Initial commit

1 files changed, 147 insertions, 0 deletions

diff --git a/pages/functional-analysis-basics/the-fundamental-four/hahn-banach-theorem.md b/pages/functional-analysis-basics/the-fundamental-four/hahn-banach-theorem.md
new file mode 100644
index 0000000..9d21d41
--- /dev/null
+++ b/pages/functional-analysis-basics/the-fundamental-four/hahn-banach-theorem.md
@@ -0,0 +1,147 @@
+---
+title: Hahn–Banach Theorem
+parent: The Fundamental Four
+grand_parent: Functional Analysis Basics
+nav_order: 1
+---
+
+# {{ page.title }}
+
+In fact, there are multiple theorems and corollaries
+which bear the name Hahn–Banach.
+All have in common that
+they guarantee the existence of linear functionals
+with various additional properties.
+
+{: .definition-title }
+> Definition (Sublinear Functional)
+>
+> A functional $p$ on a real vector space $X$
+> is called *sublinear* if it is
+> {: .mb-0 }
+>
+> {: .mt-0 .mb-0 }
+> - *positive-homogenous*, that is
+>   {: .mt-0 .mb-0 }
+>
+>   $$
+>   p(\alpha x) = \alpha \, p(x) \qquad \forall \alpha \ge 0, \ \forall x \in X,
+>   $$
+>
+> - and satisfies the *triangle inequality*
+>   {: .mt-0 .mb-0 }
+>
+>   $$
+>   p(x+y) \le p(x) + p(y) \qquad \forall x,y \in X.
+>   $$
+>   {: .katex-display .mb-0 }
+
+If $p$ is a sublinear functional,
+then $p(0)=0$ and $p(-x) \ge -p(x)$ for all $x$.
+
+Every norm on a real vector space is a sublinear functional.
+
+{: .theorem-title }
+> {{ page.title }} (Basic Version)
+>
+> Let $p$ be a sublinear functional on a real vector space $X$.
+> Then there exists a linear functional $f$ on $X$ satisfying
+> $f(x) \le p(x)$ for all $x \in X$.
+
+## Extension Theorems
+
+{: .theorem-title }
+> {{ page.title }} (Extension, Real Vector Spaces)
+>
+> Let $p$ be a sublinear functional on a real vector space $X$.
+> Let $f$ be a linear functional
+> which is defined on a linear subspace $Z$ of $X$
+> and satisfies
+>
+> $$
+> f(x) \le p(x) \qquad \forall x \in Z.
+> $$
+>
+> Then $f$ has a linear extension $\tilde{f}$ to $X$ such that
+>
+> $$
+> \tilde{f}(x) \le p(x) \qquad \forall x \in X.
+> $$
+
+{% proof %}
+{% endproof %}
+
+{: .definition-title }
+> Definition (Semi-Norm)
+>
+> We call a real-valued functional $p$ on a real or complex vector space $X$
+> a *semi-norm* if it is
+> {: .mb-0 }
+>
+> {: .mt-0 .mb-0 }
+> - *absolutely homogenous*, that is
+>   {: .mt-0 .mb-0 }
+>
+>   $$
+>   p(\alpha x) = \abs{\alpha} \, p(x) \qquad \forall \alpha \in \KK \ \forall x \in X,
+>   $$
+> - and satisfies the *triangle inequality*
+>   {: .mt-0 .mb-0 }
+>
+>   $$
+>   p(x+y) \le p(x) + p(y) \qquad \forall x,y \in X.
+>   $$
+>   {: .katex-display .mb-0 }
+
+{: .theorem-title }
+> {{ page.title }} (Extension, Real and Complex Vector Spaces)
+>
+> Let $p$ be a semi-norm on a real or complex vector space $X$.
+> Let $f$ be a linear functional
+> which is defined on a linear subspace $Z$ of $X$
+> and satisfies
+>
+> $$
+> \abs{f(x)} \le p(x) \qquad \forall x \in Z.
+> $$
+>
+> Then $f$ has a linear extension $\tilde{f}$ to $X$ such that
+>
+> $$
+> \abs{\tilde{f}(x)} \le p(x) \qquad \forall x \in X.
+> $$
+
+{: .theorem-title }
+> {{ page.title }} (Extension, Normed Spaces)
+>
+> Let $X$ be a real or complex normed space
+> and let $f$ be a bounded linear functional
+> defined on a linear subspace $Z$ of $X$.
+> Then $f$ has a bounded linear extension $\tilde{f}$ to $X$ such that $\norm{\tilde{f}} = \norm{f}$.
+
+{% proof %}
+We apply the preceding theorem with $p(x) = \norm{f} \norm{x}$
+and obtain a linear extension $\tilde{f}$ of $f$ to $X$
+satisfying $\abs{\tilde{f}(x)} \le \norm{f} \norm{x}$ for all $x \in X$.
+This implies that $\tilde{f}$ is bounded and $\norm{\tilde{f}} \le \norm{f}$.
+We have $\norm{\tilde{f}} \ge \norm{f}$, because $\tilde{f}$ extends $f$.
+{% endproof %}
+
+Corollaries
+
+Important consequence: canonical embedding into bidual
+
+## Separation Theorems
+
+{: .theorem-title }
+> {{ page.title }} (Separation, Point and Closed Subspace)
+>
+> Suppose $Z$ is a closed subspace
+> of a normed space $X$ and $x$ lies in $X \setminus Z$.
+> Then there exists a bounded linear functional on $X$
+> which vanishes on $Z$ but has a nonzero value at $x$.
+
+{: .theorem-title }
+> {{ page.title }} (Separation, Convex Sets)
+>
+> TODO