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Файл:Lectures on integral calculus of functions of one variable and series theory
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15. Functional sequences and series 171
Let us transform the difference |fm(x) −f
m+p
(x)| using the last two esti-
mates:
|fm(x) −f
m+p
(x)| = |fm(x) −f (x) + f(x) −f
m+p
(x)| ≤
≤ |fm(x) −f (x)| + |f(x) − f
m+p
(x)| <
ε
2
+
ε
2
= ε.
Thus, condition (6) is satisfied and the necessity is proven.
2. Sufficiency. Given: condition (6) is satisfied. Prove: the functional
sequence {fn(x)} converges uniformly on the set E.
First, we prove that the sequence {fn(x)} converges pointwise on the set E.
We choose some value x0∈ E. From condition (6), we obtain:
∀ε > 0 ∃ N ∈ N ∀m > N ∀p ∈ N |fm(x0) −f
m+p(x0
)| < ε
This condition coincides with the Cauchy criterion condition for the convergence of the numerical sequence {fn(x0)}. It follows from this criterion
that the numerical sequence {fn(x0)} has a limit; denote it by f (x0). Since
the choice of x0∈ E is arbitrary, we obtain that the functional sequence
{fn(x0)} converges on the set E to some function f (x). It remains to prove
that the convergence to the function f(x) is uniform.
We rewrite condition (6) with a slight change in the last inequality:
∀ε > 0 ∃ N ∈ N ∀m > N ∀p ∈ N ∀ x ∈ E
|fm(x) −f
m+p
(x)| <
ε
2
. (7)
Since this condition is satisfied for any p ∈ N, we can pass to the limit
as p → ∞. Taking into account the pointwise convergence of the sequence
{fn(x)} already proved, we get that f
m+p
(x) → f (x). Therefore, the inequal-
ity |fm(x) −f
m+p
(x)| <
ε
2
, by virtue of the theorem on passing to the limit
in inequalities, takes the following form:
|fm(x) −f (x)| ≤
ε
2
< ε.
Thus, as a result of passing to the limit as p → ∞, condition (7) is
transformed as follows:
∀ε > 0 ∃ N ∈ N ∀m > N ∀x ∈ E |fm(x) −f (x)| < ε.
We obtained a version of condition (2) from the definition of uniform convergence (its only difference from (2) is the use of the symbol m instead of
the symbol n). Therefore, the sequence {fn(x)} converges uniformly on E to
f(x).
172 M. E. Abramyan. Lectures on integral calculus and series theory
Cauchy criterion for uniform convergence
of a functional series 3.13A/10:53 (03:41)
Corollary (Cauchy criterion for uniform convergence of
a functional series).
The functional series
P
∞
k=1
uk(x) converges uniformly on the set E if and
only if the following condition is satisfied:
∀ε > 0 ∃ N ∈ N ∀m > N ∀p ∈ N ∀ x ∈ E
m+p
X
k=m+1
uk(x)
< ε. (8)
Proof.
Consider the sequence of partial sums of the initial series:
{Sn(x)} =
P
n
k=1
uk(x). The difference |S
m+p
(x) − Sm(x)| for this
sequence can be written as follows:
|S
m+p
(x) −Sm(x)| =
m+p
X
k=1
uk(x) −
m
X
k=1
uk(x)
=
m+p
X
k=m+1
uk(x)
.
Thus, condition (6) of the Cauchy criterion for uniform convergence of the
functional sequence {Sn(x)} coincides with condition (8). Therefore, condi-
tion (8) holds if and only if the sequence {Sn(x)} converges uniformly on E.
It remains to note that the uniform convergence of the series
P
∞
k=1
uk(x)
is equivalent to the uniform convergence of the sequence of its partial sums
{Sn(x)}.
Tests of uniform convergence of functional series
Weierstrass test 3.13A/14:34 (12:01)
Theorem (Weierstrass test for uniform convergence
of a functional series).
Let the functional series
P
∞
k=1
uk(x) be defined on the set E and the
following condition holds for its terms:
∃N0∈ N ∀ k > N0∀x ∈ E |uk(x)| ≤ ak. (9)
If the numerical series
P
∞
k=1
akconverges, then the initial functional series
P
∞
k=1
uk(x) converges absolutely and uniformly on E. In English, this test is
also called the Weierstrass M-test.
15. Functional sequences and series 173
Proof.
Let us write the condition for the Cauchy criterion for a convergent nu-
merical series
P
∞
k=1
ak:
∀ε > 0 ∃ N00∈ N ∀ m > N00∀p ∈ N
m+p
X
k=m+1
ak< ε. (10)
In the last inequality, we do not use the sign of the absolute value operation,
since, by virtue of (9), all terms akare non-negative.
