----- > [!proposition] Proposition. (FFA 7.7) > Let $X,Z$ be vector spaces with *algebraic* [[dual vector space|dual spaces]] $X',Z'$. Let $F \subset X'$, $G \subset Z'$ and consider maps $\begin{align} X &\xrightarrow{ T} Z \\ F\subset X' & \xleftarrow{T'} Z' \supset G \end{align}$ where $T'$ is the algebraic [[dual map]] of $T$. Then > > 1. A functional $f \in X'$ is $\sigma(X, F)$-[[continuous]] iff $f \in F$. > 2. A [[linear map]] $T:X \to Z$ is $\sigma(X, F)$-to-$\sigma(Y,G)$-[[continuous]] iff $T'(G) \subset F$. ^proposition > [!note] Remark. > If $f \in F$ then $f$ is $\sigma(X, F)$-continuous as we have *defined* $\sigma(X,F)$ to be the [[comparable topologies|coarsest]] [[topological space|topology]] making this so. The converse, that these are *all the functionals* that can be $\sigma(X,F)$-continuous, is often taken for granted but it does need to be checked and that is what we are doing here. ^note 1. Take $f \in X'$ weak continuous, look at $f ^{-1}(\mathbb{B}_{\mathbb{F}}^{\circ})$ 2. Nestle a weak-basic open set inside 3. Appeal to result on sheet 1 to get **1.** If $f \in F$ then $f$ is $\sigma(X, F)$-continuous as we have defined $\sigma(X,F)$ to be the [[comparable topologies|coarsest]] [[topological space|topology]] making this so. The converse, that these are *all the functionals* that can be $\sigma(X,F)$-continuous, is often taken for granted but it does need to be checked and that is what we are doing here. So let $f \in X'$ be $\sigma(X,F)$-continuous, WTS $f \in F$. Consider the $\sigma(X,F)$-open set $f ^{-1}(\mathbb{B}_{\mathbb{F}}^{\circ})=\{ x \in X: |f(x)| < 1 \},$ being open, it contains another $\sigma(X,F)$-basic open set of the form $U_{f_{1},\dots,f_{n}, \varepsilon}(0)=\{ x \in X : |f_{1}(x)| < \varepsilon, \dots, |f_{n}(x)|<\varepsilon\} \subset U_{f, 1}(0).$ for some $f_{1},\dots,f_{n} \in F$, $\varepsilon>0$. Now if $x \in \bigcap_{i=1}^{n} \operatorname{ker }f_{i} \subset U_{f_{1},\dots,f_{n}, \varepsilon}(0) \subset U_{f,1}(0)$, then also any scaling $\lambda x \in \bigcap_{i=1}^{n} \operatorname{ker }f_{i}$, and as a consequence $x \in \operatorname{ker }f$: if not, then we'd be able to obtain $\lambda$ big enough that $|f(\lambda x)|\geq 1$ which contradicts $x \in U_{f,1}(0)$. Hence $\bigcap_{i=1}^{n} \operatorname{ker }f_{i} \subset \operatorname{ker } f$ and by [[FFA Sheet 1]]/[[orthogonal complement|annihilator]] this means $f \in \operatorname{span}(f_{1},\dots,f_{n})$ and thus $f \in F$. **2.** By the [[universal property]] of [[initial topology]], $T:X \to Z$ is [[continuous]] if and only if $g \circ T$ is continuous for all $g \in G$. Now, $g \circ T=T'g\in X'$, which by $(1)$ we know is continuous if and only if in fact $T'g \in F \subset X'$. The result follows. > [!proposition] Corollary. ([[there are no weakly continuous functionals past the generating set]]) > Let $T:X \to Y$ be a [[linear map|linear map]] between [[norm|normed]] [[vector space|spaces]]. Then $T$ is [[continuous]] wrt the norm [[topological space|topologies]] on $X$ and $Y$ if and only if it is [[continuous]] wrt the [[weak topology|weak topologies]] on $X$ and $Y$. ^proposition > [!proof]- Proof. ([[there are no weakly continuous functionals past the generating set]]) > Recall that [[weakly convergent sequences in normed spaces are bounded]]. > > For one direction, use part 2 of the proposition above: if $T$ is norm-continuous and linear, then $T'(Y^{*})\subset X^{*}$ and so $T$ is weak continuous by the proposition part 2. > > For other direction, Suppose not. Then for each $M>0$ there exists $x \in X$ such that $\|Tx\| > M \|x\|$. Let $M_{n}=n^{2}$, then $(y_{n})$ be a sequence such that $\|Ty_{n}\| \geq n^{2} \|y_{n}\|$. Define $x_{n}:= \|Ty_{n}\| y_{n}$; then $\|x_{n}\| \leq \frac{1}{n^{2}}$ with $\|Tx_{n}\| \geq 1$. Define $u_{n}:=n x_{n}$. Then $\|u_{n}\| \leq \frac{1}{n}$ and so $\|u_{n}\| \to 0$. Now $Tu_{n} \to T(0)=0$, [[the sequential continuity lemma|using]] that $T$ is weakly continuous. By [[weakly convergent sequences in normed spaces are bounded]], $Tu_{n}=n Tx_{n}$ is a [[bounded set|bounded]] sequence. Of course, $\|nTx_{n}\| \geq n$, so this is a contradiction. > > > > > > [[uniform boundedness principle]] > > > > > > [[the sequential continuity lemma|We'll show]] $T$ is [[the sequential continuity lemma|sequentially continuous]]. Let $(x_{n})$ be a [[sequence]] in $X$ [[converge|converging]] to some $x \in X$. Then $\begin{align} > \|Tx_{n}- Tx\| &= \|T(x_{n}-x)\| = \max_{\| f\|=1} |f(x_{n }-x)|, > \end{align}$ > where we have used [[norm|computing norm with linear functionals]] (a corollary of [[Hahn-Banach Extension Theorem|HBE]]). By weak continuity, $f(x_{n}) \to f(x)$ for all $f \in X^{*}$, thus $f(x_{n}-x) \to 0$ for all $f \in X^{*}$. We conclude $\|Tx_{n}-Tx\| \to 0$ and thus $Tx_{n} \to Tx$ in norm. Thus $T$ is norm-continuous. ----- #### ---- #### 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 > ```