---
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