----- > [!proposition] Proposition. ([[characterizing bounded linear functionals]]) > Suppose $V$ is a [[normed vector space]] and $\varphi:V \to \mathbb{F}$ is a nonzero [[dual vector space|linear functional]]. Then the following are equivalent: > 1. $\varphi$ is a [[operator norm|bounded linear functional]]; > 2. $\varphi$ is a [[continuous]] [[linear functional]]; > 3. $\operatorname{ker }\varphi$ is a [[closed set|closed subspace]] of $V$; > 4. $\overline{\operatorname{ker } \varphi} \neq V$, i.e., $\operatorname{ker } \varphi$ [[closure|is not]] [[dense]] [[kernel of a linear map|in]] $V$. > Note that [[every infinite-dimensional normed vector space has a discontinuous linear functional]]. > [!note] Remark. > It is always true that $(2) \implies (3)$ for general [[linear map|linear maps]]. But for general linear maps $(3) \nRightarrow (2)$. (E.g. Example 6.45 gives a discontinuous (unbounded) linear map whose kernel is $\{ 0 \}$.) ^note > [!proof]- Proof. ([[characterizing bounded linear functionals]]) > That $(1) \iff (2)$ is a special case of [[characterizing continuity of linear maps|continuous iff bounded for linear maps between normed vector spaces]]. That $(2) \implies (3)$ follows from the fact that $\{ 0 \}$ is closed in $\mathbb{F}$ and $\operatorname{ker }\varphi= \varphi ^{-1} (\{ 0 \})$. We will prove $(3) \implies (1)$ via the contrapositive. So suppose $\varphi$ is not [[operator norm|bounded]]. Then there exists a [[sequence]] $f_{1},f_{2},\dots$ in $V$ such that $\|f_{k}\| \leq 1$ and $|\varphi(f_{k})| \geq k$ for each $k \in \mathbb{N}$. Now $\frac{f_{1}}{\varphi(f_{1})}-\frac{f_{k}}{\varphi(f_k)} \in \operatorname{ker } \varphi$ for each $k \in \mathbb{N}$ and $\lim_{k \to \infty} \left( \frac{f_{1}}{\varphi(f_{1})}- \frac{f_{k}}{\varphi(f_{k})} \right)=\frac{f_{1}}{\varphi(f_{1})}.$ > But $\varphi\left( \frac{f_{1}}{\varphi(f_{1})} \right)=1 \not \in \operatorname{ker } \varphi$. Thus $\operatorname{ker } \varphi$ is not [[closed set|closed]]: here is a [[limit point]] not inside. We have now shown $(1) \iff (2) \iff (3)$. It is easy to see $(3) \implies (4)$: if $\operatorname{ker } \varphi$ is closed, then $\overline{\operatorname{ker } \varphi}=\operatorname{ker } \varphi$; this cannot equal all of $V$ because we assumed $\varphi \neq 0$. To complete the proof, we shall show $(4) \implies (3)$. We will do this contrapositively: suppose $\operatorname{ker }\varphi$ is not closed in $V$. From the properties in [[norm]] we know that because $\operatorname{ker } \varphi$ is a [[linear subspace]] of $V$, so is $\overline{\operatorname{ker }\varphi}$. Let $f \in \overline{\operatorname{ker } \varphi}-\operatorname{ker } \varphi$. Suppose $g \in V$. Then $g=\left( g- \frac{\varphi(g)}{\varphi(f)}f \right)+\frac{\varphi(g)}{\varphi(f)}f.$ > The first term is in $\operatorname{ker } \varphi$[^1] and hence in $\overline{\operatorname{ker } \varphi}$; the second term is also as a scalar multiple of an element in the [[linear subspace]] $\overline{\operatorname{ker } \varphi}$. Thus $g \in \overline{\operatorname{ker } \varphi}$. But $g$ is arbitrary, so $V \subset \overline{\operatorname{ker } \varphi}$. Thus $V= \overline{\operatorname{ker } \varphi}$. [^1]: $\varphi\left( g- \frac{\varphi(g)}{\varphi(f)}f \right)=\varphi(g)- \frac{\varphi(g)}{\varphi(f)} \varphi(f)=0$. ----- #### ---- #### 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 > ```