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A series of functions
 
A series of functions
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u0953101.png" /></td> <td valign="top" style="width:5%;text-align:right;">(1)</td></tr></table>
+
$$ \tag{1 }
 +
\sum _ {n = 1 } ^  \infty  a _ {n} ( x),\ \
 +
x \in X,
 +
$$
  
with (in general) complex terms, such that for every <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u0953102.png" /> there is an <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u0953103.png" /> (independent of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u0953104.png" />) such that for all <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u0953105.png" /> and all <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u0953106.png" />,
+
with (in general) complex terms, such that for every $  \epsilon > 0 $
 +
there is an $  n _  \epsilon  $(
 +
independent of $  x $)  
 +
such that for all $  n > n _  \epsilon  $
 +
and all $  x \in X $,
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u0953107.png" /></td> </tr></table>
+
$$
 +
| s _ {n} ( x) - s ( x) |  < \epsilon ,
 +
$$
  
 
where
 
where
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u0953108.png" /></td> </tr></table>
+
$$
 +
s _ {n} ( x)  = \sum _ {k = 1 } ^ { n }  a _ {k} ( x)
 +
$$
  
 
and
 
and
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u0953109.png" /></td> </tr></table>
+
$$
 +
s ( x)  = \sum _ {k = 1 } ^  \infty  a _ {k} ( x).
 +
$$
  
In other words, the sequence of partial sums <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531010.png" /> is a uniformly-convergent sequence. The definition of a uniformly-convergent series is equivalent to the condition
+
In other words, the sequence of partial sums $  s _ {n} ( x) $
 +
is a uniformly-convergent sequence. The definition of a uniformly-convergent series is equivalent to the condition
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531011.png" /></td> </tr></table>
+
$$
 +
\lim\limits _ {n \rightarrow \infty } \
 +
\sup _ {x \in X } \
 +
| r _ {n} ( x) |  = 0,
 +
$$
  
which denotes the [[Uniform convergence|uniform convergence]] to zero on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531012.png" /> of the sequence of remainders
+
which denotes the [[Uniform convergence|uniform convergence]] to zero on $  X $
 +
of the sequence of remainders
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531013.png" /></td> </tr></table>
+
$$
 +
r _ {n} ( x)  = \
 +
\sum _ {k = n + 1 } ^  \infty  a _ {k} ( x),\ \
 +
n = 1, 2 \dots
 +
$$
  
 
of the series (1).
 
of the series (1).
Line 27: Line 62:
 
Example. The series
 
Example. The series
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531014.png" /></td> </tr></table>
+
$$
 +
\sum _ {n = 1 } ^  \infty 
 +
 
 +
\frac{z  ^ {n} }{n! }
 +
  = \
 +
e  ^ {z}
 +
$$
  
is uniformly convergent on each bounded disc of the complex plane, but is not uniformly convergent on the whole of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531015.png" />.
+
is uniformly convergent on each bounded disc of the complex plane, but is not uniformly convergent on the whole of $  \mathbf C $.
  
The [[Cauchy criteria|Cauchy criterion]] for uniform convergence of a series gives a condition for the uniform convergence of the series (1) on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531016.png" /> without using the sum of the series. A sufficient condition for the uniform convergence of a series is given by the [[Weierstrass criterion (for uniform convergence)|Weierstrass criterion (for uniform convergence)]].
+
The [[Cauchy criteria|Cauchy criterion]] for uniform convergence of a series gives a condition for the uniform convergence of the series (1) on $  X $
 +
without using the sum of the series. A sufficient condition for the uniform convergence of a series is given by the [[Weierstrass criterion (for uniform convergence)|Weierstrass criterion (for uniform convergence)]].
  
