> [!definition] > > Let $\cx$ be a [[Normed Vector Space|normed space]], an **affine hyperplane** is a set of the form > $ > H = [f = \alpha] = f^{-1}(\alpha) > $ > where $f: \cx \to \real$ is a [[Linear Functional|linear functional]] that does not vanish identically. > [!theorem] > > The hyperplane $[f = \alpha]$ is [[Closed Set|closed]] if and only if $f$ is [[Bounded Linear Map|continuous]]. > > *Proof*. > > ### Forward > > Suppose that $f$ is continuous, then for any convergent [[Sequence|sequence]] $\seq{x_i} \subset [f = \alpha]$ with $x_i \to x$, $f(x_i) \to f(x) = \alpha$, and $[f = \alpha]$ is closed. > > > ### Backwards > > Leveraging the linearity of $f$, if the plane is closed, then we can find a convex neighbourhood above/under the plane for any point outside of it. If any point in there were to have a value "past the plane", then we can find somewhere in between that *should* be in the plane. However, the neighbourhood is supposed to be above/under the plane. Therefore this cannot happen. > > Once we have continuity at one point on the linear functional, we can translate it back to the origin by simply shaving off the value of $f$ at that point. > > ![[hyperplane_closed.png|300]] > > Now suppose that $[f = \alpha]$ is closed. As $f$ does not vanish identically, $[f = \alpha]^c \ne \emptyset$. Let $x \in [f = \alpha]^c$ and suppose that $f(x) < \alpha$, then there exists $\varepsilon > 0$ such that $B(x, \varepsilon) \subset [f = \alpha]^c$. > > We claim that $f(y) < \alpha$ for all $y \in B(x, \varepsilon)$. Suppose that there exists $z \in B(x, \varepsilon)$ such that $f(z) > \alpha$, then in the line segment > $ > \bracs{(1 - t)x + tz: t \in [0, 1]} \subset B(x, \varepsilon) > $ > there exists $t \in [0, 1]$ such that $f((1 - t)x + tz) = \alpha$, namely > $ > t = \frac{\alpha - f(x)}{f(z) - f(x)} > $ > which is in $[0, 1]$ since $0 \le \alpha - f(x) \le f(z) - f(x)$. This contradicts the fact that $B(x, \varepsilon) \subset [f = \alpha]^c$. Therefore $f(z) < \alpha$ for all $z \in B(x, \varepsilon)$. > > Let $x \in [f = \alpha]^c$ such that $f(x) < \alpha$ and $\varepsilon > 0$ such that ${B(x, \varepsilon)} \subset [f = \alpha]^c$. Let $y \in B(x, \varepsilon)$, then $y = x + y'$. Since > $ > \begin{align*} > f(x + y') &= f(x) + f(y') \\ > f(y') &< \alpha - f(x) \\ > \abs{f(y')} &< \alpha - f(x) > \end{align*} > $ > we have > $ > \abs{f(x + y')} \le \abs{f(x)} + \abs{y'} < \abs{f(x)} + \abs{\alpha - f(x)} > $ > and $f$ is bounded on $B(x, \varepsilon)$. Since $f$ is bounded on an open set, $f$ is continuous. > [!definition] > > Let $A, B \subset \cx$ be two subsets. The hyperplane $[f = \alpha]$ **separates** $A$ and $B$ if > $ > f(x) \le \alpha \forall x \in A \quad f(x) \ge \alpha \forall x \in B > $ > and **strictly separates** $A$ and $B$ if there exists some $\varepsilon > 0$ such that > $ > f(x) \le \alpha - \varepsilon \forall x \in A \quad f(x) \ge \alpha + \varepsilon \forall x \in B > $ > [!theorem] > > Let $\cx$ be a normed space and $C \subset \cx$ be an open [[Convexity|convex]] set with $0 \in C$. For every $x \in \cx$ define > $ > p(x) = \inf\bracs{\alpha > 0: \frac{x}{\alpha} \in C} > $ > as the **gauge** of $C$. Then > 1. $p$ is a [[Sublinear Functional|sublinear functional]]. > 2. There exists $M \ge 0$ such that $0 \le p(x) \le M\norm{x}$ for all $x \in \cx$ > 3. $C = \bracs{x \in \cx: p(x) < 1}$. > > *Proof*. (1, linear) First let $\lambda > 0$, and $x \in \cx$, then > $ > \begin{align*} > \lambda \bracs{\alpha > 0: \frac{x}{\alpha} \in C} &= \bracs{\lambda\alpha > 0: \frac{x}{\alpha} \in C} \\ > &= \bracs{\lambda \alpha > 0: \frac{\lambda x}{\lambda\alpha} \in C} \\ > &= \bracs{\alpha > 0: \frac{\lambda x}{\alpha} \in C} > \end{align*} > $ > therefore $p(x) = \lambda p(x)$. > > (2) Since $C$ is open, we can find a neighbourhood centred at $0$ such that any vectors of a certain length will be inside $C$. Using this we can create an upper bound on the scale-down factor in terms of $\norm{x}$. > > Let $r > 0$ such that $\ol{B(0, r)} \subset C$, then $r\frac{x}{\norm{x}} \in C$ for any $x \ne 0$, and we have > $ > p(x) \le \frac{1}{r}\norm{x} \quad \forall x \in E > $ > > (3) Let $x \in C$, then since $C$ is open $\exists \varepsilon > 0: B(x, \varepsilon) \subset C$. Choose $\delta > 0$ such that $(1 + \delta)\norm{x} < \norm{x} + \varepsilon$, then $(1 + \delta)x \in C$ and $p(x) \le \frac{1}{1 + \delta} < 1$. > > ![[convex_subadditivity.png|300]] > > (1, sublinear) Let $x, y \in \cx$ and $\alpha, \beta > 0$ be estimates of $p(x)$ and $p(y)$ such that $\frac{x}{\alpha}$ and $\frac{y}{\beta}$ are in $C$. As $C$ is convex, $\frac{tx}{\alpha} + \frac{(1 - t)y}{\beta} \in C$ for all $t \in [0, 1]$. Let $t_0 \in [0, 1]$ such that > $ > \frac{x + y}{\alpha + \beta} = \frac{t_0 x}{\alpha} + \frac{(1 - t_0)y}{\beta} > $ > then $\frac{x + y}{\alpha + \beta} \in C$, and $\alpha + \beta \ge p(x + y)$. Since this holds for all $\alpha > p(x)$ and $\beta > p(y)$, $p(x + y) \le p(x) + p(y)$. > [!theorem] > > Let $C \subset \cx$ be a non-empty open convex set and $x \not\in C$. Then there exists a continuous [[Linear Functional|linear functional]] $f \in \cx^*$ such that $f(y) < f(x)$ for all $y \in C$. In other words, $[f = f(x)]$ separates $\bracs{x}$ and $C$. > > *Proof*. The gauge serves to separate points inside the set from ones outside of it. Extending an indicator function for a point outside of $C$ using the gauge as a constraint allows the function to also separate the point from $C$. > > First suppose that $0 \in C$, and let $g: \real x \to \real$ with $tx \mapsto t$, then since $x \not\in C$, $p(x) > 1$, and $g(tx) = t \le tp(x) = p(tx)$. By the [[Hahn-Banach Theorem|Hahn-Banach theorem]], there is an extension $f$ of $g$ to $\cx$ such that $f(x) = 1$ and $f(y) \le p(y)$ for all $y \in \cx$. > > As $f(y) \le p(y) \le M\norm{y}$ for some $M > 0$, $f$ is continuous, and $[f = 1]$ is closed. With $f(y) \le p(y) < 1$ for all $y \in C$, $[f = 1]$ is a closed hyperplane that separates $x$ and $C$. > > Suppose that $0 \not\in C$. Choose any $y \in C$ so $0 \in C - y$. Then there exists $f: \cx \to \real$ such that $f(x - y) = 1$ and $f(z) < 1$ for all $z \in C - y$. For any $z \in C$, > $ > f(z) = f(z - y) + f(y) < f(x - y) + f(y) = f(x) > $ > and $f$ separates $x$ from $C$.