> [!theoremb] Theorem > > Let $\cx, \cy$ be [[Normed Vector Space|normed spaces]] and $\alg \subset L(\cx, \cy)$ be a collection of [[Bounded Linear Map|bounded linear maps]]. > 1. If there exists a [[Baire Category Theorem|non-meagre]] subset $U$ of $\cx$ such that $\sup_{T \in \alg}\norm{Tx} < \infty$ for all $x \in U$, then $\sup_{T \in \alg}\norm{T} < \infty$. > 2. If $\cx$ is a [[Banach Space]] and $\sup_{T \in \alg}\norm{Tx} < \infty$ for all $x \in \cx$, then $\sup_{T \in \alg}\norm{T} < \infty$. > > *Proof*. > > # Bounding Operators on an Open Set > > To bound the norm of the linear operators it's sufficient to bound them on an [[Open Set|open set]]. Let $\alg$ be a collection of linear maps from $\cx$ to $\cy$. If there exists an open set $U$ such that > $ > \norm{Tx} \le M \quad \forall x \in U, T \in \alg > $ > then $\alg \subset L(\cx, \cy)$ and there exists $M' > 0$ such that $\norm{T} \le M'$ for all $T \in \alg$. > > *Proof*. Since all maps in $\alg$ are bounded on $U$, they are all [[Bounded Linear Map|bounded]] and therefore continuous. > > Let $x \in U$ and $\varepsilon > 0$ such that $\ol{B(x, \varepsilon)} \subset U$. Let $y \in \cx$ such that $\norm{y} = 1$, then for any $T \in \alg$, > $ > \begin{align*} > \varepsilon \norm{Ty} = \norm{T\varepsilon y} &\le \norm{T(\varepsilon y + x)} + \norm{T(-x)} \\ > &= \norm{T(\varepsilon y + x)} + \norm{T(x)} \\ > \varepsilon \norm{Ty} &\le 2M \\ > \norm{Ty} &\le \frac{2M}{\varepsilon} > \end{align*} > $ > This means that $\norm{T} = \sup_{\norm{y} = 1}\norm{Ty} \le \frac{2M}{\varepsilon}$, yielding $\sup_{T \in \alg}\norm{T} \le 2M/\varepsilon = M'$. > > > # Creating an Open Set > > When a non-meagre set is decomposed into a countable union, at least one of its parts must be dense in some open subset (its closure has a non-empty interior). If the countable union is made out of [[Closed Set|closed sets]], then one of them must have a non-empty interior, which is the open subset that we need. > > Let $\alg \subset L(\cx, \cy)$ be a collection of bounded linear maps, and $E \subset \cx$ be a non-meagre set such that $\sup_{T \in \alg}\norm{Tx} < \infty$ for all $x \in E$. Then there exists a non-empty open set $U$ and $M > 0$ such that $\norm{Tx} \le M$ for all $T \in M$ > > *Proof*. Let $n \in \nat$ and define > $ > E_n = \bracs{x \in \cx: \sup_{T \in \alg}\norm{Tx} \le n} = \bigcap_{T \in \alg}\bracs{x \in \cx: \norm{Tx} \le n} > $ > then each $E_n$ is closed. Let $x \in E$, then $\sup_{T \in \alg}\norm{Tx} < \infty$, and there exists $n \in \nat$ such that $\sup_{T \in \alg}\norm{Tx} \le n$, meaning that $x \in E_n$. > > Since $\bigcup_{n \in \nat}E_n \supset E$, there exists $n \in \nat$ such that $E_n^o \ne \emptyset$. Let $U = E_n^o$, then $\norm{Tx} \le n$ for all $x \in U$ and $T \in \alg$. > > > # Bounding the Operators > > Let $\alg \subset L(\cx, \cy)$ be a collection of bounded linear maps, and $E \subset \cx$ be a non-meagre set such that $\sup_{T \in \alg}\norm{Tx} < \infty$ for all $x \in E$. Then there exists $M > 0$ such that $\norm{T} \le M$ for all $T \in \alg$. > > *Proof*. Since $E$ is non-meagre and $\sup_{T \in \alg}\norm{Tx} < \infty$, there exists an open set $U$ and $M > 0$ such that $\norm{Tx} \le M$ for all $x \in U$ and $T \in \alg$. This means that there exists $M'$ such that $\norm{T} \le M'$ for all $T \in \alg$, and all operators are bounded by the same constant. > > > # Banach Space > > Let $\cx$ be a Banach space. If $\sup_{T \in \alg}\norm{Tx} < \infty$ for all $x \in \cx$, then there exists $M > 0$ such that $\norm{T} \le M$ for all $T \in \alg$. > > *Proof*. Since $\cx$ is a Banach space, it is non-meagre by the [[Baire Category Theorem]]. Applying the previous steps yields the result.