If we now put N = max{N0, N00} and use condition (9), then, for any
m > N , p ∈ N, and x ∈ E, we get
m+p
X
k=m+1
|uk(x)| ≤
m+p
X
k=m+1
ak< ε.
So, we have obtained that the following condition follows from condi-
tions (9) and (10):
∀ε > 0 ∃ N ∈ N ∀m > N ∀p ∈ N ∀ x ∈ E
m+p
X
k=m+1
|uk(x)| < ε.
By the Cauchy criterion for uniform convergence of a functional series, this
condition is sufficient for the uniform convergence of the series
P
∞
k=1
|uk(x)|.
We proved that the initial series
P
∞
k=1
uk(x) converges absolutely on E.
It remains to note that the convergence of the initial series is uniform.
Indeed, the condition of the Cauchy criterion of uniform convergence is also
satisfied for the initial series, since the following estimate holds:
m+p
X
k=m+1
uk(x)
≤
m+p
X
k=m+1
|uk(x)| < ε.
Example.
Consider the series
P
∞
k=1
sin kx
k
2
. For its common term, the following esti-
mate is true:
∀k ∈ N ∀x ∈ R
sin kx
k
2
≤
1
k
2
.
Since the numerical series
P
∞
k=1
1
k
2
converges, we conclude, by the Weier-
strass test, that the functional series
P
∞
k=1
sin kx
k
2
converges absolutely and
uniformly on the entire real axis R.
174 M. E. Abramyan. Lectures on integral calculus and series theory
Dirichlet’s test and Abel’s test 3.13A/26:35 (07:50)
Theorem (Dirichlet’s test for uniform convergence of
a functional series).
Let the functional series
P
∞
k=1
uk(x)vk(x) be defined on E and the follow-
ing conditions are satisfied for it:
1) ∃M ∀ n ∈ N ∀x ∈ E
P
n
k=1
uk(x)
≤ M ;
2) vk(x)E⇒ 0 as k → ∞, the numerical sequence {vk(x)} is monotone for
any x ∈ E.
Then the series
P
∞
k=1
uk(x)vk(x) converges uniformly on E (the conver-
gence is, generally speaking, conditional).
Proof4.
Introducing the auxiliary notation Un(x) =
P
n
k=1
uk(x) and carrying out
the same reasoning as at the beginning of the proof of Dirichlet’s test for
the convergence of a numerical series, we can obtain the following estimate,
which holds for any m, p ∈ N and x ∈ E:
m+p
X
k=m+1
uk(x)vk(x)
≤ 2M (|v
m+1
(x)| + |v
m+p
(x)|).
Now we use the condition that the sequence {vk(x)} converges uniformly
on the set E to 0, and write this condition in the following form:
∀ε > 0 ∃ N ∈ N ∀m > N ∀p ∈ N ∀ x ∈ E |v
m+p
(x)| <
ε
4M
.
For
P
m+p
k=m+1
uk(x)vk(x)
, we finally get
m+p
X
k=m+1
uk(x)vk(x)
≤ 2M (|v
m+1
(x)| + |v
m+p
(x)|) <
< 2M
ε
4M
+
ε
4M
= ε.
Thus, we have proved that the condition for uniform convergence on E
from the Cauchy criterion is satisfied for the initial series:
∀ε > 0 ∃ N ∈ N ∀m > N ∀p ∈ N ∀ x ∈ E
m+p
X
k=m+1
uk(x)vk(x)
< ε.
We also give the formulation of Abel’s test for the uniform convergence of
a functional series (for the proof, see, for example, [18, Ch. 9, Sec. 42.4]).
4
There is no proof of this theorem in video lectures.
15. Functional sequences and series 175
Theorem (Abel’s test for uniform convergence of a func-
tional series).
Let the functional series
P
∞
k=1
uk(x)vk(x) be defined on E and the follow-
ing conditions are satisfied for it:
1) the series
P
n
k=1
uk(x) converges uniformly on E;
2) the sequence {vk(x)} is monotone for any x ∈ E and is also uniformly
bounded:
∃M > 0 ∀k ∈ N ∀x ∈ E |vk(x)| ≤ M .
Then the series
P
∞
k=1
uk(x)vk(x) converges uniformly on E (the conver-
gence is, generally speaking, conditional).
16. Properties of uniformly converging
sequences and series
Continuity of the uniform limit
Formulation of the theorem
on the continuity of the uniform limit 3.13A/34:25 (09:02)
Theorem (on the continuity of the uniform limit of a func-
tional sequence with continuous elements).
Let the functional sequence {fn(x)} converge uniformly on the segment
[a, b] to the function f (x). Suppose that the function fn(x) is continuous on
[a, b] for any n ∈ N. Then the limit function f (x) is also continuous on [a, b].
Remarks.
