Uniform Convergence.  Linköping Universitycourses.mai.liu.se/GU/TATA57/Dokument/Uniform...
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Uniform Convergence.
1 Introduction.
In this course we study amongst other things Fourier series. The Fourier series for aperiodic function f(x) with period 2π is defined as the series
a02
+∞∑k=1
(ak cos kx+ bk sin kx) ,
where the coefficients ak, bk are defined as
ak =1
π
∫ π−πf(x) cos kx dx, bk =
1
π
∫ π−πf(x) sin kx dx,
with k = 0, 1, . . . (note that this means that b0 = 0).
This is an example of a functional series, which is a series whose terms are functions:
∞∑k=0
uk(x).
As usual with series, we define the above infinite sum as a limit:
∞∑k=0
uk(x) = limN→∞
N∑k=0
uk(x),
providing the limit exists. Note that different values of x will, in general, give differentlimits, if they exist.
In this lecture we shall look at functional series, and functional sequences, and weshall consider first the question of convergence. To deal with this, we consider two typesof convergence: pointwise convergence and uniform convergence. There are threemain results: the first one is that uniform convergence of a sequence of continuousfunctions gives us a continuous function as a limit. The second main result is Weierstrass’ Majorant Theorem, which gives a condition that guarantees that a functionalseries converges to a continuous function. The third result is that integrals of a sequenceof functions which converges uniformly to a limit function f(x) also converge with thelimit being the integral of f(x). These results are not only good for your mental health,they are also important tools in our later discussion of Fourier series, and that is thereason for looking at them.
2 Pointwise Convergence.
We are familiar with the power series
∞∑k=0
xk =1
1− xfor x < 1.
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This statement says that for each x ∈ ] − 1, 1[ the power series in the lefthand sideconverges to the number 1/(1− x). If we put
fn(x) =n∑k=0
xk, f(x) =1
1− x,
then we can rephrase this as
fn(x)→ f(x) as n→∞ for each x ∈ ]− 1, 1[.
We give this type of convergence a name: pointwise convergence. Note that we havefirst defined a sequence of functions fn by putting
fn(x) =n∑k=0
xk
for each n = 0, 1, 2, . . . .
Definition 2.1 (Pointwise convergence.) Suppose {fn(x) : n = 0, 1, 2, . . . } is a sequence of functions defined on an interval I. We say that fn(x) converges pointwiseto the function f(x) on the interval I if
fn(x)→ f(x), as n→∞, for each x ∈ I.
We call the function f(x) the limit function.
Example 2.1 fn(x) = x− 1n . Then fn(x) converges pointwise to x for each x ∈ R:
fn(x)− x =1
n→ 0 as n→∞.
Example 2.2 fn(x) = e−nx on [1, 3]. For each x ∈ [1, 3] we have nx → ∞ as n → ∞
and therefore fn(x)→ 0 as n→∞ for each x ∈ [1, 3]. Thus fn(x) converges pointwise tof(x) = 0 for each x ∈ [1, 3].
Example 2.3 fn(x) = e−nx on [0, 3]. For each 0 < x ≤ 3 we have nx → ∞ as n → ∞
and therefore fn(x) → 0 as n → ∞ for each 0 < x ≤ 3. However, at x = 0 we havefn(0) = 1 for all n. Thus fn(x) converges pointwise to the function f(x) defined byf(0) = 1, f(x) = 0 for each 0 < x ≤ 3. This is not a continuous function, despite thefact that each function fn(x) is continuous.
The last example shows what can happen with pointwise convergence: the limit function may fail to be continuous, even though all functions in the sequence are continuous.
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Example 2.4 Let the sequence fn be defined as
fn(x) =nx
(nx+ 1)3x ∈ [0,∞[.
Then fn(0) = 0 and for each fixed x > 0
fn(x) =n2x
(nx+ 1)3
=n2x
n3(x+ 1n)3
=1
n
x
(x+ 1n)3→ 0 as n→∞.