A series <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531017.png" /> is called regularly convergent on a set <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531018.png" /> if there is a convergent numerical series <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531019.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531020.png" />, such that for all <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531021.png" /> and all <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531022.png" />,
+
A series $  \sum _ {n = 1 }  ^  \infty  a _ {n} ( x) $
 +
is called regularly convergent on a set $  X $
 +
if there is a convergent numerical series $  \sum \alpha _ {n} $,  
 +
$  \alpha _ {n} \geq  0 $,  
 +
such that for all $  n = 1, 2 \dots $
 +
and all $  x \in X $,
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531023.png" /></td> </tr></table>
+
$$
 +
| a _ {n} ( x) |  \leq  \alpha _ {n} ,
 +
$$
  
that is, if (1) satisfies the conditions of the Weierstrass criterion for uniform convergence. On the strength of this criterion, a regularly-convergent series on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531024.png" /> is uniformly convergent on that set. In general, the converse is false; however, for every series that is uniformly convergent on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531025.png" /> the successive terms can be collected into finite groups so that the series thus obtained is regularly convergent on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531026.png" />.
+
that is, if (1) satisfies the conditions of the Weierstrass criterion for uniform convergence. On the strength of this criterion, a regularly-convergent series on $  X $
 +
is uniformly convergent on that set. In general, the converse is false; however, for every series that is uniformly convergent on $  X $
 +
the successive terms can be collected into finite groups so that the series thus obtained is regularly convergent on $  X $.
  
 
There are criteria for the uniform convergence of series analogous to Dirichlet's and Abel's criteria for the convergence of series of numbers. These tests for uniform convergence first occurred in papers of G.H. Hardy. If in a series
 
There are criteria for the uniform convergence of series analogous to Dirichlet's and Abel's criteria for the convergence of series of numbers. These tests for uniform convergence first occurred in papers of G.H. Hardy. If in a series
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531027.png" /></td> <td valign="top" style="width:5%;text-align:right;">(2)</td></tr></table>
+
$$ \tag{2 }
 +
\sum a _ {n} ( x) b _ {n} ( x)
 +
$$
  
the functions <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531028.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531029.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531030.png" /> defined on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531031.png" />, are such that the sequence <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531032.png" /> is monotone for each <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531033.png" /> and converges uniformly to zero on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531034.png" />, while the sequence of partial sums <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531035.png" /> of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531036.png" /> are uniformly bounded on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531037.png" />, then (2) is uniformly convergent on this set.
+
the functions $  a _ {n} ( x) $
 +
and $  b _ {n} ( x) $,
 +
$  n = 1, 2 \dots $
 +
defined on $  X $,  
 +
are such that the sequence $  \{ a _ {n} ( x) \} $
 +
is monotone for each $  x \in X $
 +
and converges uniformly to zero on $  X $,  
 +
while the sequence of partial sums $  \{ B _ {n} ( x) \} $
 +
of $  \sum b _ {n} ( x) $
 +
are uniformly bounded on $  X $,  
 +
then (2) is uniformly convergent on this set.
  
If the sequence <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531038.png" /> is uniformly bounded on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531039.png" /> and is monotone for each fixed <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531040.png" />, while the series <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531041.png" /> is uniformly convergent on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531042.png" />, then (2) is also uniformly convergent on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531043.png" />.
+
If the sequence $  \{ a _ {n} ( x) \} $
 +
is uniformly bounded on $  X $
 +
and is monotone for each fixed $  x \in X $,  
 +
while the series $  \sum b _ {n} ( x) $
 +
is uniformly convergent on $  X $,  
 +
then (2) is also uniformly convergent on $  X $.
  
 
==Properties of uniformly-convergent series.==
 
==Properties of uniformly-convergent series.==
If two series <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531044.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531045.png" /> are uniformly convergent on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531046.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531047.png" />, then the series
+
If two series $  \sum a _ {n} ( x) $
 +
and $  \sum b _ {n} ( x) $
 +
are uniformly convergent on $  X $
 +
and $  \lambda , \mu \in \mathbf C $,
 +
then the series
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531048.png" /></td> </tr></table>
+
$$
 +
\sum \lambda a _ {n} ( x) + \mu b _ {n} ( x)
 +
$$
  
is also uniformly convergent on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531049.png" />.
+
is also uniformly convergent on $  X $.
  
If a series <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531050.png" /> is uniformly convergent on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531051.png" /> and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531052.png" /> is bounded on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531053.png" />, then <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531054.png" /> is also uniformly convergent on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531055.png" />.
+
If a series $  \sum a _ {n} ( x) $
 +
is uniformly convergent on $  X $
 +
and $  b ( x) $
 +
is bounded on $  X $,  
 +
then $  \sum b ( x) a _ {n} ( x) $
 +
is also uniformly convergent on $  X $.
  