1. If the convergence is not uniform, then the statement of the theorem
may not hold. Earlier we gave an example of a sequence of continuous functions {xn}. This sequence is not uniformly convergent on the segment [0, 1],
but converges pointwise on this segment. The limit of this (not uniformly
converging) sequence is a function that has a discontinuity of the first kind
at point 1.
2. This theorem means that, for a uniformly converging sequence of
continuous functions, we can swap two limit operations: as n → ∞ and
as x → x0, where x0is some point of the segment [a, b]. Indeed, due to
the continuity of the functions fn(x) and the limit function f (x), we have
lim
x→x
0
fn(x) = fn(x0), lim
x→x
0
f(x) = f (x0); due to the convergence of the
functional sequence, we have lim
n→∞fn
(x) = f(x), therefore the following
chain of equalities holds:
lim
n→∞
lim
x→x
0
fn(x) = lim
n→∞
fn(x0) = f(x0) = lim
x→x
0
f(x) = lim
x→x
0
lim
n→∞
fn(x).
We emphasize that the indicated inversion of the limits can be performed
only if the initial sequence of continuous functions is uniformly convergent,
since only uniform convergence ensures the continuity of the limit function.
16. Properties of uniformly converging sequences and series 177
Proof of the theorem
on the continuity of the uniform limit 3.13B/00:00 (13:55)
It is enough for us to prove that the limit function f (x) is continuous at
some arbitrarily chosen point x0∈ [a, b].
Let us give the definition of the function f (x) continuous at the point x
0
in the language ε–δ:
∀ε > 0 ∃ δ > 0 ∀x ∈ [a, b], |x −x0| < δ, |f(x) −f(x0)| < ε. (1)
Now we write down the conditions that the initial sequence {fn(x)} sat-
isfies. Firstly, this sequence converges uniformly on the set [a, b] to the function f(x). We write this fact in the language ε–N for the previously selected
value ε:
∃N ∈ N ∀n > N ∀x ∈ [a, b] |fn(x) −f (x)| <
ε
3
. (2)
We choose some value n0> N . For this value n0and for any x ∈ [a, b],
by virtue of (2), the estimate holds:
|f
n
0
(x) −f (x)| <
ε
3
. (3)
Since the previously selected point x0also belongs to the segment [a, b],
a similar estimate is fulfilled for it:
|f
n
0
(x0) −f (x0)| <
ε
3
. (4)
Secondly, by condition, the functions fn, n ∈ N, are continuous at the
point x0. Therefore, the function f
n
0
is also continuous at the point x0. We
write this fact in the language ε–δ for the previously selected value ε:
∃δ > 0 ∀x ∈ [a, b], |x −x0| < δ, |f
n
0
(x) −f
n
0
(x0)| <
ε
3
. (5)
So, we got that, for the chosen value ε > 0, there exists a value δ > 0 (specified in condition (5)) such that estimates (3), (4), and (5) are simultaneously
fulfilled for all x ∈ [a, b], |x − x0| < δ.
We transform the expression |f (x) −f (x0)| using these three estimates:
|f(x) −f (x0)| =
= |f(x) −f
n
0
(x) + f
n
0
(x) −f
n
0
(x0) + f
n
0
(x0) −f (x0)| ≤
≤ |f (x) −f
n
0
(x)| + |f
n
0
(x) −f
n
0
(x0)| + |f
n
0
(x0) −f (x0)| <
<
ε
3
+
ε
3
+
ε
3
= ε.
Thus, we have shown that condition (1) is satisfied. Therefore, the limit
function f is continuous at an arbitrary point x0of the segment [a, b].
178 M. E. Abramyan. Lectures on integral calculus and series theory
Corollary for functional series 3.13B/13:55 (09:55)
Corollary (on the continuity of the sum of a uniformly
converging functional series with continuous terms).
Let the series
P
∞
k=1
uk(x) uniformly converge on the segment [a, b] to the
function S(x). Let all terms of the series uk(x) be continuous functions on the
segment [a, b]. Then the sum of the series S(x) is also a continuous function
on this segment.
Remark.
This corollary can also be interpreted in terms of a swap of two limit
operations. Namely, we choose the point x0∈ [a, b] and consider the following
limit: lim
x→x
0
P
∞
k=1
uk(x). This limit is understood as the limit of the sum
of a converging series: lim
x→x
0
S(x).
Since, according to the corollary, the sum S(x) is a continuous function
on [a, b], the last limit is equal to the value of the function S at the point x0.
But the value of S(x0) is the sum of the series
P
∞
k=1
uk(x0). If we now use
the continuity of the functions uk, the last series can be written in the form
P
∞
k=1
lim
x→x
0
uk(x).
