----- > [!proposition] Proposition. ([[norm and weak continuity coincide for linear maps]]) > 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. ([[norm and weak continuity coincide for linear maps]]) > Recall that [[weakly convergent sequences in normed spaces are bounded]]. > > Since weak topology is [[comparable topologies|coarser]] than norm topology, it is clear that $T$ being norm-continuous implies $T$ being weak-continuous. Conversely suppose $T$ is weak-continuous. [[characterizing continuity of linear maps|We must show]] $T$ is [[operator norm|bounded]]. > > 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 > ```