Continuity of the sum of a series. In the study of the sum of a series of functions, the notion of  "point of uniform convergence"  turns out to be useful. Let <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531056.png" /> be a topological space and let the series (1) converge on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531057.png" />. A point <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531058.png" /> is called a point of uniform convergence of (1) if for any <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531059.png" /> there are a neighbourhood <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531060.png" /> of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531061.png" /> and a number <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531062.png" /> such that for all <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531063.png" /> and all <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531064.png" /> the inequality <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531065.png" /> holds.
+
Continuity of the sum of a series. In the study of the sum of a series of functions, the notion of  "point of uniform convergence"  turns out to be useful. Let $  X $
 +
be a topological space and let the series (1) converge on $  X $.  
 +
A point $  x _ {0} \in X $
 +
is called a point of uniform convergence of (1) if for any $  \epsilon > 0 $
 +
there are a neighbourhood $  U = U ( x _ {0} ) $
 +
of $  x _ {0} $
 +
and a number $  n _  \epsilon  $
 +
such that for all $  x \in U $
 +
and all $  n > n _  \epsilon  $
 +
the inequality $  | r _ {n} ( x) | < \epsilon $
 +
holds.
  
If <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531066.png" /> is a compact set, then in order that the series (1) be uniformly convergent on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531067.png" /> it is necessary and sufficient that each point <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531068.png" /> is a point of uniform convergence.
+
If $  X $
 +
is a compact set, then in order that the series (1) be uniformly convergent on $  X $
 +
it is necessary and sufficient that each point $  x \in X $
 +
is a point of uniform convergence.
  
If <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531069.png" /> is a topological space, the series (1) is convergent on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531070.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531071.png" /> is a point of uniform convergence of (1), and there are finite limits
+
If $  X $
 +
is a topological space, the series (1) is convergent on $  X $,  
 +
$  x _ {0} $
 +
is a point of uniform convergence of (1), and there are finite limits
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531072.png" /></td> </tr></table>
+
$$
 +
\lim\limits _ {x \rightarrow x _ {0} }  a _ {n} ( x)  = c _ {n} ,\ \
 +
n = 1, 2 \dots
 +
$$
  
then the numerical series <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531073.png" /> converges, the sum <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531074.png" /> of (1) has a limit as <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531075.png" />, and, moreover,
+
then the numerical series $  \sum c _ {n} $
 +
converges, the sum $  s ( x) $
 +
of (1) has a limit as $  x \rightarrow x _ {0} $,  
 +
and, moreover,
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531076.png" /></td> <td valign="top" style="width:5%;text-align:right;">(3)</td></tr></table>
+
$$ \tag{3 }
 +
\lim\limits _ {x \rightarrow x _ {0} }  s ( x)  = \
 +
\lim\limits _ {x \rightarrow x _ {0} } \
 +
\sum a _ {n} ( x) = \
 +
\sum c _ {n} ,
 +
$$
  
that is, under the assumptions made on (1) it is possible to pass term-by-term to the limit in the sense of formula (3). Hence it follows that if (1) converges on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531077.png" /> and its terms are continuous at a point of uniform convergence <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531078.png" />, then its sum is also continuous at that point:
+
that is, under the assumptions made on (1) it is possible to pass term-by-term to the limit in the sense of formula (3). Hence it follows that if (1) converges on $  X $
 +
and its terms are continuous at a point of uniform convergence $  x _ {0} \in X $,  
 +
then its sum is also continuous at that point:
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531079.png" /></td> </tr></table>
+
$$
 +
\lim\limits _ {x \rightarrow x _ {0} }  s ( x)  = \sum
 +
\lim\limits _ {x \rightarrow x _ {0} }  a _ {n} ( x)  = \
 +
\sum a _ {n} ( x _ {0} )  = s ( x _ {0} ).
 +
$$
  
Therefore, if a series of continuous functions converges uniformly on a topological space, then its sum is continuous on that space. When <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531080.png" /> is a compactum and the terms of (1) are non-negative on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531081.png" />, then uniform convergence of (1) is also a necessary condition for the continuity on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531082.png" /> of the sum (see [[Dini theorem|Dini theorem]]).
+
Therefore, if a series of continuous functions converges uniformly on a topological space, then its sum is continuous on that space. When $  X $
 +
is a compactum and the terms of (1) are non-negative on $  X $,  
 +
then uniform convergence of (1) is also a necessary condition for the continuity on $  X $
 +
of the sum (see [[Dini theorem|Dini theorem]]).
  
