----- > [!proposition] Proposition. ([[weakly convergent sequences in normed spaces are bounded]]) > 1. Let $X$ be a [[norm|normed vector space]] and suppose $(x_{n})$ is a [[sequence]] in $X$ [[converge|converging]] [[weak topology|weakly]] to $x \in X$. Then $(x_{n})$ is [[bounded set|bounded]]. > 2. Let $X$ be a [[Banach space]] and $f_{n} \to f$ weak$^{*}$ly in $X^{*}$. Then $(f_{n})$ is bounded. Fails if we don't assume [[complete|completeness]].[^5] ^proposition [^5]: Consider $X=C_{c}(0,1)$ [[Lp-norm|with]] $L^{2}$-norm. Define $f_{n}$ by $f_{n}(x)=n \int _{0}^{1/n} x(t) \, dt$. Then $f_{n} \in X^{*}$ with $\|f_{n}\|=\sqrt{ n }$. For all $x \in X$, $f_{n} (x) \to 0$ (using [[compact]] [[support]] of $x$), [[initial topology|thus]] $(f_{n})$ converges weak$^{*}$ly to the zero functional. But $f_{n}$ is not bounded. Notice that this implies that being weak$^{*}$ [[compact]] does not imply being [[bounded set|bounded]] (because a convergent [[sequence]] together with its limit is always [[compact]] as a set). > [!proof]+ Proof. ([[weakly convergent sequences in normed spaces are bounded]]) > (Really we only need that $(x_{n})$ is such that $\big(f(x_{n})\big)$ is bounded for all $f \in X^{*}$.) > > [[double dual of a finite-dimensional vector space is naturally isomorphic to that space|bidual]] > > We want to show $\sup_{n \in \mathbb{N}} \|x_{n}\|<\infty$. Recall that we have a [[natural transformation|natural]] [[topological embedding|embedding]] $\begin{align} > X &\hookrightarrow X^{* *} \\ > x & \mapsto \_ (x) > \end{align}$ > and that this embedding is [[Lipschitz continuous|isometric]]. Thus it suffices to show $\sup_{n \in \mathbb{N}} \| \_ (x)\|<\infty$. Since $x_{n} \to x$ weakly, $f(x_{n}) \to f(x)$ for all $f \in X^{*}$, and so the [[sequence]] $\big( f(x_{n}) \big)$ is [[bounded set|bounded]], i.e., $\sup_{n \in \mathbb{N}} \|f(x_{n})\|<\infty$ for all $f \in X^{*}$. Now the [[uniform boundedness principle]] applied with $\mathscr{A}:= \{ T_{n}:= \_(x_{n}) \}_{n \in \mathbb{N}}$ (and recall continuous duals are [[Banach space|Banach]]). This finishes. ^proof ----- #### ---- #### 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 > ```