> [!theoremb] Theorem > > Let $\cx$ be a [[Vector Space|vector space]] over $\real$, $p$ a [[Sublinear Functional|sublinear functional]] on $\cx$. Let $\cm$ be a subspace on $\cx$, and $f$ be a [[Linear Functional|linear functional]] on $\cm$ such that $f(x) \le p(x)$ for all $x \in \cm$. Then there is an extension $F$ of $f$ on $\cx$ such that $F(x) \le p(x)$ for all $x \in \cx$. > > *Proof*. > > # Single-Dimensional Extensions > > First let $x \in \cx \setminus \cm$, then $\real x \cap \cm = \bracs{0}$ and any vector $y \in \real x + \cm$ can be written uniquely as $\lambda x + m$ for some $\lambda \in \real$ and $m \in \cm$. Given a linear functional $f$ on $\cm$, $f$ can be extended to be a linear functional $g$ on $\real x + \cm$ by > $ > g_\alpha(y) = g_\alpha(\lambda x + m) = \lambda \alpha + f(m) \quad \alpha \in \real > $ > where $g|_{\cm} = f$. > > # Bounded by Sublinear Functional > > Let $x \in \cx \setminus \cm$ and $g_\alpha: \real x + \cm$ with $\lambda x + m \mapsto \lambda \alpha + f(m)$. Then there exists an $\alpha \in \real$ such that $g_\alpha(y) \le p(y)$ for all $y \in \cx$. > > *Proof*. Let $y_1, y_2 \in \cm$, then > $ > f(y_1) + f(y_2) \le p(y_1 + y_2) \le p(y_1 - x) + p(y_2 + x) > $ > and > $ > f(y_1) - p(y_1 - x) \le p(y_2 + x) - f(y_2) > $ > meaning that > $ > \underbrace{\sup_{y \in \cm}\braks{f(y) - p(y - x)}}_{a} > \le > \underbrace{\inf_{y \in \cm}[p(y + x) - f(y)]}_{A} > $ > Let $\alpha \in [a, A]$, then for any $\lambda > 0$ and $y \in \cm$, using the fact that $\alpha \le A$, > $ > \begin{align*} > g_\alpha\paren{\lambda x + y} &= \lambda \alpha + f(y) \\ > &= \lambda\braks{\alpha + f\paren{\frac{y}{\lambda}}} \\ > &\le \lambda\braks{p\paren{\frac{y}{\lambda} + x} - f\paren{\frac{y}{\lambda}} + f\paren{\frac{y}{\lambda}}} \\ > &= \lambda p\paren{\frac{y}{\lambda} + x} \\ > &= p(y + \lambda x) > \end{align*} > $ > Now let $\lambda = -\mu < 0$, then using the fact that $\alpha \ge a$, > $ > \begin{align*} > g_\alpha(\lambda x + y) &= f(y) - \mu\alpha \\ > &= \frac{1}{\mu}\braks{f\paren{\frac{y}{\mu}} - \alpha} \\ > &\le \frac{1}{\mu}\braks{f\paren{\frac{y}{\mu}} - f\paren{\frac{y}{\mu}} + p\paren{\frac{y}{\mu} - x}}\\ > &= \frac{1}{\mu}p\paren{\frac{y}{\mu} - x} \\ > &= p(y - \mu x) \\ > &= p(y + \lambda x) > \end{align*} > $ > and we have $g_\alpha(\lambda x + y) \le p(\lambda x + y)$ for any $\lambda \in \real$ and $y \in \cm$. > > > # Union of Subspaces > > Let > $ > \cf = \bracs{g: \cx \to \real: g|_\cm = f, g(x) \le p(x)\ \forall x} > $ > be the family of all linear extensions of $f$ to subspaces of $\cx$ that are bounded by $p$. [[Partial Order|Order]] $\cf$ by $g \le h$ if $h$ is an extension of $g$, then $\le$ is a partial order[^1]. > > Let $C$ be a chain in $\cf$. Denote $\text{Dom}(g)$ as the domain of $g$, and let > $ > \cn = \bigcup_{g \in C}\text{Dom}({g}) > $ > then $\cn$ is a subspace of $\cm$: For any $x, y \in \cn$ and $\lambda \in \real$, there exists $g_1, g_2 \in C$ such that $x \in \text{Dom}(g_1)$, $y \in \text{Dom}(g_2)$. Since $C$ is a chain, assume without loss of generality