Removing the intermediate transformations, we obtain the following relation:
lim
x→x
0
∞
X
k=1
uk(x) =
∞
X
k=1
lim
x→x
0
uk(x). (6)
This relation means that in the case of a uniformly converging series with
continuous terms, the limit as x → x0can be moved under the infinite sum
sign or taken out of it. This makes it possible in some cases to more simply
calculate the limit of the sum of a uniformly converging series: instead of
finding an expression for the sum of the initial functional series and passing
to the limit in this expression (i. e., performing the actions in the left-hand
side of (6)), it often turns out to be easier to find the limit for the terms
of this series and then find the sum of the resulting numerical series (i. e.,
perform the actions in the right-hand side of (6)).
Note also that a special justification of reversing the limit sign and the
summation sign is required only in the case of infinite sums. If we replace
the series with a finite sum of the form
P
n
k=1
uk(x) in relation (6), then the
validity of this relation will immediately follow from the fact that the finite
sum of continuous functions is a continuous function.
16. Properties of uniformly converging sequences and series 179
Proof.
By definition, the function S(x) is the limit of a sequence of partial sums
Sn(x) =
P
n
k=1
uk(x). The uniform convergence of the series means that the
sequence {Sn(x)} converges to S(x) uniformly on [a, b].
In addition, the partial sum Sn(x) is a continuous function as a finite sum
of continuous functions uk(x).
So, all the conditions of the previous theorem are satisfied for the sequence
{Sn(x)}: it converges uniformly on [a, b] to S(x) and its elements are continuous functions on this segment. Applying this theorem, we obtain that the
limit function S(x) is also continuous on [a, b].
Integration of functional sequences and series
Formulation of the theorem
on the integration of a functional sequence 3.13B/23:50 (03:57)
Theorem (on the integration of a uniformly converging
functional sequence).
Let the functional sequence {fn(x)} converge uniformly on the segment
[a, b] to the function f (x). Let the function fn(x) be continuous on [a, b] for
any n ∈ N. Let x0∈ [a, b].
Then the functional sequence
n
R
x
x
0
fn(t) dtois defined on the segment [a, b]
and converges uniformly on this segment to the function
R
x
x
0
f(t) dt:
Z
x
x
0
fn(t) dt
[a,b]
⇒
Z
x
x
0
f(t) dt, n → ∞. (7)
Remarks.
1. All the integrals mentioned in the theorem exist, since the integrands
are continuous: the functions fn(x) are continuous by condition, the limit
function f (x) is continuous by virtue of the previous theorem.
2. The result of the theorem can be reformulated in terms of reversing the
limit operation and the integration operation:
lim
n→∞
Z
x
x
0
fn(t) dt =
Z
x
x
0
f(t) dt =
Z
x
x
0
lim
n→∞
fn(t) dt.
Proof of the theorem
on the integration of a functional sequence 3.13B/27:47 (06:14)
Let us write the condition for uniform convergence of the initial sequence
{f
n
(x)} in the language ε–N:
180 M. E. Abramyan. Lectures on integral calculus and series theory
∀ε > 0 ∃ N ∈ N ∀n > N ∀x ∈ [a, b]
|fn(x) −f (x)| <
ε
b − a
. (8)
For definiteness, we assume that the estimate x0< x holds.
Consider the difference
R
x
x
0
fn(t) dt −
R
x
x
0
f(t) dt
and transform it as fol-
lows:
Z
x
x
0
fn(t) dt −
Z
x
x
0
f(t) dt
=
Z
x
x
0
fn(t) −f (t)dt
≤
≤
Z
x
x
0
|fn(t) −f (t)| dt.
We use estimate (8), which is valid for any t, since t ∈ [x0, x] ⊂ [a, b], and
the corollary of the theorem on integration of a positive continuous function:
Z
x
x
0
|fn(t) −f (t)| dt <
ε
b − a
· (x − x0) ≤
ε
b − a
· (b − a) = ε.
So, we have proved that the following condition is satisfied for the sequence
n
R
x
x
0
fn(t) dto:
∀ε > 0 ∃ N ∈ N ∀n > N ∀x ∈ [a, b]
Z
x
x
0
fn(t) dt −
Z
x
x
0
f(t) dt
< ε.
This means that the sequence
n
R
x
x
0
fn(t) dtoconverges uniformly on [a, b]
and the limit relation (7) holds.
Formulation of the corollary
on the integration of a functional series 3.14A/00:00 (08:28)
Corollary (on the integration of a uniformly converging
functional series).
Let the series
P
∞
k=1
uk(x) converge uniformly on the segment [a, b] to the
function S(x). Let all terms of the series uk(x) be continuous functions on the
segment [a, b]. Then the series
P
∞
k=1
R
x
x
0
uk(t) dt converges uniformly on [a, b]
to the integral
R
x
x
0
S(t) dt for any point x0∈ [a, b].
Remarks.
1. All the integrals mentioned in the corollary exist because the integrands
are continuous: the functions uk(x) are continuous by condition, the limit
function S(x) is continuous by virtue of the corollary of the theorem on the
continuity of the uniform limit.
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