So that fn(x)→ f(x) = 0 pointwise on [0,∞[.Then for x > − 1
nwe have
f ′n(x) =n2(1− 2nx)
(nx+ 1)4
and we see that for x > 0 we have f ′n(x)→ 0 as n→∞ whereas f ′n(0) = n2 →∞. Herewe see that f ′n → f ′ only on for x > 0. This shows that differentiability is not alwaysrespected by pointwise convergence.
The last two examples then lead us to pose the question: what extra condition (otherthan just pointwise convergence) can guarantee that the limit function is also continuousor differentiable? The answer to this is given by the concept of uniform convergence.
3 Uniform convergence
We define for a realvalued (or complexvalued) function f on a nonempty set I thesupremum norm of f on the set I:
‖f‖I = supx∈If(x).
Note that if f is a bounded function on I then
supx∈If(x) = sup{ f(x) : x ∈ I}
exists, by the socalled supremum axiom. Observe that
f(x) ≤ ‖f‖I for all x ∈ I,
and that f(x) takes on values which are arbitrarily near ‖f‖I . In particular ‖f‖I =the largest value of f(x) whenever such a value exists (such as when I is a closed,bounded interval and f(x) is a continuous function on I).
The supremum norm has the following properties for functions f and g on a set I:
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‖f‖I ≥ 0 and ‖f‖I = 0⇔ f(x) = 0 for all x ∈ I
‖cf‖ = c · ‖f‖I for any constant c
‖f + g‖I ≤ ‖f‖I + ‖g‖I (triangle inequality)
‖f‖J ≤ ‖f‖I when J is a subset of I.
The proof of these properties is left as an exercise for the interested reader.
Now we come to the definition of uniform convergence:
Definition 3.1 A sequence of functions fn(x) defined on an set I is said to convergeuniformly to f(x) on I if
‖fn − f‖I → 0 as n→∞.
We write this as
limn→∞
fn = f uniformly on I
or as
fn → f uniformly on I as n→∞.
Uniform convergence implies pointwise convergence, however there are sequenceswhich converge pointwise but not uniformly. Indeed we have
fn(x)− f(x) ≤ supx∈Ifn(x)− f(x) = ‖fn − f‖I ,
so that
fn → f uniformly on I as n→∞
=⇒ fn(x)− f(x) → 0 for each x ∈ I
=⇒ fn → f pointwise on I.
We record this as a result:
Lemma 3.1 If the sequence of functions fn(x) converges uniformly to f(x) on theinterval I, then fn(x) converges pointwise to f(x).
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This Lemma says that the limit function obtained through uniform convergence (if thisoccurs) is the same as the limit function obtained from pointwise convergence. Or: iffn(x) converges to f(x) uniformly, then it must converge to f(x) pointwise. This thentells us how to go about testing for uniform convergence: first, obtain the pointwiselimit f(x) and then see if we have uniform convergence to f(x).
Example 3.1 fn(x) = e−nx on [1, 3]. We have seen above that fn(x) converges pointwise
to f(x) = 0 for each x ∈ [1, 3]. Then we have fn(x)− f(x) = fn(x) and we then have
‖fn − f‖ = supx∈[1,3]
fn(x)
= supx∈[1,3]
e−nx
= supx∈[1,3]
e−nx
= e−n → 0 as n→∞.
Thus we have uniform convergence in this case. Note that the last step follows from theobservation that e−nx is strictly decreasing for x ≥ 0 with n ≥ 0, so that e−n ≥ e−nx forall x ≥ 1.