In the general case, a necessary and sufficient condition for the continuity of the sum of a series (1) that converges on a topological space <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531083.png" />, and whose terms are continuous on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531084.png" />, is [[Quasi-uniform convergence|quasi-uniform convergence]] of the sequence of partial sums <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531085.png" /> to the sum <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531086.png" /> (the Arzelà–Aleksandrov theorem).
+
In the general case, a necessary and sufficient condition for the continuity of the sum of a series (1) that converges on a topological space $  X $,  
 +
and whose terms are continuous on $  X $,  
 +
is [[Quasi-uniform convergence|quasi-uniform convergence]] of the sequence of partial sums $  s _ {n} ( x) $
 +
to the sum $  s ( x) $(
 +
the Arzelà–Aleksandrov theorem).
  
The answer to the question of the existence of points of uniform convergence for a convergent series of functions that are continuous on an interval is given by the Osgood–Hobson theorem: If (1) converges at each point of an interval <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531087.png" /> and the terms <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531088.png" /> are continuous on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531089.png" />, then there is an everywhere-dense set in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531090.png" /> of points of uniform convergence of the series (1). Hence it follows that the sum of any series of continuous functions, convergent in some interval, is continuous on a dense set of points of the interval. At the same time there exists a series of continuous functions, convergent at all points of an interval, such that the points at which it converges non-uniformly form an everywhere-dense set in the interval in question.
+
The answer to the question of the existence of points of uniform convergence for a convergent series of functions that are continuous on an interval is given by the Osgood–Hobson theorem: If (1) converges at each point of an interval $  [ a, b] $
 +
and the terms $  a _ {n} ( x) $
 +
are continuous on $  [ a, b] $,
 +
then there is an everywhere-dense set in $  [ a, b] $
 +
of points of uniform convergence of the series (1). Hence it follows that the sum of any series of continuous functions, convergent in some interval, is continuous on a dense set of points of the interval. At the same time there exists a series of continuous functions, convergent at all points of an interval, such that the points at which it converges non-uniformly form an everywhere-dense set in the interval in question.
  
Term-by-term integration of uniformly-convergent series. Let <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531091.png" />. If the terms of the series
+
Term-by-term integration of uniformly-convergent series. Let $  X = [ a, b] $.  
 +
If the terms of the series
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531092.png" /></td> <td valign="top" style="width:5%;text-align:right;">(4)</td></tr></table>
+
$$ \tag{4 }
 +
\sum a _ {n} ( x),\ \
 +
x \in [ a, b],
 +
$$
  
are Riemann (Lebesgue) integrable on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531093.png" /> and (4) converges uniformly on this interval, then its sum <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531094.png" /> is also Riemann (Lebesgue) integrable on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531095.png" />, and for any <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531096.png" /> the equality
+
are Riemann (Lebesgue) integrable on $  [ a, b] $
 +
and (4) converges uniformly on this interval, then its sum $  s ( x) $
 +
is also Riemann (Lebesgue) integrable on $  [ a, b] $,
 +
and for any $  x \in [ a, b] $
 +
the equality
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531097.png" /></td> <td valign="top" style="width:5%;text-align:right;">(5)</td></tr></table>
+
$$ \tag{5 }
 +
\int\limits _ { a } ^ { x }  s ( t)  dt  = \
 +
\int\limits _ { a } ^ { x }
 +
\left [ \sum a _ {n} ( t) \right ]  dt  = \
 +
\sum \int\limits _ { a } ^ { x }  a _ {n} ( t) dt
 +
$$
  
holds, where the series on the right-hand side is uniformly convergent on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531098.png" />.
+
holds, where the series on the right-hand side is uniformly convergent on $  [ a, b] $.
  
In this theorem it is impossible to replace the condition of uniform convergence of (4) by convergence on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u09531099.png" />, since there are series, even of continuous functions and with continuous sums, that converge on an interval and for which (5) does not hold. At the same time there are various generalizations. Below some results for the [[Stieltjes integral|Stieltjes integral]] are given.
+
In this theorem it is impossible to replace the condition of uniform convergence of (4) by convergence on $  [ a, b] $,
 +
since there are series, even of continuous functions and with continuous sums, that converge on an interval and for which (5) does not hold. At the same time there are various generalizations. Below some results for the [[Stieltjes integral|Stieltjes integral]] are given.
  
