---- Here, $\mathbb{F}$ denotes the [[field]] $\mathbb{R}$ or $\mathbb{C}$. > [!theorem] Theorem. ([[Hahn-Banach Extension Theorem]]) > Suppose $X$ is a [[real numbers|real]] [[vector space]] and $Y$ a proper [[linear subspace]] of $X$. Suppose $p:X \to \mathbb{R}$ is a [[sublinear functional]] and $g:Y \to \mathbb{R}$ is a [[dual vector space|linear functional]] on $Y$ such that $g \leq p |_{Y}$. > > Then there exists a [[linear functional]] $f:X \to \mathbb{R}$ such that $f |_{Y}=g$ and $f \leq p$. > > ```tikz > \usepackage{tikz-cd} > \usepackage{amsmath} > \usepackage{amsfonts} > \begin{document} > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRAE0QBfU9TXfIRQBGclVqMWbABrdeIDNjwEiZYePrNWiEAB1dAWzo4AFgCMzwAEpdu4mFADm8IqABmAJwgGkZEDggkUX86LAY2EwgIAGs5dy8fRD8ApAAmak0pHUcAAn0cGAAPHGAchiwwGDoPW2oGOjMYBgAFfmUhEA8sRxMcOJBPbyDqFMR0kEawKCQAdmDM7T1dBhgARxy0PN0C4tK4JjNyyuquHP7BxODR8aPFqDo4EwcQDMlF-SKsOBw4HLctnYlMoVKo1F4geqNFptQRsCrYWB2LhAA > \begin{tikzcd} > Y \arrow[r, hook] \arrow[d, "g \text{ linear}"'] & X \arrow[ld, "\leq p \text{ sublinear} ", bend left=71] \arrow[ld, "\exists f \text{ linear}" description, dashed] \\ > \mathbb{R} & > \end{tikzcd} > \end{document} > ``` > > [!specialization] Corollary. ([[Hahn-Banach Extension Theorem|Hahn-Banach for Banach Spaces]]) > Suppose $X$ is a [[seminorm|(semi)]][[norm|normed]] [[vector space]], $Y$ is a [[linear subspace]] of $X$, and $g:Y \to \mathbb{F}$ is a [[operator norm|bounded]] [[linear map|linear]] [[dual vector space|functional]]. Then $g$ can be extended to a [[operator norm|bounded]] [[linear map|linear]] [[dual vector space|functional]] $f$ on $X$ whose norm equals $\|g\|$. ^theorem > [!proof]- Proof of Corollary. > **Real case.** Assume $g$ is bounded, i.e., $\|g\|<\infty$. Applying the Theorem with $p:= \|g\| \, \|\cdot\|$, we obtain[^1] $f \in X^{\vee}$ such that $f |_{Y}=g$ and $|f| \leq \|g\|\, \|\cdot\|$. Thus $\|f\| \leq \|g\|$. Of course $\|g\| \leq \|f\|$ monotonicity of [[supremum]]. Thus $f$ is bounded and indeed $\|f\|=\|g\|$ as claimed. [^1]: We can pass from $f$ to $|f|$ because since $p$ is a [[seminorm]] $f(x)\leq p(x)$ implies $-f(x)=f(-x) \leq p(-x)=p(x)$. So $|f(x)|\leq p(x)$. > [!note] Remark. > The upper bound is what makes this analytically useful (otherwise the result is just an exercise with [[basis|Hamel bases]]). ^note > [!proof]- Proof. ([[Hahn-Banach Extension Theorem]]) > ~ ---- #### ----- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` [[successive approximations lemma]]