that $g_1 \le g_2$, then $\text{Dom}(g_2) \supset \text{Dom}(g_1)$ and $\lambda x, x + y \in \text{Dom}(g_1) \subset \cn$. > > > # Maximal Extension > > Let $x \in \cn$, then there exists $g \in C$ such that $x \in \text{Dom}(g)$. Define $F(x) = g(x)$. First verify that $F$ is well-defined: let $g_1, g_2 \in C$ such that $x \in \text{Dom}(g_1), \text{Dom}(g_2)$. Since $C$ is a chain, assume without loss of generality that $g_1 \le g_2$, then $g_2$ is an extension of $g_1$ and $g_1(x) = g_2(x)$. Therefore the value of $F$ on $x$ is independent of the choice of $g$. > > Now, let $x, y \in \cn$ and $\lambda \in \real$, then there exists $g_1, g_2 \in C$ such that $x \in \text{Dom}(g_1)$ and $y \in \text{Dom}(g_2)$. Assume again that $g_2 \ge g_1$, and we have > $ > \begin{align*} > F(\lambda x + y) &= g_2(\lambda x + y) \\ > &= \lambda g_2(x) + g_2(y) \\ > &= \lambda F(x) + F(y) \\ > F(x) &= g_2(x) \le p(x) > \end{align*} > $ > that $F \in \cf$ being an upper bound of $C$. > > Since every chain $C$ in $\cf$ has an upper bound, by [[Zorn's Lemma]], $\cf$ has a maximal element $F$. If $\text{Dom}(F) \subsetneq \cx$, then $F$ can be extended to $\text{Dom}(F) + \real x$ for some $x \in \cx \setminus \text{Dom}(F)$, which contradicts the fact that $F$ is a maximal element of $\cf$. Therefore $\text{Dom}(F) = \cx$, and $F$ is an extension of $f$ such that $F(x) \le p(x)$ for all $x \in \cx$. > [!theorem]- On [[Normed Vector Space|Normed Spaces]] > > Let $\cx$ be a real vector space, $\norm{\cdot}$ be a seminorm on $\cx$, $\cm$ be a subspace of $\cx$, and $f$ be a linear functional on $\cm$ such that $\abs{f(x)} \le \norm{x}$ for all $x \in \cm$. Then there is a linear extension $F$ of $f$ to $\cx$ such that $\abs{F(x)} \le \norm{x}$ for all $x \in \cx$. > > *Proof*. Let $x \in \cx$, then > $ > f(x) \le \norm{x} \Leftrightarrow f(-x) \le \norm{x} \Leftrightarrow \abs{f(x)} \le \norm{x} > $ > Since the seminorm is sublinear, applying the Hahn-Banach theorem yields the desired extension. > [!theorem] Complex Version > > Let $\cx$ be a complex vector space, $p$ a seminorm on $\cx$, $\cm$ a subspace of $\cx$, and $f$ a complex linear functional on $\cm$ such that $\abs{f(x)} \le p(x)$ for $x \in \cm$. Then there exists an extension $F$ of $f$ to $\cx$ such that $\abs{F(x)} \le p(x)$ for all $x \in \cx$. > > *Proof*. Let $u = \re{f}$, then there is a linear extension $U$ of $u$ to $\cx$ with $\abs{U(x)} \le p(x)$. Let $F = U(x) - iU(ix)$, then $F$ is a linear extension of $f$. For any non-zero $x \in \cx$, let $\alpha = \ol{\sgn F(x)}$, then > $ > \begin{align*} > F(x) = F(\alpha x) = \alpha F(x) = U(x) \le p(x) > \end{align*} > $ > [!theorem] > > Let $\cx$ be a [[Normed Vector Space|normed space]] over $\complex$, then > 1. If $\cm$ is a [[Closed Set|closed]] subspace of $\cx$ and $x \in \cx \setminus \cm$, there exists $f \in \cx^*$ such that $f(x) \ne 0$ and $f|_M = 0$. If $\delta = \inf_{y \in \cm}\norm{x - y}$, then $f$ can be taken to satisfy $\norm{f} = 1$ and $f(x) = \delta$. > 2. If $x \in \cx \setminus \bracs{0}$, then there