $a = b$, so that $a = b \in A \cap B$. Thus $x = ta = tb \in t(A \cap B)$. On the otherhand, if $x \in t(A \cap B)$ then $x = tc$ for some $c \in A \cap B$. So $c \in A$ and $c \in B$ so that $x = tc \in (tA) \cap (tB)$.)

• Hence:

(6)

\begin{align} \quad p_{A \cap B} (x) \leq \sup \{ \lambda, \mu \} \end{align}

• Taking the infimum over all $\lambda, \mu > 0$ with $x \in \lambda A$ and $x \in \mu B$, and we get that:

(7)

\begin{align} \quad p_{A \cap B} (x) \leq \sup \{ p_A(x), p_B(x) \} \quad (\star \star) \end{align}

• So combining $(\star)$ and $(\star \star)$ we see that $p_{A \cap B} = \sup \{ p_A, p_B \}$. $\blacksquare$

• Proof of (4): Let $x \in E$. Since $\{ x : p(x) < 1 \} \subseteq B \subseteq \{ x : p(x) \leq 1 \}$, for each $\lambda > 0$, we have that:

(8)

\begin{align} \quad \lambda \{ x : p_A(x) < 1 \} \subseteq \lambda B \subseteq \lambda \{ x : p_A(x) \leq 1 \} \end{align}

• Or equivalently:

(9)

\begin{align} \quad \{ x : p_A(x) < \lambda \} \subseteq \lambda B \subseteq \{ x : p_A(x) \leq \lambda \} \quad (\dagger) \end{align}

• So if $\lambda > 0$ is such that $x \in \lambda B$ then from the inclusion $(\dagger)$ above, $p_A(x) \leq \lambda$. Taking the infimum over all such $\lambda$ yields:

(10)

\begin{align} \quad p_A(x) \leq p_B(x) \quad (\star) \end{align}

• Now if $\lambda > 1$ then observe that $\{ x : p_A(x) \leq 1 \} \subseteq \{ x : p_A(x) < \lambda \}$. Thus, from the inclusion at $(\dagger)$ we have that for all $\lambda > 1$:

(11)

\begin{align} \quad A \subseteq \{ x : p_A(x) \leq 1 \} \subseteq \{ x : p_A(x) < \lambda\} \subseteq \lambda B \end{align}

• So for all $\mu > 0$ we have that $\mu A \subseteq \mu \lambda B$. Thus, for all $\lambda > 0$ and for all $n \in \mathbb{N}$ we have that $\lambda A \subseteq \left ( \lambda + \frac{1}{n} \right ) B$. Hence if $\lambda > 0$ is such that $x \in \lambda A$ then $x \in \left ( \lambda + \frac{1}{n} \right ) B$ so that:

(12)

\begin{align} \quad p_B(x) \leq \lambda + \frac{1}{n} \end{align}

• Taking the limit as $n \to \infty$ and then taking the infimum over all such $\lambda$ yields:

(13)

\begin{align} \quad p_B(x) \leq p_A(x) \quad (\star \star) \end{align}

• Combining $(\star)$ and $(\star \star)$ shows us that if $B$ is an absolutely convex and absorbent subset of $E$ such that $\{ x : p_A(x) < 1 \} \subseteq B \subseteq \{ x : p_A(x) \leq 1 \}$ then $p_A = p_B$ on $E$. $\blacksquare$