---
title: Hahn–Banach Theorem
parent: The Fundamental Four
grand_parent: Functional Analysis Basics
nav_order: 1
---

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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.

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    Table of contents
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{% definition Sublinear Functional %}
A functional $p$ on a real vector space $X$
is called *sublinear* if it is
{: .mb-0 }

{: .mt-0 .mb-0 }
- *positively homogenous*, that is
  {: .mt-0 .mb-0 }

  $$
  p(\alpha x) = \alpha \, p(x) \qquad \forall \alpha \ge 0, \ \forall x \in X,
  $$

- and *subadditive*, that is
  {: .mt-0 .mb-0 }

$$
p(x+y) \le p(x) + p(y) \qquad \forall x,y \in X.
$$
{% enddefinition %}

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 * Hahn–Banach Theorem (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 $x \in X$.
{% endtheorem %}

## Extension Theorems

{% theorem * Hahn–Banach Theorem (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

$$
\tilde{f}(x) \le p(x) \qquad \forall x \in X.
$$
{% endtheorem %}

{% proof %}
{% endproof %}

{% 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 }

  $$
  p(x+y) \le p(x) + p(y) \qquad \forall x,y \in X.
  $$
{% enddefinition %}

{% theorem * Hahn–Banach Theorem (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

$$
\abs{\tilde{f}(x)} \le p(x) \qquad \forall x \in X.
$$
{% endtheorem %}

{% theorem * Hahn–Banach Theorem (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 $\norm{\tilde{f}} = \norm{f}$.
{% endtheorem %}

{% 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

{% theorem * Hahn–Banach Theorem (Existence of Functionals) %}
Let $X$ be a real or complex normed space
and let $x$ be a nonzero element of $X$.
Then there exists a bounded linear functional $f$ on $X$
with $f(x) = \norm{x}$ and $\norm{f} = 1$.
{% endtheorem %}

{% proof %}
On the linear subspace $\KK x \subset X$ spanned by $x$
we define a functional $f_0$ by $f_0(\alpha x) = \alpha \norm{x}$ for $\alpha \in \KK$.
It is easy to check that $f_0$ is linear and bounded with norm $\norm{f_0} = 1$.
By the Hahn–Banach Extension Theorem for Normed Spaces,
there exists a bounded linear functional $f$ on $X$ extending $f_0$ with identical norm.
Hence we have $f(x) = f_0(x) = \norm{x}$ and $\norm{f} = \norm{f_0} = 1$.
{% endproof %}

Recall that for a normed space $X$ we denote its (topological) dual space by $X'$.

{% corollary %}
For every element $x$ of a real or complex normed space $X$ one has

$$
\norm{x} = \sup_{f \in X' \setminus \braces{0}} \frac{\abs{f(x)}}{\norm{f}} 
$$

and the supremum is attained.
{% endcorollary %}

{% corollary %}
The elements of a real or complex normed space $X$
are separated by the elements of its dual $X'$.
{% endcorollary %}

## Separation Theorems

{% theorem * Hahn–Banach Theorem (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 $f$ on $X$
vanishing on $Z$ and with nonzero value $f(x) = \dist{x,Z}$.
{% endtheorem %}

{% proof %}
Since $Z$ is a closed subspace of $X$,
the quotient vector space $X/Z$ becomes a normed space
with the quotient norm given by

$$
\norm{y + Z} = \dist{y,Z} = \inf_{z \in Z} \norm{y-z} \quad \forall y \in X.
$$

Moreover, the canonical mapping $\pi : X \to X/Z$, $y \mapsto y+Z$, is bounded.
Given a $x \in X$ that does not lie in $Z$, the null space of $\pi$,
we see that $\pi(x)$ is a nonzero element of $X/Z$.
By Hahn–Banach, there exists a bounded linear functional $g$ on $X/Z$
with $g(\pi(x)) = \norm{x} = \dist{x,Z} \ne 0$.
Now the composition $f = g \circ \pi$ is a bounded functional on $X$
with the desired properties.
{% endproof %}

{% theorem * Hahn–Banach Theorem (Separation, Convex Sets) %}
TODO
{% endtheorem %}