# 泛函中的重要定理之 Hahn-Banach 定理

Hahn-Banach Theorem is one of the core theorems of linear functional analysis. The Hahn-Banach Theorem in Vector Space is also called Analytic Form of Hahn-Banach Theorem. Two corollaries are especially important: Hahn-Banach Theorem in Normed Vector Space, and Geometric Form of Hahn-Banach Theorem.

## Analytic Form

Let $X$ be a real vector space, and $p$ is a sublinear functional in $X$, that is, $p:X\rightarrow\mathbb{R}$ is a function satisfies the following properties:

$$\begin{eqnarray}p(\alpha x) = \alpha p(x),\quad \forall \alpha>0 \text{ and } x\in X,\\p(x+y) \leqslant p(x)+p(y)\quad \forall x,y\in X.\end{eqnarray}$$
And let $Y$ be a subspace of $X$, $l:Y\rightarrow\mathbb{R}$ is a linear functional in $Y$ which satisfies
$$l(y)\leqslant q(y),\quad \forall y\in Y.$$
Then there exists a linear functional $\widetilde{j}:X\rightarrow\mathbb{R}$, such that
$$\widetilde{l}(y)=l(y),\quad \forall y\in Y.\quad \text{ and } \widetilde{l}(y)\leqslant p(x),\quad \forall x\in X.$$

[证明.]
Two steps to complete this proof.

Step (i). Add one element to $Y$ and form a larger vector space, and proof for this space.

Assume $Y\subsetneqq X$, Given $x_0\in X-Y$ (Obviously, $x_0\neq 0$), define the subspace of $X$

$$\text{Dom } f:= {(\alpha x_0 +y)\mid \alpha\in \mathbb{R}, y\in Y},$$

then $Y\subset \text{Dom } f$.

Claim 1. Exists linear functional $f:\text{Dom } f\rightarrow\mathbb{R}$, such that

$$f(y)=l(y)~ \forall y\in Y\quad \text{and} \quad f(x)\leqslant p(x)~ \forall x\in \text{Dom } f.$$

The problem can be reduced to find a real number $\lambda:=f(x_0)$, such that

$$f(\alpha x_0 +y) = f(\alpha x_0) + f(y) = \alpha \lambda +l(y)\leqslant p(\alpha x_0 +y), \quad \forall \alpha \in \mathbb{R}, y\in Y.$$

This inequality holds for $\alpha=0$, thus $\lambda$ only need to satisfy:

$$\begin{eqnarray} \lambda\leqslant\alpha ^{-1}[p(\alpha x_0 +y) - l(y)] = p(x_0 + \alpha^{-1} y) - l(\alpha^{-1} y) ,\quad \forall\alpha>0, y\in Y;\\ \lambda\geqslant\alpha ^{-1}[p(\alpha x_0 +y) - l(y)] = -p(-x_0 + \alpha^{-1} y) + l(-\alpha^{-1} y),\quad \forall\alpha<0, y\in Y; \end{eqnarray}$$

By linearity of $l$ and sublinearity of $p$, we have
$$-p(-x_0 + u) + l(u)\leqslant p(x_0 + v) - l(v),\quad \forall u,v \in Y.$$

Let
$$a:=\sup_{u\in Y}{-p(-x_0 + u) + l(u)}\leqslant b:=\inf_{v\in Y}{p(x_0 + v) - l(v)},$$
and chose $\lambda$ that satisfies $a<\lambda<b$ will done.

Step (ii). functionals that satisfy previous property form a set with appropriate order, Zorn’s Lemma gives the existence of $\widetilde{l}$.

All the linear functionals $f:\text{Dom } f\rightarrow\mathbb{R}$ defined on subspaces of $X$ which contains $Y$ and satisfy the following properties form a set, we denote it by $\mathscr{F}$:
$$f(y)=l(y)~ \forall y\in Y\quad \text{and} \quad f(x)\leqslant p(x)~ \forall x\in \text{Dom } f.$$

$\mathscr{F}\neq \varnothing$ since $l\in\mathscr{F}$. And $\mathscr{F}$ is partially ordered with relation $\preccurlyeq$:
$$\text{Dom }f_1 \subset \text{Dom }f_2, f_1(x)=f_2(x),\quad \forall x\in \text{Dom }f_1.$$

Given a totally ordered subset $\mathscr{E}$ of $\mathscr{F}$, let
$$\text{Dom } g := \bigcup_{f\in \mathscr{E}}\text{Dom } f,$$
then it is a subspace of $X$.

Claim 2. $\forall x\in \text{Dom }g$,
$$g(x):=f(x),\quad \forall f\in \mathscr{E} \text{ s.t } x\in \text{Dom }f$$
clearly define a linear functional $g:\text{Dom }g\rightarrow \mathbb{R}$, and $\forall x\in \text{Dom }g$, there is $g(x)\leqslant p(x)$.

Let $x\in \text{Dom }g$ such that $x\in \text{Dom }f_1$ and $x\in \text{Dom }f_2$, and $f_1, f_2\in \mathscr{E}$, without loss of generality, assume $f_1 \preccurlyeq f_2$. Then $g(x) = f_1(x)=f_2(x)\leqslant p(x)$.

If $x_1\in \text{Dom }f_1, x_2\in \text{Dom }f_2$, then $f_1 \preccurlyeq f_2$ implies $(x_1+x_2)\in \text{Dom }f_2$, thus $g(x_1 +x_2) = f_2(x_1 +x_2) = f_2(x_1) +f_2(x_2) = g(x_1) +g(x_2)$; and $\forall \alpha \in \mathbb{R}$, $g(\alpha x_1) = f_1(\alpha x_1) = \alpha f_1(x) = \alpha g(x_1)$. [This shows that g(x) is a linear functional indeed!]

