The Gauge of an Absolutely Convex and Absorbent Set

# The Gauge of an Absolutely Convex and Absorbent Set

Proposition 1: Let $E$ be a vector space.(1) If $p$ is a seminorm on $E$ then for each $\alpha > 0$, the sets $\{ x \in E : p(x) < \alpha \}$ and $\{ x \in E : p(x) \leq \alpha \}$ are absolutely convex and absorbent.(2) If $A$ is an absolutely convex and absorbent subset of $E$, then the function $p$ defined for all $x \in E$ by $p(x) = \inf \{ \lambda : \lambda > 0 \: \mathrm{and} \: x \in \lambda A \}$ is a seminorm on $E$ such that $\{ x \in E : p(x) < 1 \} \subseteq A \subseteq \{ x \in E : p(x) \leq 1 \}$. |

**Proof of (1):**Let $p$ be a seminorm on $E$ and let $\alpha > 0$.

- Let $a, b \in \{ x \in E : p(x) < \alpha \}$ and let $\lambda, \mu \in \mathbf{F}$ be such that $|\lambda| + |\mu| \leq 1$. Then $p(a), p(b) < \alpha$. Since $p$ is a seminorm on $A$ we have that:

\begin{align} \quad p(\lambda a + \mu b) \leq p(\lambda a) + p(\mu b) = |\lambda| p(a) +|\mu| p(b) < (|\lambda| + |\mu|) \alpha \leq \alpha \end{align}

- So $\lambda a + \mu b \in \{ x \in E : p(x) < \alpha \}$ and thus $\{ x \in E : p(x) < \alpha \}$ is absolutely convex.

- Now, we will show that $\{ x \in E : p(x) < \alpha$ is absorbent. Observe that for all $\mu \in \mathbf{F}$, $\mu \neq 0$:

\begin{align} \quad \mu A = \mu \{ x : p(x) < \alpha \} = \{ x : p(x) < |\mu| \alpha \} \end{align}

- (Indeed, if $y \in \mu \{ x : p(x) < \alpha \}$ then there exists an $x \in E$ such that $y = \mu x$ and $p(x) < \alpha$. Then for $p(y) = p(\mu x) = |\mu| p(x) < |\mu| \alpha$. On the other hand, if $y \in \{ x : p(x) < |\mu| \alpha \}$ then $p(y) < |\mu| \alpha$. Setting $x := \frac{y}{\mu}$ we see that $y = \mu x$, and $p(x) = p \left ( \frac{y}{\mu} \right ) = \frac{1}{|\mu|} p(y) < \frac{1}{|\mu|} \cdot |\mu| \alpha = \alpha$.)

- So let $e \in E$. If $p(e) = 0$ then for any $\lambda > 0$, if $\mu \in \mathbf{F}$ is such that $|\mu| \geq \lambda$ then $e \in \mu \{ x : p(x) < |\mu| \alpha \}$ since $p(e) = 0 < |\mu| \alpha$.

- If $p(e) \neq 0$, then take $\lambda > 0$ such that $p(e) < \lambda \alpha$. Then if $\mu \in \mathbf{F}$ is such that $|\mu| \geq \lambda$ then:

\begin{align} p(e) < \lambda \alpha \leq |\mu| \alpha \end{align}

- so that $e \in \mu \{ x : p(x) < \alpha \}$. So $\{ x : p(x) < \alpha \}$ is absorbent.

- A similar argument shows that the sets $\{ x : p(x) \leq \alpha \}$ are also absolutely convex and absorbent. $\blacksquare$

**Proof of (2):**Let $A$ be absolutely convex and absorbent.

- Since $A$ is absorbent, for each $x \in E$ there exists a $\lambda_x > 0$ such that if $\mu \in \mathbf{F}$ is such that $|\mu| \geq \lambda_x$ then $x \in \mu A$. Thus $p(x) \leq \lambda_x$ for each $x \in E$, and so $p : X \to [0, \infty)$.

- We now verify that $p$ satisfies properties (1)-(3) of a seminorm.

- Clearly, $p(x) \geq 0$ for all $x \in E$ since $p$ is an infimum of nonnegative numbers.

- Suppose that $\mu \in \mathbf{F}$ is such that $\mu \neq 0$. Observe that $\mu x \in \lambda A$ if and only if $x \in \frac{\lambda}{|\mu|} A$ since $A$ is balanced. Thus:

\begin{align} \quad p(\mu x) = \inf \{ \lambda : \lambda > 0 \: \mathrm{and} \: \mu x \in \lambda A \} = |\mu| \inf \left \{ \frac{\lambda}{|\mu|} : \lambda > 0 \: \mathrm{and} \: x \in \frac{\lambda}{|\mu|} A \right \} = |\mu| \inf \{ \lambda : \lambda > 0 \: \mathrm{and} \: x \in \lambda A \} = |\mu| p(x) \end{align}

- Lastly, let $x, y \in E$ and suppose that $\lambda, \mu > 0$ are such that $x \in \lambda A$ and $y \in \mu A$. Then:

\begin{align} \quad x + y \in \lambda A + \mu A \end{align}

- But since $\lambda, \mu > 0$ and since $A$ is convex, we have by one of the propositions on the Properties of Convex Sets of Vectors page that $\lambda A + \mu A \subseteq (\lambda + \mu) A$. Thus $x + y \in (\lambda + \mu)A$. So by the definition of $p(x+ y)$ being an infimum, we have that

\begin{align} \quad p(x + y) \leq \lambda + \mu \end{align}

- Since this holds for all $\lambda, \mu > 0$ with $x \in \lambda A$ and $y \in \mu A$, we conclude that $p(x + y) \leq p(x) + p(y)$.

- Thus, $p$ is a seminorm on $E$.

- Lastly, observe that if $y \in \{ x \in E : p(x) < 1 \}$ then $p(y) < 1$. By the definition of $p(y)$, there must exist an $\epsilon > 0$ such that $p(y) + \epsilon \leq 1$ and $y \in (p(y) + \epsilon) A$. But since $A$ is balanced, $(p(y) + \epsilon)A \subseteq A$. Thus $y \in A$. So $\{ x \in E : p(x) < 1 \} \subseteq A$.

- Then if $y \in A$ then clearly $y \in 1A$. So $p(y) \leq 1$. So $y \in \{ x \in E : p(x) \leq 1 \}$ and thus $A \subseteq \{ x \in E : p(x) \leq 1 \}$.

- Hence $\{ x \in E : p(x) < 1 \} \subseteq A \subseteq \{ x \in E : p(x) \leq 1 \}$. $\blacksquare$

Definition: Let $E$ be a vector space. If $A \subseteq E$ is an absolutely convex and absorbent subset of $E$, then the Gauge of $A$ is the seminorm $p : E \to [0, \infty)$ defined for all $x \in E$ by $p(x) = \inf \{ \lambda : \lambda > 0 \: \mathrm{and} \: x \in \lambda A \}$. |

*Sometimes we will use the notation "$p_A$" in place of just "$p$" when we are taking about the gauge of an absolutely convex and absorbent set $A$.*