Hahn-Banach theorem

Let (X, \Vert\cdot\Vert_X) a normed space and (X', \Vert\cdot \Vert_{X'}) its dual.

Let $U\subseteq X$, $u^{\prime }:U\to \mathbb{R}$ a continuous linear functional. Then there exists an extention $x^{\prime }:X\to \mathbb{R}$ a continuous linear functional such that $x^{\prime }(u)=u^{\prime }(u)$ for all $u\in U$ and $\Vert x^{\prime }{\Vert }_{X^{\prime }}=\Vert u^{\prime }{\Vert }_{U^{\prime }}$.

Applications §

Let $(X,\Vert \cdot {\Vert }_X)$ be a normed space.

1. For all $x\in X$, $x\ne 0$ there is a linear operator $x^{\prime }\in X^{\prime }$ with $\Vert x^{\prime }{\Vert }_{X^{\prime }}=1$ and $x^{\prime }(x)=\Vert x{\Vert }_X$. Define a linear functional on the subspace $U=span\{x\}$ as $u^{\prime }(\lambda x):=\lambda \Vert x{\Vert }_X$ using Hahn-Banach we know there exists a continuous linear functional $x^{\prime }\in X^{\prime }$, let’s check:

$$x^{\prime }(x)=u(x)=\Vert x{\Vert }_X\Vert x^{\prime }{\Vert }_{X^{\prime }}=\Vert u^{\prime }{\Vert }_{U^{\prime }}=\sup \{|u^{\prime }(u)|\mid \Vert u{\Vert }_X=1\}$$

since $|u'(u)| = |\lambda| \Vert x \Vert_X$ so if $u$ is a unit vector $\lambda = \frac{1}{\Vert x \Vert_X}$. 

2. $(X^{\prime },\Vert \cdot {\Vert }_{X\to \mathbb{R}})$ separates points of $X$: for any $x_1,x_2\in X$ such that $x_1\ne x_2$ we can always find two linear operators $l_1,l_2\in X^{\prime }$ so that $l_1(x_1)\ne l_2(x_2)$. Just use the previous result on the (non-zero) vector $x:=x_2-x_1$, define the functional $x^{\prime }\in X^{\prime }$ as above. Then $x^{\prime }(x)=\Vert x{\Vert }_X\ne 0$. By linearity $$ x’(x) = x’(x_2-x_1) = x’(x_2) - x’(x_1) \neq 0

3. For all $x \in X$ we can compute its norm by "reversing" the operator norm:

\Vert x \Vert_X = \sup{ |x’(x)| \mid x’ \in X’, \Vert x’ \Vert_{X’} = 1}

by the $\sup$ definition of the operator norm

\Vert x’\Vert_{X’} \geq \frac {|x’(x)|}{\Vert x \Vert_X}

$$andweseethat$$

\sup{ |x’(x)| \mid x’ \in X’, \Vert x’ \Vert_{X’} = 1} \leq \Vert x \Vert_X $$ for the other inequality, we use $1.$, so we know there exists a functional such that $x^{\prime }(x)=\Vert x{\Vert }_X$ with $\Vert x^{\prime }{\Vert }_{X^{\prime }}=1$, so that

$$\Vert x{\Vert }_X\le \underset{\Vert x^{\prime }{\Vert }_{X^{\prime }}=1}{\sup }|x^{\prime }(x)|$$

and we get the desired equality. Indeed the $\sup$ is a $\max$.