Hahn–Banach Theorem

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 (Sublinear Functional)

A functional $p$ on a real vector space $X$ is called sublinear if it is

• positively homogenous, that is

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

• and subadditive, that is

$$p(x+y)\le p(x)+p(y)\forall x,y\in X.$$

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.

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 $x\in X$.

Extension Theorems

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

$$f(x)\le p(x)\forall x\in Z.$$

Then $f$ has a linear extension $\tilde{f}$ to $X$ such that

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

Definition (Semi-Norm)

We call a real-valued functional $p$ on a real or complex vector space $X$ a semi-norm if it is

• absolutely homogenous, that is

  $$p(\alpha x)=|\alpha |p(x)\forall \alpha \in \mathbb{K}\forall x\in X,$$

• and satisfies the triangle inequality

  $$p(x+y)\le p(x)+p(y)\forall x,y\in X.$$

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

$$|f(x)|\le p(x)\forall x\in Z.$$

Then $f$ has a linear extension $\tilde{f}$ to $X$ such that

$$|\tilde{f}(x)|\le p(x)\forall x\in X.$$

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 $\Vert \tilde{f}\Vert =\Vert f\Vert$.

Proof   We apply the preceding theorem with $p(x)=\Vert f\Vert \Vert x\Vert$ and obtain a linear extension $\tilde{f}$ of $f$ to $X$ satisfying $|\tilde{f}(x)|\le \Vert f\Vert \Vert x\Vert$ for all $x\in X$. This implies that $\tilde{f}$ is bounded and $\Vert \tilde{f}\Vert \le \Vert f\Vert$. We have $\Vert \tilde{f}\Vert \ge \Vert f\Vert$, because $\tilde{f}$ extends $f$. $\square$

Corollaries

Important consequence: canonical embedding into bidual

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 $f(x)=\Vert x\Vert$ and $\Vert f\Vert =1$.

Proof   On the linear subspace $\mathbb{K}x\subset X$ spanned by $x$ we define a functional $f_0$ by $f_0(\alpha x)=\alpha \Vert x\Vert$ for $\alpha \in \mathbb{K}$. It is easy to check that $f_0$ is linear and bounded with norm $\Vert f_0\Vert =1$. 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 $f(x)=f_0(x)=\Vert x\Vert$ and $\Vert f\Vert =\Vert f_0\Vert =1$. $\square$

Recall that for a normed space $X$ we denote its (topological) dual space by $X^{\prime }$.

For every element $x$ of a real or complex normed space $X$ one has

$$\Vert x\Vert =\underset{f\in X^{\prime }\setminus \{0\}}{\sup }\frac{|f(x)|}{\Vert f\Vert }$$

and the supremum is attained.

The elements of a real or complex normed space $X$ are separated by the elements of its dual $X^{\prime }$.

Separation Theorems

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 $f(x)=\mathrm{dist}(x,Z)$.

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

$$\Vert y+Z\Vert =\mathrm{dist}(y,Z)=\underset{z\in Z}{\inf }\Vert y-z\Vert \forall y\in X.$$

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 $g(\pi (x))=\Vert x\Vert =\mathrm{dist}(x,Z)\ne 0$. Now the composition $f=g\circ \pi$ is a bounded functional on $X$ with the desired properties. $\square$

Hahn–Banach Theorem (Separation, Convex Sets)

TODO