# Uniformly-convergent series

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.

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Uniformly-convergent series.

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