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Difference between revisions of "Normal convergence"

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m (latex details)
 
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$$ \tag{1 }
 
$$ \tag{1 }
f  =  \sum _ { k= } 1 ^  \infty  u _ {k}  $$
+
f  =  \sum _ { k= 1} ^  \infty  u _ {k}  $$
  
 
formed by bounded mappings  $  u _ {k} :  X \rightarrow Y $
 
formed by bounded mappings  $  u _ {k} :  X \rightarrow Y $
 
from a set  $  X $
 
from a set  $  X $
 
into a normed space  $  Y $,  
 
into a normed space  $  Y $,  
such that the series with positive terms  $  \sum _ {k=} 1 ^  \infty  \| u _ {k} \| $
+
such that the series with positive terms  $  \sum _ {k= 1}  ^  \infty  \| u _ {k} \| $
 
formed by the norms of the mappings,
 
formed by the norms of the mappings,
  
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converges.
 
converges.
  
Normal convergence of the series (1) implies absolute and uniform convergence of the series  $  \sum _ {k=} 1 ^  \infty  u _ {k} ( x) $
+
Normal convergence of the series (1) implies absolute and uniform convergence of the series  $  \sum _ {k= 1}  ^  \infty  u _ {k} ( x) $
 
consisting of elements of  $  Y $;  
 
consisting of elements of  $  Y $;  
 
the converse is not true. For example, if  $  u _ {k} :  \mathbf R \rightarrow \mathbf R $
 
the converse is not true. For example, if  $  u _ {k} :  \mathbf R \rightarrow \mathbf R $
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and  $  u _ {k} ( x) = 0 $
 
and  $  u _ {k} ( x) = 0 $
 
for  $  x \in \mathbf R \setminus  [ k, k+ 1] $,  
 
for  $  x \in \mathbf R \setminus  [ k, k+ 1] $,  
then the series  $  \sum _ {k=} ^  \infty  u _ {k} ( x) $
+
then the series  $  \sum _ {k= 1 } ^  \infty  u _ {k} ( x) $
converges absolutely, whereas  $  \sum _ {k=} 1 ^  \infty  \| u _ {k} \| = \sum _ {k=} 1 ^  \infty  1 / k $
+
converges absolutely, whereas  $  \sum _ {k= 1}  ^  \infty  \| u _ {k} \| = \sum _ {k= 1}  ^  \infty  1 / k $
 
diverges.
 
diverges.
  
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$$  
 
$$  
 
\int\limits _ { I } f ( t)  d t  = \  
 
\int\limits _ { I } f ( t)  d t  = \  
\sum _ { k= } 1 ^  \infty  \int\limits _ { I } u _ {k} ( t)  d t .
+
\sum _ { k= 1 }^  \infty  \int\limits _ { I } u _ {k} ( t)  d t .
 
$$
 
$$
  
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====References====
 
====References====
<table><TR><TD valign="top">[1]</TD> <TD valign="top">  N. Bourbaki,  "Elements of mathematics. General topology" , Addison-Wesley  (1966)  (Translated from French)</TD></TR><TR><TD valign="top">[2]</TD> <TD valign="top">  N. Bourbaki,  "Elements of mathematics. Functions of a real variable" , Addison-Wesley  (1976)  (Translated from French)</TD></TR><TR><TD valign="top">[3]</TD> <TD valign="top">  L. Schwartz,  "Cours d'analyse" , '''1''' , Hermann  (1967)</TD></TR></table>
+
<table>
 +
<TR><TD valign="top">[1]</TD> <TD valign="top">  N. Bourbaki,  "Elements of mathematics. General topology" , Addison-Wesley  (1966)  (Translated from French)</TD></TR><TR><TD valign="top">[2]</TD> <TD valign="top">  N. Bourbaki,  "Elements of mathematics. Functions of a real variable" , Addison-Wesley  (1976)  (Translated from French)</TD></TR><TR><TD valign="top">[3]</TD> <TD valign="top">  L. Schwartz,  "Cours d'analyse" , '''1''' , Hermann  (1967)</TD></TR>
 +
</table>

Latest revision as of 07:35, 5 January 2024


Convergence of a series

$$ \tag{1 } f = \sum _ { k= 1} ^ \infty u _ {k} $$

formed by bounded mappings $ u _ {k} : X \rightarrow Y $ from a set $ X $ into a normed space $ Y $, such that the series with positive terms $ \sum _ {k= 1} ^ \infty \| u _ {k} \| $ formed by the norms of the mappings,

$$ \| u _ {k} \| = \ \sup \{ {\| u _ {k} ( x) \| } : {x \in X } \} , $$

converges.

Normal convergence of the series (1) implies absolute and uniform convergence of the series $ \sum _ {k= 1} ^ \infty u _ {k} ( x) $ consisting of elements of $ Y $; the converse is not true. For example, if $ u _ {k} : \mathbf R \rightarrow \mathbf R $ is the real-valued function defined by $ u _ {k} ( x) = ( \sin \pi x ) / k $ for $ k \leq x \leq k + 1 $ and $ u _ {k} ( x) = 0 $ for $ x \in \mathbf R \setminus [ k, k+ 1] $, then the series $ \sum _ {k= 1 } ^ \infty u _ {k} ( x) $ converges absolutely, whereas $ \sum _ {k= 1} ^ \infty \| u _ {k} \| = \sum _ {k= 1} ^ \infty 1 / k $ diverges.

Suppose, in particular, that each $ u _ {k} : \mathbf R \rightarrow Y $ is a piecewise-continuous function on a non-compact interval $ I \subset \mathbf R $ and that (1) converges normally. Then one can integrate term-by-term on $ I $:

$$ \int\limits _ { I } f ( t) d t = \ \sum _ { k= 1 }^ \infty \int\limits _ { I } u _ {k} ( t) d t . $$

Let $ f: I \times A \rightarrow Y $, where $ I \subset \mathbf R $ is an interval, have left and right limits at each point of $ I $. Then the improper integral

$$ \tag{2 } \int\limits _ { I } f ( t ; \lambda ) d t ,\ \ \lambda \in A , $$

is called normally convergent on $ A $ if there exists a piecewise-continuous positive function $ g : \mathbf R \rightarrow \mathbf R $ such that: 1) $ \| f( x ; \lambda ) \| \leq g ( x) $ for any $ x \in I $ and any $ \lambda \in A $; and 2) the integral $ \int _ {I} g ( t) d t $ converges. Normal convergence of (2) implies its absolute and uniform convergence; the converse is not true.

References

[1] N. Bourbaki, "Elements of mathematics. General topology" , Addison-Wesley (1966) (Translated from French)
[2] N. Bourbaki, "Elements of mathematics. Functions of a real variable" , Addison-Wesley (1976) (Translated from French)
[3] L. Schwartz, "Cours d'analyse" , 1 , Hermann (1967)
How to Cite This Entry:
Normal convergence. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Normal_convergence&oldid=54826
This article was adapted from an original article by E.D. Solomentsev (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article