Example 3.2 fn(x) = xe−nx on I = [0,∞[. Here the interval is unbounded. First we
look at pointwise convergence: fn(0) = 0 and for x > 0 we have that fn(x)→ 0 as n→∞.Thus fn(x)→ 0 pointwise on I. We now need to investigate uniform convergence. Sincethe limit function f(x) = 0 we have
‖fn − f‖ = supx∈[0,∞[
xe−nx
= supx∈[0,∞[
xe−nx
because fn(x) ≥ 0 for x ≥ 0. Now, we have f ′n(x) = (1 − nx)e−nx for x > 0 (observethat you should never differentiate on closed intervals), and we see that f ′n(x) = 0 whenx = 1/n. Further, f ′n(x) > 0 for 0 < x < 1/n, and f
′n(x) < 0 for x > 1/n, so we conclude
that fn(x) has a maximum at x = 1/n and hence
‖fn − f‖ = supx∈[0,∞[
xe−nx
= fn(1
n)
=1
ne→ 0 as n→∞.
So we see that the sequence of functions fn(x) = xe−nx converges uniformly to 0 on the
interval I = [0,∞[.
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4 Uniform convergence and continuity.
We now come to two important results. The first is the following.
Theorem 4.1 Suppose fn(x) is a sequence of continuous functions on an interval Iand suppose also that fn(x) converges uniformly to f(x) on the interval I. Then the limitfunction f(x) is also continuous.
Proof: We need to show that f(x)→ f(a) when x→ a for x, a ∈ I. First we note thatfor any n = 0, 1, 2, . . . we have
f(x)− f(a) = (f(x)− fn(x)) + (fn(x)− fn(a)) + (fn(a)− f(a))
≤ f(x)− fn(x)+ fn(x)− fn(a)+ fn(a)− f(a).
by the triangle inequality. Now, as we have already noted,
f(x)− fn(x) ≤ ‖f − fn‖I , f(a)− fn(a) ≤ ‖f − fn‖I ,so we now have, for any n,
f(x)− f(a) ≤ 2‖f − fn‖I + fn(x)− fn(a). (4.1)We shall now see that we may make the righthand side as small as we like. To do thiswe note that the inequality (4.1) is true for an arbitrary value of n.
Now choose a positive number � > 0, as small as we like. We know that
‖fn − f‖I → 0 as n→∞.Therefore there must be an N > 0 for which we have ‖fn − f‖I < �/3 for all n ≥ N .Then choose and fix such a value of n, say n = N .
We also have the fact that fN(x) is continuous, so for any choice of the number � > 0there is an interval centered on a so that
fN(x)− fN(a) < �/3whenever x belongs to that interval. Formally, this is described as follows: since
fN(x)→ fN(a) as x→ athere is, for any � > 0 a corresponding number δ > 0 so that
fN(x)− fN(a) < �/3 whenever x− a < δ.With all this put into equation (4.1), we obtain:
f(x)− f(a) ≤ 2‖f − fN‖I + fN(x)− fN(a)
≤ 2 �3
+�
3= �
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if x − a < δ. Thus for each � > 0, chosen as small as we like, we may always choose aδ > 0 so that
f(x)− f(a) < � whenever x− a < δ.This means that f(x)→ f(a) when x→ δ. Hence f(x) is continuous.
This result is very useful as a quick test for absence of uniform convergence: if
(i) fn(x), n = 0, 1, 2, . . . is a sequence of continuous functions (on some interval);
(ii) fn(x) converges pointwise to f(x);
(iii) f(x) is not continuous;
(iv) Then fn(x) does not converge uniformly to f(x).
Example 4.1 fn(x) = e−nx on [0, 3] is a sequence of continuous functions, converging
pointwise to f(x) defined by
f(x) =
{1 for x = 0
0 for 0 < x ≤ 3,
which is not continuous, and so, by Theorem 4.1, the sequence does not converge uniformly to f(x).
The second result we mention is the following, which is of great use in integratingseries:
Theorem 4.2 Suppose that fn(x) is sequence of continuous functions which convergesuniformly to a continuous function f(x) on a bounded interval [a, b]. Then we have
limn→∞
∫ ba
fn(x)dx =
∫ ba
limn→∞
fn(x)dx =
∫ ba
f(x)dx.