If <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u095310100.png" /> is an increasing function on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u095310101.png" />, the <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u095310102.png" /> are integrable functions relative to <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u095310103.png" /> and (4) converges uniformly on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u095310104.png" />, then the sum <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u095310105.png" /> of (4) is Stieltjes integrable relative to <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u095310106.png" />,
+
If $  g ( x) $
 +
is an increasing function on $  [ a, b] $,  
 +
the $  a _ {n} ( x) $
 +
are integrable functions relative to $  g ( x) $
 +
and (4) converges uniformly on $  [ a, b] $,
 +
then the sum $  s ( x) $
 +
of (4) is Stieltjes integrable relative to $  g ( x) $,
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u095310107.png" /></td> </tr></table>
+
$$
 +
\int\limits _ { a } ^ { x }  s ( t)  dg ( t)  = \
 +
\sum \int\limits _ { a } ^ { x }  a _ {n} ( t)  dg ( t),
 +
$$
  
and the series on the right-hand side converges uniformly on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u095310108.png" />.
+
and the series on the right-hand side converges uniformly on $  [ a, b] $.
  
 
Formula (5) has been generalized to functions of several variables.
 
Formula (5) has been generalized to functions of several variables.
  
Conditions for term-by-term differentiation of series in terms of uniform convergence. If the terms of (4) are continuously differentiable on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u095310109.png" />, if (4) converges at some point of the interval and the series of derivatives of the terms of (4) is uniformly convergent on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u095310110.png" />, then the series (4) itself is uniformly convergent on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u095310111.png" />, its sum <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u095310112.png" /> is continuously differentiable and
+
Conditions for term-by-term differentiation of series in terms of uniform convergence. If the terms of (4) are continuously differentiable on $  [ a, b] $,
 +
if (4) converges at some point of the interval and the series of derivatives of the terms of (4) is uniformly convergent on $  [ a, b] $,
 +
then the series (4) itself is uniformly convergent on $  [ a, b] $,
 +
its sum $  s ( x) $
 +
is continuously differentiable and
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u095310113.png" /></td> <td valign="top" style="width:5%;text-align:right;">(6)</td></tr></table>
+
$$ \tag{6 }
 +
{
 +
\frac{d}{dx}
 +
} s ( x)  = \
 +
{
 +
\frac{d}{dx}
 +
} \sum a _ {n} ( x)  = \
 +
\sum {
 +
\frac{d}{dx}
 +
} a _ {n} ( x).
 +
$$
  
In this theorem the condition of uniform convergence of the series obtained by term-by-term differentiation cannot be replaced by convergence on <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u095310114.png" />, since there are series of continuously-differentiable functions, uniformly convergent on an interval, for which the series obtained by term-by-term differentiation converges on the interval, but the sum of the original series is either not differentiable on the whole interval in question, or it is differentiable but its derivative is not equal to the sum of the series of derivatives.
+
In this theorem the condition of uniform convergence of the series obtained by term-by-term differentiation cannot be replaced by convergence on $  [ a, b] $,
 +
since there are series of continuously-differentiable functions, uniformly convergent on an interval, for which the series obtained by term-by-term differentiation converges on the interval, but the sum of the original series is either not differentiable on the whole interval in question, or it is differentiable but its derivative is not equal to the sum of the series of derivatives.
  
 
In this way, the presence of the property of uniform convergence of a series, in much the same way as absolute convergence (see [[Absolutely convergent series|Absolutely convergent series]]), permits one to transfer to these series certain rules of operating with finite sums: for uniform convergence — term-by-term passage to the limit, term-by-term integration and differentiation (see (3)–(6)), and for absolute convergence — the possibility of permuting the order of the terms of the series without changing the sum, and multiplying series term-by-term.
 
