----- > [!proposition] Proposition. ([[linear functional characterization of linear subspace closure]]) > Suppose $U$ is a [[linear subspace]] of a [[norm|normed]] [[vector space]] $V$ and $h \in V$. Then $h \in \overline{U}$ if and only [[orthogonal complement|if]] $\varphi \in U^{\circ} \implies \varphi(h)=0$ for all $\varphi \in V^{\vee}$. (That is, iff $\varphi \in (U^{\circ})_{\circ}$.) ^proposition > [!proposition] A quantitative strengthening of the above. > If $Y$ is a [[linear subspace]] of $X$ and $x_{0} \in X$, then there exists $f \in Y^{\circ}$ such that $\|f\| \leq 1$ [[distance from point to set|and]] $f(x_{0})=\text{dist}(x_{0}, Y)$. > > In particular, $\overline{Y}=(Y^{\circ})_{\circ}$ and $Y$ is [[dense]] in $X$ iff $Y^{\circ}=(0)$. > (Later we'll see $(Y_{\circ})^{\circ}=\overline{Y}^{\sigma(X^{*}, X)}$) > [!proof]- Proof. > Let $Y \leq X$ and $x_{0} \in X$. Have a [[seminorm]] $\begin{align} > p: X &\to \mathbb{R} \\ > x & \mapsto \text{dist}(x, Y). > \end{align}$ > > [[Hahn-Banach Extension Theorem|Hahn-Banach]] (Corollary) allows us to extend the [[dual vector space|linear functional]] $g:\span(x_{0}) \to \mathbb{R}$ determined by $x_{0} \mapsto \text{dist}(x_{0}, Y)=p(x_{0})$ to a [[dual vector space|linear functional]] $f:X \to \mathbb{R}$ such that $\|f\|=\|g\|=\text{dist}(x_{0},Y)$ and $|f(x)|\leq p(x)\leq \|x\|$[^1] for all $x \in X$. Thus we see $\|f\| \leq 1$ and $f(y) \leq p(y)=\text{dist}(y, Y)=0$ for all $y \in Y$, hence $f \in Y^{\circ}$. > > *In particular...* these checks follow from the [[distance from point to set|fact]] that $\text{dist}(x, Y)=0$ [[closure|iff]] $x \in \overline{Y}$. To be super explicit, write $\begin{align} > (Y^{\circ})_{\circ} &= \{ x_{0} \in X: \varphi (Y)=0 \implies \varphi(x_{0})=0 \text{ for all } \varphi \in X^{\vee} \} \\ > &\text{ and } \\ > \overline{Y} &= \{ x_{0} \in X: \text{dist}(x_{0} ,Y)=0 \}. > \end{align}$ > We have to show two inclusions. > $\subset$. Let $x_{0} \in (Y^{\circ})_{\circ}$. Take $\varphi=f$ as define above. Then $f(Y) \leq \text{dist}(Y,Y)=0$, hence $f(x_{0})=0$. But $f(x_{0})=\text{dist}(x_{0}, Y)$. > > $\supset$. Suppose $x_{0} \in X$ is such that $\text{dist}(x_{0}, Y)=0$ and $\varphi \in X^{\vee}$ is such that $\varphi(Y)=0$. That $\varphi(x_{0})=0$ then follows from [[continuous|continuity]] of $\varphi$: since $x_{0}$ can be made arbitrarily close to $Y$, $\varphi(x_{0})$ can be made arbitrarily close to $\varphi(y)=0$, hence equals zero. > > Moreover, if $\overline{Y}=X$ then we have $(Y^{\circ})_{\circ}=X$, implying $\varphi(Y)=0 \implies \varphi(X)=0$ for all $\varphi \in X^{\vee}$. The only $\varphi$ satisfying this condition is $\varphi=0$. Conversely, we have $\overline{Y}=(Y^{\circ})_{\circ}=(0)_{\circ}=X$. > > [^1]: $f\leq p$ implies $|f|\leq p$ because $-f(x)=f(-x)\leq p(-x)=p(x)$ for all $x$, using that $f$ is [[linear map|linear]] and $p$ is a [[seminorm]]. > [!proof]- Proof. ([[linear functional characterization of linear subspace closure]]) > First suppose $h \in \overline{U}$. [[dual vector space|If]] $\varphi \in V^{\vee}$ with $\varphi (U)=0$, then $\varphi(h)=0$ by the [[continuous|continuity]] of $\varphi$.[^1] This proves one direction. > For the converse, contrapositively suppose $h \not \in U$. Define $\psi:U + \span(h) \to \mathbb{F}$ by $\psi(f + \alpha h)=\alpha$ for $f \in U$ and $\alpha \in \mathbb{F}$. Then $\psi$ is a [[dual vector space|linear functional]] on $U+\span(h)$ with $\operatorname{ker }\psi=U$ and $\psi(h)=1$. [[closure|Because]] $h \not \in \overline{U}$, $\overline{\operatorname{ker } \psi} \neq U+\span(h)$. Thus [[characterizing bounded linear functionals]] says that $\psi$ is a [[operator norm|bounded]] [[linear map|linear]] functional on $U+\span(h)$. The [[Hahn-Banach Extension Theorem]] implies that $\psi$ can be extended to a bounded linear functional $\varphi$ on $V$. Thus we have witnessed the existence of $\varphi \in V^{\vee}$ satisfying $\varphi(U)=0$ but $\varphi(h) \neq 0$. ----- #### [^1]: To be precise: $h \in \overline{U}$ means that either $h \in U$ [[closure is set together with limit points|or]] $h$ is a [[limit point]] of $U$. If $h \in U$ then $\varphi(h)=0$ and we're done, so assume $h$ is a [[limit point]] of $U$. Since we're working in the setting of normed vector spaces, there is a [[sequence]] $(h_{n})$ of elements in $U$ converging to $h$. [[the sequential continuity lemma|sequential continuity|By]] ([[metric-space continuity|sequential]]) continuity, then: $\varphi(h)= \lim_{n \to \infty} \underbrace{ \varphi(h_{n}) }_{ 0 }=0.$ ---- #### 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 > ```