---- Given a linear functional $\ell$ defined on a subspace $Y$ of a vector space $X$, when can we extend $\ell$ to a functional defined on $X$? Asked like this, the answer is "always" (using [[basis|Hamel bases]] and the [[Axiom of Choice]]), so a more useful version of this question imposes some control on $g$ and asks for extensions retaining "control". For example, if $X$ is a [[norm|normed space]] we might ask to extend *[[operator norm|bounded]]* [[linear functional|linear functionals]] $\ell$ without increasing the [[operator norm|norm]]. We use a slightly more general version of "control" we use for extension in this course is [[sublinear functional|sublinearity]] of $\ell$. (In fact our proof holds even more generally, see discussion later.) > [!theorem] Theorem. ([[Hahn-Banach Extension Theorem]], real case) > 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 $\ell:Y \to \mathbb{R}$ is a [[dual vector space|linear functional]] on $Y$ such that $\ell \leq p |_{Y}$. > > Then there exists a [[linear functional]] $f:X \to \mathbb{R}$ such that $f |_{Y}=\ell$ 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, "\ell \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} > ``` > > [!note] Remark. > HBE is distinguished by its algebraic nature. There is nothing even [[topological space|topological]] happening, in contrast to the other [[the principle results of functional analysis (PROFA)|PROFAS]] which depend on true analytic notions such as completeness. ^note (complex case is still todo, he skipped in lecture but it's in the notes) The analytic content of the HBE proof lay in the lemma below (the rest is then just [[Zorn's lemma]]). > [!proposition] Lemma. (One-step HBE) > Let $x_{0} \in X-Y$, $\ell:Y \to \mathbb{R}$ [[linear functional|linear]] with $\ell \leq p |_{Y}$. Then there exists $\tilde{\ell}:Y \oplus \langle x_{0} \rangle \to \mathbb{R}$ [[linear functional|linear]] such that $\tilde{\ell} |_{Y}=\ell$ and $\tilde{\ell} \leq p |_{Y\oplus \langle x_{0} \rangle}$. ^proposition > [!proof]- Proof of Lemma. > > Let $x_{0} \in X-Y$, $\ell:Y \to \mathbb{R}$ linear with $\ell \leq p |_{Y}$. Being linear, $\tilde{\ell}$ will be determined by $c:=\tilde{\ell}(x_{0})$. Indeed, $\tilde{\ell}( z \overset{!}{=} \overbrace{ y }^{ \in Y } + \overbrace{ t }^{ \in \mathbb{R} } x _{0} )={\ell} (y) + t c$ for all $z \in Y \oplus \langle x_{0} \rangle$ . We are looking to pick a $c$ so that the domination ${\ell}(y) + t c \overset{?}{\leq} p(y + t x_{0}) \, \, (*)$ > holds for all $y \in Y$, $t \in \mathbb{R}$. > > Casework on the sign of $t$. Observe that, in general, $p(\lambda x)=(-\lambda)p(-x)$ for $\lambda<0$. > > **Case $t=0$.** Done no matter $c$ because $\ell \leq p |_{Y}$ by hypothesis. > > **Case $t>0$.** Divide $(*)$ by $t$ and use positive homogeneity of $p$ to obtain that we desire ${\ell\left( \frac{y}{t} \right)} + c \overset{?}{\leq} p\left( \frac{y}{t} + x_{0} \right)$ > to hold for all $y \in Y$, $t>0$. Of course this is just requiring $ c \leq p(y + x_{0})-\ell(y) \text{ for all } y \in Y$ > for all $y \in Y$ (making the absorbtion $y \leftarrow \frac{y}{t}$ (which still ranges over all $Y$)). > > > **Case $t<0$.** Dividing $(*)$ by a negative number dualizes the case above, the punchline being that we require $c \geq \ell(y) - p(y-x_{0}) \text{ for all }y \in Y.$ > **Conclusion.** We conclude that any choice of $c$ satisfying $\ell(y) - p(y-x_{0}) \leq c \leq p(y+x_{0})-\ell(y) \text{ for all } y \in Y$ > suffices to define $\tilde{\ell}$. To check that such $c$ exists all we must check is that $\ell(y)-p(y-x_{0}) \overset{?}{\leq} p(y+x_{0})-\ell(y)$ for all $y \in Y$. This is true, expand $p(y-x_{0})\leq p(y)-p(-x_{0})$, the rearrange and use $p(y)=p(y+ x_{0} - x_{0})\leq p(y+x_{0}) -p(-x_{0})$ to see it. > [!specialization] Corollary. (HBE for normed spaces) > Suppose $X$ is a [[seminorm|(semi)]][[norm|normed]] [[vector space]], $Y$ is a [[linear subspace]] of $X$, and $\ell \in Y^{*}$ is a [[operator norm|bounded]] [[linear map|linear]] [[dual vector space|functional]]. Then $\ell$ can be extended to a [[operator norm|bounded]] [[linear map|linear]] [[dual vector space|functional]] $f \in X^{*}$ whose norm equals $\|g\|$. ^theorem > [!proof]- Proof of Corollary. > **Real case.** Assume $\ell$ is bounded, i.e., $\|\ell\|<\infty$. Applying the Theorem with $p:= \|\ell\| \, \|\cdot\|$, we obtain[^1] $f \in X^{\vee}$ such that $f |_{Y}=\ell$ and $|f| \leq \|\ell\|\, \|\cdot\|$. Thus $\|f\| \leq \|g\|$. Of course $\|\ell\| \leq \|f\|$ monotonicity of [[supremum]]. Thus $f$ is bounded and indeed $\|f\|=\|\ell\|$ 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]]