Editing Uniform boundedness principle (section)

==Corollaries==

{{math theorem|name=Corollary | math_statement=
If a sequence of bounded operators <math>\left(T_n\right)</math> converges pointwise, that is, the limit of <math>\left(T_n(x)\right)</math> exists for all <math>x \in X,</math> then these pointwise limits define a bounded linear operator <math>T.</math>
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The above corollary does {{em|not}} claim that <math>T_n</math> converges to <math>T</math> in operator norm, that is, uniformly on bounded sets.  However, since <math>\left\{T_n\right\}</math> is bounded in operator norm, and the limit operator <math>T</math> is continuous, a standard "<math>3\varepsilon</math>" estimate shows that <math>T_n</math> converges to <math>T</math> uniformly on {{em|compact}} sets.

{{math proof | proof =
Essentially the same as that of the proof that a pointwise convergent sequence of equicontinuous functions on a compact set converges to a continuous function.

By uniform boundedness principle, let <math>M = \max\{\sup_n \|T_n\|, \|T\|\}</math> be a uniform upper bound on the operator norms.

Fix any compact <math>K\subset X</math>. Then for any <math>\epsilon > 0</math>, finitely cover (use compactness) <math>K</math> by a finite set of open balls <math>\{B(x_i, r)\}_{i=1, ..., N}</math> of radius <math>r = \frac{\epsilon}{M}</math>

Since <math>T_n \to T</math> pointwise on each of <math>x_1, ..., x_N</math>, for all large <math>n</math>, <math>\|T_n(x_i) - T(x_i)\|\leq \epsilon</math> for all <math>i= 1,..., N</math>.

Then by triangle inequality, we find for all large <math>n</math>, <math>\forall x\in K, \|T_n(x) - T(x)\|\leq 3\epsilon</math>.
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{{math theorem | name=Corollary | math_statement=
Any weakly bounded subset <math>S \subseteq Y</math> in a normed space <math>Y</math> is bounded.
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Indeed, the elements of <math>S</math> define a pointwise bounded family of continuous linear forms on the Banach space <math>X := Y',</math> which is the [[continuous dual space]] of <math>Y.</math> 
By the uniform boundedness principle, the norms of elements of <math>S,</math> as functionals on <math>X,</math> that is, norms in the second dual <math>Y'',</math> are bounded. 
But for every <math>s \in S,</math> the norm in the second dual coincides with the norm in <math>Y,</math> by a consequence of the [[Hahn–Banach theorem]].

Let <math>L(X, Y)</math> denote the continuous operators from <math>X</math> to <math>Y,</math> endowed with the [[operator norm]]. 
If the collection <math>F</math> is unbounded in <math>L(X, Y),</math> then the uniform boundedness principle implies:
<math display=block>R = \left \{x \in X  \ : \ \sup\nolimits_{T \in F} \|Tx\|_Y = \infty \right\} \neq \varnothing.</math>

In fact, <math>R</math> is dense in <math>X.</math> The complement of <math>R</math> in <math>X</math> is the countable union of closed sets <math display="inline">\bigcup X_n.</math>
By the argument used in proving the theorem, each <math>X_n</math> is [[nowhere dense]], i.e. the subset <math display="inline">\bigcup X_n</math> is {{em|of first category}}. 
Therefore <math>R</math> is the complement of a subset of first category in a  Baire space. By definition of a Baire space, such sets (called {{em|[[Comeagre set|comeagre]]}} or {{em|residual sets}}) are dense. 
Such reasoning leads to the {{em|principle of condensation of singularities}}, which can be formulated as follows:

{{math theorem | math_statement=
Let <math>X</math> be a Banach space, <math>\left(Y_n\right)</math> a sequence of normed vector spaces, and for every <math>n,</math> let <math>F_n</math> an unbounded family in <math>L\left(X, Y_n\right).</math> Then the set
<math display=block>R := \left\{x \in X \ : \ \text{ for all } n \in \N , \sup_{T \in F_n} \|Tx\|_{Y_n} = \infty\right\}</math>
is a residual set, and thus dense in <math>X.</math>
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{{math proof| proof = 
The complement of <math>R</math> is the countable union
<math display=block>\bigcup_{n,m} \left\{x \in X \ : \ \sup_{T \in F_n} \|Tx\|_{Y_n} \leq m\right\}</math>
of sets of first category. Therefore, its residual set <math>R</math> is dense.
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