In this way, the presence of the property of uniform convergence of a series, in much the same way as absolute convergence (see [[Absolutely convergent series|Absolutely convergent series]]), permits one to transfer to these series certain rules of operating with finite sums: for uniform convergence — term-by-term passage to the limit, term-by-term integration and differentiation (see (3)–(6)), and for absolute convergence — the possibility of permuting the order of the terms of the series without changing the sum, and multiplying series term-by-term.
Line 108: Line 269:
 
The properties of absolute and uniform convergence for series of functions are independent of each other. Thus, the series
 
The properties of absolute and uniform convergence for series of functions are independent of each other. Thus, the series
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u095310115.png" /></td> </tr></table>
+
$$
 +
\sum _ {n = 0 } ^  \infty 
  
is absolutely convergent on the whole axis, since all its terms are non-negative, but obviously <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u095310116.png" /> is not a point of uniform convergence, since its sum
+
\frac{x  ^ {2} }{( 1 + x  ^ {2} )  ^ {n} }
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u095310117.png" /></td> </tr></table>
+
$$
 +
 
 +
is absolutely convergent on the whole axis, since all its terms are non-negative, but obviously  $  x = 0 $
 +
is not a point of uniform convergence, since its sum
 +
 
 +
$$
 +
s ( x)  = \
 +
\left \{
  
 
is discontinuous at this point (whereas all terms are continuous).
 
is discontinuous at this point (whereas all terms are continuous).
Line 118: Line 287:
 
The series
 
The series
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/u/u095/u095310/u095310118.png" /></td> </tr></table>
+
$$
 +
\sum _ {n = 1 } ^  \infty 
 +
 
 +
\frac{(- 1) ^ {n + 1 } }{x  ^ {2} + n }
 +
 
 +
$$
  
 
is uniformly convergent on the whole real axis but does not converge absolutely at any point.
 
is uniformly convergent on the whole real axis but does not converge absolutely at any point.
  
 
For references see [[Series|Series]].
 
For references see [[Series|Series]].

Latest revision as of 08:27, 6 June 2020


A series of functions

$$ \tag{1 } \sum _ {n = 1 } ^ \infty a _ {n} ( x),\ \ x \in X, $$

with (in general) complex terms, such that for every $ \epsilon > 0 $ there is an $ n _ \epsilon $( independent of $ x $) such that for all $ n > n _ \epsilon $ and all $ x \in X $,

$$ | s _ {n} ( x) - s ( x) | < \epsilon , $$

where

$$ s _ {n} ( x) = \sum _ {k = 1 } ^ { n } a _ {k} ( x) $$

and

$$ s ( x) = \sum _ {k = 1 } ^ \infty a _ {k} ( x). $$

In other words, the sequence of partial sums $ s _ {n} ( x) $ is a uniformly-convergent sequence. The definition of a uniformly-convergent series is equivalent to the condition

$$ \lim\limits _ {n \rightarrow \infty } \ \sup _ {x \in X } \ | r _ {n} ( x) | = 0, $$

which denotes the uniform convergence to zero on $ X $ of the sequence of remainders

$$ r _ {n} ( x) = \ \sum _ {k = n + 1 } ^ \infty a _ {k} ( x),\ \ n = 1, 2 \dots $$

of the series (1).

Example. The series

$$ \sum _ {n = 1 } ^ \infty \frac{z ^ {n} }{n! } = \ e ^ {z} $$

is uniformly convergent on each bounded disc of the complex plane, but is not uniformly convergent on the whole of $ \mathbf C $.

The Cauchy criterion for uniform convergence of a series gives a condition for the uniform convergence of the series (1) on $ X $ without using the sum of the series. A sufficient condition for the uniform convergence of a series is given by the Weierstrass criterion (for uniform convergence).

A series $ \sum _ {n = 1 } ^ \infty a _ {n} ( x) $ is called regularly convergent on a set $ X $ if there is a convergent numerical series $ \sum \alpha _ {n} $, $ \alpha _ {n} \geq 0 $, such that for all $ n = 1, 2 \dots $ and all $ x \in X $,

$$ | a _ {n} ( x) | \leq \alpha _ {n} , $$

that is, if (1) satisfies the conditions of the Weierstrass criterion for uniform convergence. On the strength of this criterion, a regularly-convergent series on $ X $ is uniformly convergent on that set. In general, the converse is false; however, for every series that is uniformly convergent on $ X $ the successive terms can be collected into finite groups so that the series thus obtained is regularly convergent on $ X $.