By construction of $g$, it is an upper bound of $\mathscr{E}$. By Zorn’s Lemma, $\mathscr{F}$ has a maximum element $\widetilde{l}$, it is defined on a subspace $\text{Dom }\widetilde{l}\subset X$.

Claim 3. $\text{Dom }\widetilde{l} = X$, and $\widetilde{l}\in \mathscr{F}$ satisfies every property that we required.

Prove this claim by contradiction. Suppose that $\widetilde{l}\subsetneqq X$, then repeat step (i), a linear functional $\widetilde{f}:\text{Dom }\widetilde{f}\rightarrow\mathbb{R}$ would exist, and it satisfies all the related properties, and $\widetilde{l}\preccurlyeq\widetilde{f}$. Hence the claim holds.

Theorem is now proved.

Note. Norm and Semi-norm are examples of sublinear functional, but the concept of sublinear functional is more general, since $p(\alpha x) = |\alpha|p(x)$ is only required for $\alpha > 0$.

Let $X$ be a complex vector space, and $p:X\rightarrow\mathbb{R}$ is a semi-norm in $X$. And let $Y$ be a subspace of $X$, $l:Y\rightarrow\mathbb{C}$ is a linear functional in $Y$ which satisfies

$$|l(y)|\leqslant q(y),\quad \forall y\in Y.$$

Then there exists a linear functional $\widetilde{j}:X\rightarrow\mathbb{C}$, such that

$$\widetilde{l}(y)=l(y),\quad \forall y\in Y,\quad \text{ and } \left|\widetilde{l}(y)\right|\leqslant p(x),\quad \forall x\in X.$$

## Hahn-Banach Theorem in Normed Vector Space

Denote the dual space of a normed space $X$ by $X^*$.

Let $X$ be a normed vector space, $Y$ is a subspace of $X$, and let $l:Y\rightarrow \mathbb{K}$ is a continuous linear functional. Then there exists a continuous linear functional $\widetilde{l}:X\rightarrow\mathbb{K}$ that satisfies

$$\widetilde{l}(y) = l(y) \quad \forall y\in Y, \text{ and } \|\widetilde{l}\| _{X^*} = \|l\| _{Y^*}.$$

Some notions need to be clarified first: $\|\cdot\|$ is the norm in $X$; $\|\cdot\| _{X^{*}}$ is the norm in $X^{*}$; so is $\|\cdot\| _{Y^{*}}$; and $|\cdot|$ is the norm of complex numbers.

$\forall x\in X$, let $p(x):=\|l\|_{Y^*}\|x\|$. $p:X\rightarrow\mathbb{C}$ is a norm (while $l\neq 0$), thus a sublinear functional. As $l$ is continuous, then

$$|l(y)|\leqslant\|l\|_{Y^*}\|x\|\quad \forall y\in Y.$$

By Theorem\ref{THM:HB-Complex}, there exists a linear functional $\widetilde{l}:X\rightarrow\mathbb{C}$ that satisfies

$$\widetilde{l}(y)=l(y),\quad \forall y\in Y,\quad \text{ and } \left|\widetilde{l}(x)\right|\leqslant p(x) = \|l\|_{Y^*}\|x\|,\quad \forall x\in X.$$

Hence $\widetilde{l}(x)$ is continuous and

$$\|l\| _{Y ^*} = \sup _{y\in Y, y\neq 0}\frac{|l(y)|}{\|y\|}\leqslant \sup _{x\in X, x\neq 0}\frac{|l(x)|}{\|x\|} = \left\| \widetilde{l} \right\| _{X ^*}\leqslant \|l\| _{Y^*}.$$

The real normed space case can be proved in similar way.

Remark. (1) In the case of Hilbert Space, Hahn-Banach Theorem can be proved in a simpler way without using the Axiom of Choice, Furthermore, it the uniqueness of expension is given.

(2) In general, norm-preserving extension is not necessarily unique. several examples can be found in \cite{ref1}\cite{ref2}. But with certain conditions, the uniqueness can be guaranteed (Corollary\ref{COR:HB-TF}).

## Geometric Form

Several preparatory knowledge to be mentioned first.

Let $X$ be a normed vector space. A Hyperplane is a set defined as

$$H:=x\in X\mid f(x)=\alpha},$$

in which $f$ is a linear functional that is not always zero, and $\alpha\in \mathbb{R}$. $[f=\alpha]$ is called the equation of hyperplane $H$.

Let $A,B\subset X$. $A, B$ are said to be separated by hyperplane $[f=\alpha]$ if

$$f(x)\leqslant\alpha,\quad \forall x\in A,\quad \text{and } f(x)\geqslant \alpha,\quad \forall x\in B.$$

$A, B$ are said to be strictly separated by hyperplane $[f=\alpha]$ if $\exists \varepsilon>0$, such that
$$f(x)\leqslant\alpha-\varepsilon,\quad \forall x\in A,\quad \text{and } f(x)\geqslant \alpha+\varepsilon,\quad \forall x\in B.$$ Let $A\subset X$, $B\subset X$ are two nonempty convex set. If $A$ is an open set, then there exists a hyperplane which separates $A$ and $B$.

Let $A\subset X$, $B\subset X$ are two nonempty convex set. If $A$ is an closed set and $B$ is a compact set, then there exists a hyperplane which strictly separates $A$ and $B$.

Note. 更多关于 Hahn-Banach 型定理以及 “凸空间”的内容和结果将在另一份笔记中加以陈述.（这些内容非常重要且有意义.）

Zengfk

2022-07-15

2022-09-15