Proof: The proof is quite simple:
∣∣∣∣∫ ba
fn(x)dx−∫ ba
f(x)dx
∣∣∣∣ = ∣∣∣∣∫ ba
(fn(x)− f(x))dx∣∣∣∣
≤∫ ba
fn(x)− f(x)dx
≤∫ ba
‖fn − f‖dx
= ‖fn − f‖∫ ba
1 dx
= ‖fn − f‖(b− a)→ 0 as n→∞.
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Thus: ∫ ba
fn(x)dx→∫ ba
f(x)dx
as n→∞ if fn → f uniformly on I, which is what we wanted to prove.
Remark 4.1 Theorem 4.2 is proved here for continuous functions so that the integralsexist. It is however possible to replace the word continuous by the word integrable, andthe theorem is still true.
Example 4.2 If fn(x) =1
1 + x2 + x4
n
calculate
limn→∞
∫ 10
fn(x)dx.
In this example, it is easy to see that fn(x) → f(x) =1
1 + x2pointwise on [0, 1]. Then
we have
f(x)− fn(x) =1
n· x
4
(1 + x2)(1 + x2 + x4
n)
and it is difficult to calculate supx∈[0,1]
f(x) − fn(x) (you should try to do this, in order to
see how difficult it is). The best way to deal with this is to use the fact that (1 + x2)(1 +x2 + x
4
n) ≥ 1 on [0, 1] so that
0 ≤ x4
(1 + x2)(1 + x2 + x4
n)≤ x4 ≤ 1
for x ∈ [0, 1], and then we have f(x)−fn(x) ≤1
nfor all x ∈ [0, 1], from which we deduce
that
‖f − fn‖ = supx∈[0,1]
f(x)− fn(x) ≤1
n→ 0 as n→∞,
and then we have uniform convergence of fn → f , so that
limn→∞
∫ 10
fn(x)dx =
∫ 10
f(x)dx =π
4
by Theorem 4.2.
In the last calculation we have given an example of proving uniform convergence of asequence fn(x) defined on an interval I without computing the value of supx∈I fn(x)−f(x). It is important to realise that this is a useful way of avoiding very complicatedcalculations. All one has to do is show that supx∈I fn(x)− f(x) → 0 as n → ∞, whichcan be done either by first computing the value of supx∈I fn(x)− f(x), or by obtainingan inequality which then leads directly to supx∈I fn(x)− f(x) → 0 as n→∞.
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We now prove a result about uniform convergence and differentiability: it tells usunder which conditions the limit function f(x) is differentiable whenever the functions ofthe sequence fn(x) are differentiable.
Theorem 4.3 Suppose that {fn(x);n = 0, 1, 2, . . . } is a sequence of functions on aninterval I and satisfying the following conditions:
(i) fn(x) is differentiable on I for each n = 0, 1, 2, . . .
(ii) fn(x)converges pointwise to f(x) on I
(iii) f ′n(x) is continuous for each n and f′n → g converges uniformly on I where g(x)
is a continuous function on I.
Then the limit function f(x) is differentiable and f ′(x) = g(x).
Proof: First note that
fn(x)− fn(a) =∫ xa
f ′n(t)dt
for each fn(x) and for each choice of x, a ∈ I. Because fn(x) converges pointwise to f(x)for all x ∈ I, the lefthand side converges to f(x) − f(a) as n → ∞. Also, f ′n → guniformly on I so by Theorem 4.2 we have that∫ x
a
f ′n(t)dt→∫ xa
g(t)dt,
and we then find that
f(x)− f(a) =∫ xa
g(t)dt.
Now, g(t) is continuous, so that, by the Fundamental Theorem of Calculus,
d
dx
∫ xa
g(t)dt = g(x)
so that f(x) must be differentiable and f ′(x) = g(x).This result is very useful, as we shall see, in examining the differentiability of functional
series.
5 Applications to functional series.
Definition 5.1 A functional series is a series
∞∑k=0
uk(x)
where each term of the series uk(x) is a function on an interval I.
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We can also define pointwise convergence for functional series:
Definition 5.2 The functional series
∞∑k=0
uk(x)
is pointwise convergent for each x ∈ I if the limit
∞∑k=0
uk(x) = limN→∞
N∑k=0
uk(x)
exists for each x ∈ I.