There are criteria for the uniform convergence of series analogous to Dirichlet's and Abel's criteria for the convergence of series of numbers. These tests for uniform convergence first occurred in papers of G.H. Hardy. If in a series

$$ \tag{2 } \sum a _ {n} ( x) b _ {n} ( x) $$

the functions $ a _ {n} ( x) $ and $ b _ {n} ( x) $, $ n = 1, 2 \dots $ defined on $ X $, are such that the sequence $ \{ a _ {n} ( x) \} $ is monotone for each $ x \in X $ and converges uniformly to zero on $ X $, while the sequence of partial sums $ \{ B _ {n} ( x) \} $ of $ \sum b _ {n} ( x) $ are uniformly bounded on $ X $, then (2) is uniformly convergent on this set.

If the sequence $ \{ a _ {n} ( x) \} $ is uniformly bounded on $ X $ and is monotone for each fixed $ x \in X $, while the series $ \sum b _ {n} ( x) $ is uniformly convergent on $ X $, then (2) is also uniformly convergent on $ X $.

Properties of uniformly-convergent series.

If two series $ \sum a _ {n} ( x) $ and $ \sum b _ {n} ( x) $ are uniformly convergent on $ X $ and $ \lambda , \mu \in \mathbf C $, then the series

$$ \sum \lambda a _ {n} ( x) + \mu b _ {n} ( x) $$

is also uniformly convergent on $ X $.

If a series $ \sum a _ {n} ( x) $ is uniformly convergent on $ X $ and $ b ( x) $ is bounded on $ X $, then $ \sum b ( x) a _ {n} ( x) $ is also uniformly convergent on $ X $.

Continuity of the sum of a series. In the study of the sum of a series of functions, the notion of "point of uniform convergence" turns out to be useful. Let $ X $ be a topological space and let the series (1) converge on $ X $. A point $ x _ {0} \in X $ is called a point of uniform convergence of (1) if for any $ \epsilon > 0 $ there are a neighbourhood $ U = U ( x _ {0} ) $ of $ x _ {0} $ and a number $ n _ \epsilon $ such that for all $ x \in U $ and all $ n > n _ \epsilon $ the inequality $ | r _ {n} ( x) | < \epsilon $ holds.

If $ X $ is a compact set, then in order that the series (1) be uniformly convergent on $ X $ it is necessary and sufficient that each point $ x \in X $ is a point of uniform convergence.

If $ X $ is a topological space, the series (1) is convergent on $ X $, $ x _ {0} $ is a point of uniform convergence of (1), and there are finite limits

$$ \lim\limits _ {x \rightarrow x _ {0} } a _ {n} ( x) = c _ {n} ,\ \ n = 1, 2 \dots $$

then the numerical series $ \sum c _ {n} $ converges, the sum $ s ( x) $ of (1) has a limit as $ x \rightarrow x _ {0} $, and, moreover,

$$ \tag{3 } \lim\limits _ {x \rightarrow x _ {0} } s ( x) = \ \lim\limits _ {x \rightarrow x _ {0} } \ \sum a _ {n} ( x) = \ \sum c _ {n} , $$

that is, under the assumptions made on (1) it is possible to pass term-by-term to the limit in the sense of formula (3). Hence it follows that if (1) converges on $ X $ and its terms are continuous at a point of uniform convergence $ x _ {0} \in X $, then its sum is also continuous at that point:

$$ \lim\limits _ {x \rightarrow x _ {0} } s ( x) = \sum \lim\limits _ {x \rightarrow x _ {0} } a _ {n} ( x) = \ \sum a _ {n} ( x _ {0} ) = s ( x _ {0} ). $$

Therefore, if a series of continuous functions converges uniformly on a topological space, then its sum is continuous on that space. When $ X $ is a compactum and the terms of (1) are non-negative on $ X $, then uniform convergence of (1) is also a necessary condition for the continuity on $ X $ of the sum (see Dini theorem).

In the general case, a necessary and sufficient condition for the continuity of the sum of a series (1) that converges on a topological space $ X $, and whose terms are continuous on $ X $, is quasi-uniform convergence of the sequence of partial sums $ s _ {n} ( x) $ to the sum $ s ( x) $( the Arzelà–Aleksandrov theorem).

