---- > [!theorem] Theorem. ([[uniform boundedness principle]]) > Suppose $V$ is a [[Banach space]], $W$ is a [[norm|normed]] [[vector space]], and $\mathscr{A}$ is a family of [[operator norm|bounded]] [[linear map|linear maps]] $V \to W$ such that $\sup_{T \in \mathscr{A}} \|Tf\|<\infty \text{ for every } f \in V.$ Then $\sup_{T \in \mathscr{A}} \|T\|<\infty.$ ^theorem > [!proof]- Proof. ([[uniform boundedness principle]]) > Let $\mathscr{A}$ be a family of bounded linear maps $V \to W$ such that for all $f \in V$, $\sup_{T \in \mathscr{A}}\|Tf\|<\infty$. Then $V=\bigcup_{n=1}^{\infty} \underbrace{ \{ f \in V: \|Tf\| \leq n \text{ for all }T \in \mathscr{A} \} }_{ V_{n} }.$ > [[characterizing continuity of linear maps|Because each]] $T \in \mathscr{A}$ is [[continuous]], $V_{n}$ is a [[closed set|closed subset]] of $V$ for all $n \in \mathbb{N}$.[^1] [[Baire category theorem|Baire's Theorem]] says that $V$ is a [[Baire space]], meaning that one of the $V_{n}$ does not have empty [[topological interior|interior]]: there exist $n \in \mathbb{N}$, $h \in V$, and $r > 0$ such that $B_{r}(h) \subset V_{n}$, that is, such that $\|f - h\|< r \implies \|Tf\| \leq n \text{ for all }T \in \mathscr{A}.$ > Now suppose $g \in V$ with $\|g\|<1$. Thus $rg+h \in B_{r}(h)$. Hence if $T \in \mathscr{A}$, then $\|T(rg+h)\| \leq n$, which implies that $\|Tg\|=\|\frac{T(rg+h)}{r}- \frac{Th}{r}\| \leq \frac{\|T(rg+h)\|}{r} + \frac{\|Th\|}{r} \leq \frac{n+\|Th\|}{r}.$ > Hence for all $T \in \mathscr{A}$, $\|T\| \leq \frac{n+\|Th\|}{r}$. Taking the [[supremum]] on both sides, we get > $\sup_{T \in \mathscr{A}} \|T\| \leq \frac{n + \overbrace{ \sup_{T \in \mathscr{A}} \|Th\| }^{ < \infty, \text{ by hypothesis} }}{r}<\infty,$ > completing the proof. [^1]: Justification: for each fixed $T \in \mathscr{A}$, the set $F_{n, T}:=\{ f\in V: \|Tf\| \leq n \}$ is closed because if $\varphi:V \to [0, \infty)$ is given by $\varphi(f):=\|Tf\|$, then $h ^{-1}([0,n])=\{ f \in V: 0 \leq \varphi(f) \leq n \}=F_{n, T}$. Since $[0,n]$ is a closed subset of $[0,\infty)$, its preimage under the continuous map $\varphi$ (the composition $\|\cdot\| \circ T$ of continuous maps) is a closed subset of $V$. Now $V_{n}=\bigcap_{T \in \mathscr{A}} F_{n, T}$ is closed, as an (arbitrary) intersection of closed sets. ---- #### ----- #### 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 > ```