Hahn Banach Theorem

Sublinear

Let $X$ be a normed vector space over a field $F$, $F\in \{\mathbb{R},\mathbb{C}\}$,$p:X\to \mathbb{R}$ is sublinear if it satisfies:

• $p$ is positively homogeneous : $p(\lambda x)=\lambda p(x)$, $\forall \lambda >0,x\in X$.
• $p$ satisfy triangle inequality/subadditivity: $p(x+y)\le p(x)+p(y)$, $\forall x,y\in X$

Hahn Banach Theorem (original)

Let $X$ be a normed vector space over $\mathbb{R}$ Let $Y$ be a vector subspace of $X$. Suppose $p:X\to \mathbb{R}$ is sublinear, $L:Y\to \mathbb{R}$ linear and satisfies $L(y)\le p(y)$, $\forall y\in Y$. Then $\exists l:X\to \mathbb{R}$ linear such that

\begin{cases} l(x)\leq\ p(x), &\ \forall x \in\ X \ l(y)\ =\ L(y) & \forall\ y\ \in\ Y \end{cases}

>[!remark] > >- $X$ not necessarily complete and $Y$ not necessarily closed or complete. >- $L$ and $l$ not necessarily bounded. >- Usually we need some small argument to show $l \in X^*$. >[!Corollary] > >Let $Y$ be a subspace of a normed [[Vector Spaces#^f4b63e|vector space]] $X$ over $\mathbb{R}$, every $L \in Y^*$ extend to a bounded linear functional $l \in X^*$ with $\|l\|_{X^{*}}= \|L\|_{Y^*}$ *Proof* Let $p(x) = \|L\|_{Y^*}\|x\|$, $\forall x\in X$, $p$ is sublinear and $L(y) \leq |L(y)| \leq p(y)$ $\forall y \in Y$. By Hahn-Banach theorem, we have a linear extension $l : X \to \mathbb{R}$ with $l(x) \leq p(x) = \|L\|_{Y^*}\|x\|$ $\forall x\ \in X$. Then $-l(x) =l(-x) \leq \|L\|_{Y^*}\|x\|$ gives us $|l(x)| \leq \|L\|_{Y^*}\|x\|$, so $l \in X^*$ and $\|l\|_{X^{*}}\leq \|L\|_{Y^*}$. $\|l\|_{X^{*}}\geq \|L\|_{Y^*}$ as $\sup_{\|x\|=1 ,x \in X}|l(x)| \geq \sup_{\|x\|=1 ,x \in Y}|L(x)|$. So $\|l\|_{X^{*}}= \|L\|_{Y^*}$. $\square$ >[!Corollary] > >If $X$ a normed [[Vector Spaces#^f4b63e|vector space]], $X^*$ the dual space of $X$ over $\mathbb{R}$, $\forall x_{0} \in X$, $\exists l \in X^*$ such that $$\|l\|_{X^{*}}= 1 \ \text{with} \ \|x_{0}\|_{X} = l(x_{0})$$ >In particular, we have $$\sup_{l \in X^{*},\|l\|_{X^{*}}= 1}|l(x_{0})| = \|x_{0}\|_{X}$$ *Proof* Consider $Y = \mathrm{span}\{x_{0}\}$ and define $L(ax_{0}) = a\|x_{0}\|$, we have $L \in Y^*$ and $L(y) \leq p(y)$ for $p(y) = \|y\|$. Moreover, $\|L\|_{Y^{*}} = \sup_{\|y\| = 1}|L(y)| = \|y\| = 1$. So by previous corollary, $L$ extends to $l \in X^*$ with $\|l\|_{X^{*}}= \|L\|_{Y^{*}}= 1$. $l(x_{0}) = L(x_{0}) = \|x_{0}\|_{X}$. As we always get $$|l(x_{0})| \leq \|l\|_{X^*}\|x_{0}\|_{X},\sup_{\|l\|_{X^{*}}= 1}|l(x_{0})| \leq \|x_{0}\|_{X},$$ The supremum achieved by what we showed above, so $$\sup_{l \in X^{*},\|l\|_{X^{*}}= 1}|l(x_{0})| = \|x_{0}\|_{X}$$$\square$ >[!remark] > >We can achieve the supremum above, but in general we won't achieve the supremum in $\|L\|_{X^*}$, see [[Dual Space#^05f4cb|example]].