Thus, we always define a sequence of partial sums SN(x) given as
SN(x) =N∑k=0
uk(x)
so that
S0(x) = u0(x), S1(x) = u0(x) + u1(x), S2(x) = u0(x) + u1(x) + u2(x), . . .
and if
limN→∞
SN(x)
exists for x then we say that the series
∞∑k=0
uk(x) = limN→∞
SN(x)
converges at x. It converges pointwise on the interval I if
limN→∞
SN(x)
exists for each x ∈ I.With these definitions, we deduce from Theorem 4.1 that if the functions uk(x) are all
continuous on I and if the sequence of partial sums SN(x) converges uniformly to S(x)on I, then S(x) is continuous. However, we would like an efficient way of deciding if afunctional series converges uniformly to a (unique) limit. It is not at all easy to applythe definition of uniform convergence to an infinite sum of functions, so another methodis desirable. The appropriate result is Weierstrass’ Majorant Theorem:
Theorem 5.1 Suppose that the functional series
∞∑k=0
uk(x)
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is defined on an interval I and that there is a sequence of positive constants Mk sothat
uk(x) ≤Mk, k = 0, 1, 2, . . .
for all x ∈ I. If∞∑k=0
Mk
converges, then
∞∑k=0
uk(x)
converges uniformly on I.
Proof: If the conditions are fullfilled then we immediately have, from the ComparisonTheorems for Positive Series, that, for each x ∈ I, the series
∞∑k=0
uk(x)
is convergent, so that
∞∑k=0
uk(x)
is absolutely convergent, and therefore convergent. This means that
∞∑k=0
uk(x)
is pointwise convergent on I, and we denote the limit by S(x). We now show that thepartial sums
SN(x) =N∑k=0
uk(x)
converges uniformly to S(x) on I under the conditions of the theorem. We have
S(x)− SN(x) =∞∑
k=N+1
uk(x)
(all we do is subtract the first N terms from the series). Then it follows that
S(x)− SN(x) ≤∞∑
k=N+1
uk(x) ≤∞∑
k=N+1
Mk
for each x ∈ I, since uk(x) ≤Mk for each x ∈ I according to our assumption. Then
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‖S − SN‖I ≤∞∑
k=N+1
Mk.
We also know (by assumption) that∑∞
k=0Mk converges, so we must have that∑∞
k=N+1Mk →0 as N →∞. Consequently,
‖S − SN‖I → 0 as N →∞,
and our result is proved.
Corollary 5.1 If
(i) the functional series
S(x) =∞∑k=0
uk(x) converges uniformly on interval I,
(ii) uk(x) is a continuous function on I for each k = 0, 1, 2, . . . ,
then S(x) is continuous on I.
Proof: Because a finite sum of continuous functions is again a continuous function, itfollows that the partial sums
SN(x) =N∑k=0
uk(x)
are continuous functions for N = 0, 1, 2, . . . . Then by Theorem 4.1, we have that S(x) =limN→∞ SN(x) is a continuous function.
Example 5.1 Take the functional series
∞∑k=1
sin kx
k2.
We have
uk(x) =∣∣∣∣sin kxk2
∣∣∣∣ =  sin kxk2 ≤ 1k2since  sin t ≤ 1 for all real t. We know (standard positive series) that
∞∑1
1
k2
converges (series of the form∑
1/kα converge for α > 1 and diverge for α ≤ 1). Hence,by Weierstrass’ Majorant Theorem,
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∞∑k=1
sin kx
k2
converges uniformly for all x, and by Corollary 5.1 this series is a continuous function ofx for all x ∈ R.
Remark 5.1 One advantage of Weierstrass’ Majorant Theorem is that we do not haveto calculate the value of the series at each x ∈ I in order to decide if we have uniformconvergence. However, a drawback is that the conditions of the theorem are only sufficient to establish uniform convergence, they are not absolutely necessary for uniformconvergence. In the final section of these lecture notes we give a necessary and sufficientcondition for uniform convergence.