exists $f \in \cx^*$ such that $\norm{f} = 1$ and $f(x) = \norm{x}$. > 3. The bounded linear functionals on $\cx$ separate points. > 4. If $x \in \cx$, define $\hat x: \cx^* \to \complex$ by $\hat x(f) = f(x)$. Then the map $x \mapsto \hat x$ is a linear isometry from $\cx$ to $\cx^{**}$. Moreover, the [[Topological Closure|closure]] of its image $\hat \cx = \bracs{\hat x: x \in \cx}$ is a [[Banach Space|Banach space]], known as the **completion** of $\cx$. If $\cx$ is already a Banach space, then $\ol{\hat{\cx}} = \hat \cx$. > > *Proof*. > > # Point Indicator > > ![[hahn-banach.png|300]] > > ![[hahn-banach-urysohn.png]] > > Let $\cm$ be a closed subspace of $\cx$ and $x \in \cx \setminus \cm$, then since $\cm$ is closed, $\delta = \inf_{y \in \cm}\norm{x - y} > 0$. > > Define $f: \cm + \complex x \to \complex$ by $m + \lambda x \mapsto \lambda \delta$, then since for any $\lambda \ne 0$, > $ > \begin{align*} > \abs{f(m + \lambda x)} = \abs{\lambda}\delta &\le > \abs{\lambda}\norm{x + y} &\forall y \in \cm \\ > \abs{\lambda} \delta &\le \abs{\lambda} \norm{x + \frac{m}{\lambda}} \\ > \abs{\lambda}\delta &\le \norm{\lambda x + m} > \end{align*} > $ > $f$ is a bounded linear functional on $\cm + \complex x$. > > Extend $f$ to $F$ on $\cx$ using the Hahn-Banach theorem with $p(y) = \norm{y}$, then $F(y) \le \norm{y}$ implies that $\norm{F} \le 1$. For any $\varepsilon > 0$, there exists $y \in \cm$ such that $\norm{x - y} < \delta + \varepsilon$, in which case > $ > \frac{F(x - y)}{\norm{x - y}} > \frac{\delta}{\delta + \varepsilon} > $ > where sending $\varepsilon \to 0$ yields $\delta/(\delta - \varepsilon) \to 1$. Therefore $\norm{f} \ge 1$. > > Setting $\cm = \bracs{0}$ yields the second result. > > # Separating Points > > Let $x, y \in \cx$, $x \ne y$, then there exists a continuous indicator functional on $\cx$ such that $f(x - y) \ne 0$. Therefore $f(x) \ne f(y)$ and $\cx^*$ separates points. > > > # Completion > > Let $x \in \cx$ and define $\hat x: \cx^* \to \complex$ by $f \mapsto f(x)$, then the map $\phi: \cx \to \cx^{**}$ where $x \mapsto \hat x$ is a linear isometry. Moreover, the closure of its image $\hat \cx = \bracs{\hat x: x \in \cx}$ is a Banach space. If $\cx$ is already a Banach space, then $\ol{\hat{\cx}} = \hat \cx$. > > *Proof*. Let $\lambda \in \complex$ and $x, y \in \cx$, then > $ > \paren{\widehat{\lambda x + y}}(f) = f(\lambda x + y) = \lambda f(x) + f(y) = \lambda \hat x(f) + \hat{y}(f) > $ > and $f$ is linear. Now let $x \in \cx$, then > $ > \norm{\hat x} = \sup_{\norm{f} = 1}\abs{f(x)} \le \norm{x} > $ > and since there exists $f \in \cx^*$ such that $f(x) = 1$, $\norm{\hat x} = \norm{x}$. > > Since $\cx^{**} = L(\cx^*, \complex)$ and $\complex$ is complete, $\cx^{**}$ is also complete. As $\ol{\hat \cx}$ is closed, $\ol{\hat \cx}$ is also complete. > > Lastly, suppose that $\cx$ is a Banach space and let $\seq{x_n}$ be a [[Cauchy Sequence|Cauchy sequence]] in $\cx$, then since the natural map $x \mapsto \hat x$ is continuous, $\widehat{\limv{n}x_n} = \limv{n}\hat{x}_n$. Therefore $\hat \cx$ is