Editing Uniform boundedness principle (section)

==Theorem==

{{math theorem|name=Uniform Boundedness Principle|math_statement=
Let <math>X</math> be a [[Banach space]], <math>Y</math> a [[normed vector space]] and <math>B(X,Y)</math> the space of all [[continuous linear operators]] from <math>X</math> into <math>Y</math>. Suppose that <math>F</math> is a collection of continuous linear operators from <math>X</math> to <math>Y.</math> If, for every <math>x \in X</math>,
<math display=block>\sup_{T \in F} \|T(x)\|_Y < \infty,</math>
then
<math display=block>\sup_{T \in F} \|T\|_{B(X,Y)} < \infty.</math>
}}

The first inequality (that is, <math display=inline>\sup_{T \in F} \|T(x)\| < \infty</math> for all <math>x</math>) states that the functionals in <math>F</math> are pointwise bounded while the second states that they are uniformly bounded. 
The second supremum always equals
<math display=block>\sup_{T \in F} \|T\|_{B(X,Y)} = \sup_{\stackrel{T \in F}{\|x\| \leq 1}} \|T(x)\|_Y = \sup_{T \in F} \sup_{\|x\| \leq 1} \|T(x)\|_Y</math>
and if <math>X</math> is not the trivial vector space (or if the supremum is taken over <math>[0, \infty]</math> rather than <math>[-\infty, \infty]</math>) then closed unit ball can be replaced with the unit sphere
<math display=block>\sup_{T \in F} \|T\|_{B(X,Y)} = \sup_{\stackrel{T \in F,}{\|x\| = 1}} \|T(x)\|_Y.</math>

The completeness of the Banach space <math>X</math> enables the following short proof, using the [[Baire category theorem]].

{{math proof | proof =
Suppose <math>X</math> is a Banach space and that for every <math>x \in X,</math>
<math display=block>\sup_{T \in F} \|T(x)\|_Y < \infty.</math>

For every integer <math>n \in \N,</math> let
<math display=block>X_n = \left\{x \in X \ : \ \sup_{T \in F} \|T (x)\|_Y \leq n \right\}.</math>

Each set <math>X_n</math> is a [[closed set]] and by the assumption,
<math display=block>\bigcup_{n \in \N} X_n = X \neq \varnothing.</math>

By the [[Baire category theorem]] for the non-empty [[complete metric space]] <math>X,</math> there exists some <math>m \in \N</math> such that 
<math>X_m</math> has non-empty [[Interior (topology)|interior]]; that is, there exist <math>x_0 \in X_m</math> and <math>\varepsilon > 0</math> such that
<math display=block>\overline{B_\varepsilon (x_0)} ~:=~ \left\{x \in X \,:\, \|x - x_0\| \leq \varepsilon \right\} ~\subseteq~ X_m.</math>

Let <math>u \in X</math> with <math>\|u\| \leq 1</math> and <math>T \in F.</math>
Then:
<math display=block>\begin{align}
\|T(u)\|_Y &= \varepsilon^{-1}\left\|T\left(x_0 + \varepsilon u\right) - T\left(x_0\right)\right\|_Y    & [\text{by linearity of } T ] \\
&\leq \varepsilon^{-1}\left(\left\| T (x_0 + \varepsilon u) \right\|_Y + \left\|T(x_0)\right\|_Y \right ) \\
&\leq \varepsilon^{-1}(m + m).   & [ \text{since } \ x_0 + \varepsilon u, \ x_0 \in X_m ] \\
\end{align}</math>

Taking the supremum over <math>u</math> in the unit ball of <math>X</math> and over <math>T \in F</math> it follows that
<math display=block>\sup_{T \in F} \|T\|_{B(X,Y)} ~\leq~ 2 \varepsilon^{-1} m ~ < ~ \infty.</math>
}}

There are also simple proofs not using the Baire theorem {{harv|Sokal|2011}}.