Remark 5.2 In our statement of Weierstrass’ Majorant Theorem, we have not said anything about how to find the constants Mk. Usually we take
Mk = supx∈Iuk(x),
but this is not strictly necessary: any sequence (of constants) will do provided that∑Mk
converges.
Another result of interest is the following:
Theorem 5.2 If
(i) the functional series
∞∑k=0
uk(x) converges uniformly on the interval I
(ii) uk(x) is continuous on I for each k = 0, 1, 2, . . . ,
then ∫ xa
(∞∑k=0
uk(t)
)dt =
∞∑k=0
(∫ xa
uk(t)dt
)for all a, x ∈ I. In other words, if the series of continuous functions converges uniformlyon I, then the integral of the sum is the sum of the integrals of the functions, just as inthe case of a finite sum.
Proof: Put S(t) =∑∞
k=0 uk(t) and SN(t) =∑N
k=0 uk(t), then we have SN → S uniformlyon I so that
limN→∞
∫ xa
SN(t)dt =
∫ xa
limN→∞
SN(t)dt =
∫ xa
S(t)dt,
according to Theorem 4.2. Note that since SN(t) is a finite sum of functions, we see that
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∫ xa
SN(t)dt =
∫ xa
(N∑k=0
uk(t)
)dt
=N∑k=0
(∫ xa
uk(t)dt
),
and we then find that ∫ xa
(∞∑k=0
uk(t)
)dt =
∞∑k=0
(∫ xa
uk(t)dt
).
We can also say something about the differentiability of the series∑uk(x), using
Theorem 4.3 In this case, as in the previous two theorems, we replace fn(x) by SN(t) andf(x) by S(t). Thus, we want the following:
• SN(x)→ S(x) pointwise on I
• S ′N(x)→ G(x) uniformly on I
• SN(x) is continuously differentiable for each N
and then we may conclude that S(x) is continuously differentiable with S ′(x) = G(x). Allwe need is to formulate these requirements and result as follows:
Theorem 5.3 Suppose that∞∑k=0
uk(x) satisfies the following conditions:
•∞∑k=0
uk(x) converges pointwise on I
•∞∑k=0
u′k(x) converges uniformly on I
• uk(x) is continuously differentiable for each k
Then∞∑k=0
uk(x) is continuously differentiable and
d
dx
(∞∑k=0
uk(x)
)=∞∑k=0
u′k(x).
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Proof: We have that each S ′N(x) is continuous on I and
SN(x) −→ S(x) pointwise on I
S ′N −→ G uniformly on I.
Then by Theorem 4.3, S(x) is differentiable and S ′(x) = G(x) on I. In other words:
d
dx
(∞∑k=0
uk(x)
)=∞∑k=0
u′k(x)
6 Integrals dependent on a parameter.
We shall come across integrals of the form∫ ∞0
f(x, y)dy and
∫ ∞−∞
f(x, y)dy,
examples of which are the (onesided) Laplace transform of f :
F (s) =
∫ ∞0
f(t)e−stdt,
and the Fourier transform of f :
F (ω) =1
2π
∫ ∞−∞
f(t)e−iωtdt.
These are called integrals depending on a parameter.
If we have
F (x) =
∫ ∞0
f(x, y)dy,
we would like to have
F ′(x) =
∫ ∞0
fx(x, y)dy,
whenever f(x, y) satisfies suitable conditions. This situation is analogous to the case forfunctional series (where we replace f(x, y) by un(x) and integration over y is replaced bysummation over n). With this in mind, we introduce the concept of uniform convergence of integrals.
Definition 6.1 We say that the integral
F (x) =
∫ ∞a
f(x, y)dy
converges uniformly on I if:
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(i) F (x) =
∫ ∞a
f(x, y)dy converges pointwise for each x ∈ I;
(ii) the family of functions FR defined as
FR(x) =
∫ Ra
f(x, y)dy
converges uniformly to F on I. That is, if
‖FR − F‖I −→ 0 as R→∞.