complete. > [!theoremb] Hahn-Banach, First Geometric Form > > ![[hahn-banach-1.png|300]] > > Let $\cx$ be a normed space over $\real$ and $A, B \subset X$ be non-empty [[Convexity|convex]] sets such that $A \cap B \ne \emptyset$. If one of them is [[Open Set|open]], then there exists a [[Affine Hyperplane|closed hyperplane]] that separates $A$ and $B$. > > *Proof*. Let $C = A - B$, then for any $a - b$ and $a' - b'$ in $C$, > $ > t(a - b) + (1 -t)(a' - b') = \underbrace{\braks{ta + (1 - t)a'}}_{\in A} - \underbrace{\braks{tb + (1 - t)b'}}_{\in B} > $ > and since $C = \bigcup_{y \in B}(A - y)$, $C$ is open. The collection of differences being convex allows it to be separated from $0$ with a plane. > > As $A \cap B = \emptyset$, $0 \not\in C$. By the Hahn-Banach theorem[^2], there exists $f \in \cx^*$ that separates $C$ from $0$, i.e. $f(z) < 0 \forall z \in C$. Then > $ > f(z) < f(y) \quad \forall x \in A, y \in B > $ > Let $\alpha \in [\sup_{x \in A}f(x), \inf_{y \in B} f(y)]$, then > $ > f(a) < \alpha < f(b) \quad \forall x \in A, y \in B > $ > and $[f = \alpha]$ separates $A$ from $B$. > [!theoremb] Hahn-Banach, Second Geometric Form > > ![[hahn-banach-2.png|300]] > > Let $\cx$ be a normed space over $\real$ and $A, B$ be disjoint non-empty convex subsets. If $A$ is closed and $B$ is [[Compactness|compact]], then there exists a closed hyperplane that strictly separates $A$ and $B$. > > *Proof*. Set $C = A - B$, then $C$ is convex and $0 \not\in C$. > > Let $\seq{c_n} \subset C$ be a [[Limit|convergent]] [[Sequence|sequence]] in $C$ with $c_n \to c$, write $c_n = a_n - b_n$ for some $a_n \in A$ and $b_n \in B$. Since $B$ is compact, there exists a convergent subsequence $\seq{b_{n_k}}$. This allows a subsequence $\seq{c_{n_k}}$ with $c_{n_k} = a_{n_k} - b_{n_k}$ such that $c_{n_k} \to c$, $b_{n_k} \to b \in B$, and as addition is continuous, $a_{n_k} \to a \in A$. Therefore there exists $a \in A$ and $b \in B$ such that $c = a + b$, and $C$ is closed. > > The convex set being closed allows the existence of an open neighbourhood that separates $0$ from $C$. This provides "room" for the function to interpolate between its value on $C$ and on $0$, and allows a strict separation. > > Since $0 \not\in C$, there exists $r > 0$ such that $B(0, r) \subset C^c$. Let $f \in \cx^*$ such that > $ > f(c) \le f(rz) \quad \forall c \in C, z \in B(0, 1) > $ > As $f(c) \le f(rz) = rf(z)$ for any $z \in B(0, 1)$, $f(c) \le -r\norm{f}$. Take $\varepsilon = \frac{1}{2}r\norm{f} > 0$, then > $ > \begin{align*} > f(x - y) &\le -r\norm{f} \\ > f(x) + \frac{1}{2}r\norm{f}&\le f(y) - \frac{1}{2}r\norm{f} \\ > f(x) + \varepsilon &\le f(y) - \varepsilon > \end{align*} > $ > Choose > $ > \alpha \in [\sup f(x) + \varepsilon, \inf f(y) - \varepsilon] > $ > then the hyperplane $[f = \alpha]$ strictly separates $A$ from $B$. [^1]: Any functional is an extension of itself, an extension of an extension is still an extension. Two functions being each other's extensions have the same domain and same values on them, and are therefore equal. [^2]: See [[Affine Hyperplane]].