> [!definition] > > Let $E$ be a [[Vector Space|vector space]] over $K \in \bracs{\real,\complex}$ and $M \subset E$ be a [[Absorbing|absorbing]] set. Define > $ > [\cdot]_M: E \to [0, \infty) \quad x \mapsto \inf \bracs{\lambda > 0: \lambda^{-1}x \in M} > $ > as the **Minkowski functional/gauge** of $M$, then > 1. $[\lambda x]_M = \lambda[x]_M$ for all $x \in E$ and $\lambda \ge 0$. > 2. If $M$ is [[Convexity|convex]], then $[x + y]_M \le [x]_M + [y]_M$ for all $x, y \in E$ and $[\cdot]_M$ is a [[Sublinear Functional|sublinear functional]]. > 3. If $M$ is [[Balanced Set|balanced]], then $[\lambda x]_M = \abs{\lambda}[x]_M$ for all $\lambda \in K$. In particular, if $M$ is also convex, then $[\cdot]_M$ is a [[Seminorm|seminorm]]. > 4. If $E$ is a [[Topological Vector Space|topological vector space]] and $M$ is [[Open Set|open]], then $M = \bracs{x \in E: [x]_M < 1}$. > 5. If $E$ is [[Normed Vector Space|normed]] and $M$ is open, then there exists $C \ge 0$ such that $[x]_M \le C\norm{x}$. > > With $(2), (3)$ combined, there is a correspondence between the neighbourhoods of $0$ induced by seminorms and the seminorms themselves. > > *Proof*. $(2)$: Let $x, y \in M$ and $\lambda, \mu > 0$ such that $\lambda^{-1}x, \mu^{-1}y \in M$. By convexity, $t\lambda^{-1}x + (1 - t)\mu^{-1}y \in M$ for all $t \in [0, 1]$. Let $t \in [0, 1]$ such that > $ > (\lambda + \mu)^{-1}(x + y) = t\lambda^{-1}x + (1 - t)\mu^{-1}y > $ > then $(\lambda + \mu)^{-1}(x + y) \in M$ and $\lambda + \mu \ge [x + y]_M$. As this holds for all $\lambda > [x]_M$ and $\mu > [y]_M$, $[x + y]_M \le [x]_M + [y]_M$. > > $(4)$: $M \subset \bracs{x \in E: [x]_M < 1}$ directly. Let $x \in M$, then there exists $\delta > 0$ such that $(1 + \delta)x \in M$ by continuity. In this case, $[x]_M \le (1 + \delta)^{-1} < 1$. Therefore $M \supset \bracs{x \in E: [x]_M < 1}$. > > $(5)$: Since $M$ is open, there exists $r > 0$ such that $B(0, r) \subset M$. For any $x \in E$, $x \cdot (\norm{x}^{-1}r) \in M$, so $\norm{x} \cdot r^{-1} \ge [x]_M$.