The answer to the question of the existence of points of uniform convergence for a convergent series of functions that are continuous on an interval is given by the Osgood–Hobson theorem: If (1) converges at each point of an interval $ [ a, b] $ and the terms $ a _ {n} ( x) $ are continuous on $ [ a, b] $, then there is an everywhere-dense set in $ [ a, b] $ of points of uniform convergence of the series (1). Hence it follows that the sum of any series of continuous functions, convergent in some interval, is continuous on a dense set of points of the interval. At the same time there exists a series of continuous functions, convergent at all points of an interval, such that the points at which it converges non-uniformly form an everywhere-dense set in the interval in question.

Term-by-term integration of uniformly-convergent series. Let $ X = [ a, b] $. If the terms of the series

$$ \tag{4 } \sum a _ {n} ( x),\ \ x \in [ a, b], $$

are Riemann (Lebesgue) integrable on $ [ a, b] $ and (4) converges uniformly on this interval, then its sum $ s ( x) $ is also Riemann (Lebesgue) integrable on $ [ a, b] $, and for any $ x \in [ a, b] $ the equality

$$ \tag{5 } \int\limits _ { a } ^ { x } s ( t) dt = \ \int\limits _ { a } ^ { x } \left [ \sum a _ {n} ( t) \right ] dt = \ \sum \int\limits _ { a } ^ { x } a _ {n} ( t) dt $$

holds, where the series on the right-hand side is uniformly convergent on $ [ a, b] $.

In this theorem it is impossible to replace the condition of uniform convergence of (4) by convergence on $ [ a, b] $, since there are series, even of continuous functions and with continuous sums, that converge on an interval and for which (5) does not hold. At the same time there are various generalizations. Below some results for the Stieltjes integral are given.

If $ g ( x) $ is an increasing function on $ [ a, b] $, the $ a _ {n} ( x) $ are integrable functions relative to $ g ( x) $ and (4) converges uniformly on $ [ a, b] $, then the sum $ s ( x) $ of (4) is Stieltjes integrable relative to $ g ( x) $,

$$ \int\limits _ { a } ^ { x } s ( t) dg ( t) = \ \sum \int\limits _ { a } ^ { x } a _ {n} ( t) dg ( t), $$

and the series on the right-hand side converges uniformly on $ [ a, b] $.

Formula (5) has been generalized to functions of several variables.

Conditions for term-by-term differentiation of series in terms of uniform convergence. If the terms of (4) are continuously differentiable on $ [ a, b] $, if (4) converges at some point of the interval and the series of derivatives of the terms of (4) is uniformly convergent on $ [ a, b] $, then the series (4) itself is uniformly convergent on $ [ a, b] $, its sum $ s ( x) $ is continuously differentiable and

$$ \tag{6 } { \frac{d}{dx} } s ( x) = \ { \frac{d}{dx} } \sum a _ {n} ( x) = \ \sum { \frac{d}{dx} } a _ {n} ( x). $$

In this theorem the condition of uniform convergence of the series obtained by term-by-term differentiation cannot be replaced by convergence on $ [ a, b] $, since there are series of continuously-differentiable functions, uniformly convergent on an interval, for which the series obtained by term-by-term differentiation converges on the interval, but the sum of the original series is either not differentiable on the whole interval in question, or it is differentiable but its derivative is not equal to the sum of the series of derivatives.

In this way, the presence of the property of uniform convergence of a series, in much the same way as absolute convergence (see Absolutely convergent series), permits one to transfer to these series certain rules of operating with finite sums: for uniform convergence — term-by-term passage to the limit, term-by-term integration and differentiation (see (3)–(6)), and for absolute convergence — the possibility of permuting the order of the terms of the series without changing the sum, and multiplying series term-by-term.

The properties of absolute and uniform convergence for series of functions are independent of each other. Thus, the series

$$ \sum _ {n = 0 } ^ \infty \frac{x ^ {2} }{( 1 + x ^ {2} ) ^ {n} } $$

is absolutely convergent on the whole axis, since all its terms are non-negative, but obviously $ x = 0 $ is not a point of uniform convergence, since its sum

$$ s ( x) = \ \left \{ is discontinuous at this point (whereas all terms are continuous). The series $$ \sum _ {n = 1 } ^ \infty

\frac{(- 1) ^ {n + 1 } }{x ^ {2} + n }

$$

is uniformly convergent on the whole real axis but does not converge absolutely at any point.

For references see Series.

How to Cite This Entry:
Uniformly-convergent series. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Uniformly-convergent_series&oldid=49074
This article was adapted from an original article by L.D. Kudryavtsev (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article