A test for uniform convergence of integrals is an analogy of the Weierstrass test forfunctional series:
Theorem 6.1 (Mtest) Suppose
(i) f(x, y) is continuous on I × [a,∞[
(ii) f(x, y) ≤M(y) for all x ∈ I and y ∈ [a,∞[
(iii)
∫ ∞a
M(y)dy converges.
Then
F (x) =
∫ ∞a
f(x, y)dy
converges uniformly on I.
Proof: We have
FR(x)− F (x) =∫ ∞R
f(x, y)dy
from which we obtain
FR(x)− F (x) ≤∫ ∞R
f(x, y)dy
≤∫ ∞R
M(y)dy,
by assumption. Consequently,
‖FR − F‖I ≤∫ ∞R
M(y)dy → 0 as R→∞,
because
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∫ ∞a
M(y)dy converges
implies that ∫ ∞R
M(y)dy −→ 0 as R→∞.
Thus, FR −→ F uniformly on I.Now we come to proving that if FR → F uniformly on an interval I, then F is
continuous if each FR is continuous. here, the problem is to show that FR(x) is continuous.We have the following result:
Lemma 6.1 For each finite R > 0 we have
FR(x) =
∫ Ra
f(x, y)dy
is a continuous function on [c, d] if f(x, y) is continuous on [c, d]× [a,R].
The proof of this result is given in the appendix. From this result, it follows that iff(x, y) is continuous on I× [a,∞[ then FR(x) is continuous at every x ∈ I for every choiceof R > a: choosing x0 ∈ I we can find an bounded, closed interval [c, d] so that x0 ∈ [c, d].With these remarks we have:
Theorem 6.2 If
(i) f(x, y) is continuous on I × [a,∞[
(ii) F (x) =
∫ ∞a
f(x, y)dy converges uniformly on I
Then F (x) is continuous.
Proof: From Lemma 6.1 we know that each
FR(x) =
∫ Ra
f(x, y)dy
is continuous at each x ∈ I for each R > 0. Since FR → F uniformly, it follows fromTheorem 4.1 that F is continuous at each x ∈ I. This proves the result.
We now come to a very useful result on the integration of integrals with parameters:
Theorem 6.3 (i) f(x, y) is continuous on I × [a,∞[
(ii) F (x) =
∫ ∞a
f(x, y)dy converges uniformly on I
Then for any bounded, closed interval [b, c] ⊂ I we have∫ cb
(∫ ∞a
f(x, y)dy
)dx =
∫ ∞a
(∫ cb
f(x, y)dx
)dy.
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Proof: Note that since
FR(x) =
∫ Ra
f(x, y)dy
is continuous for each R > 0, by Lemma 6.1, we have∫ cb
FR(x)dx =
∫ Ra
(∫ cb
f(x, y)dx
)dy.
(This result is known as Fubini’s Theorem). Then we have, by uniform convergence ofFR → F (Theorem 4.2) that
limR→∞
∫ cb
FR(x)dx =
∫ cb
F (x)dx,
from which we immediately have∫ cb
(∫ ∞a
f(x, y)dy
)dx =
∫ ∞a
(∫ cb
f(x, y)dx
)dy.
Finally, we come to our theorem on differentiating under the integral:
Theorem 6.4 (i) f(x, y) and fx(x, y) be continuous on I × [a,∞[
(ii) F (x) =
∫ ∞a
f(x, y)dy converges pointwise on I
(iii) G(x) =
∫ ∞a
fx(x, y)dy converges uniformly on I
Then
F ′(x) =
∫ ∞a
fx(x, y)dy.
Proof: Since G(x) =
∫ ∞a
fx(x, y)dy converges uniformly on I, and fx(x, y) is continuous
on I × [a,∞[, G(x) is a continuous function on I, by Theorem 6.2. Then for any b, x ∈ Iwe have
∫ xb
G(t)dt =
∫ xb
(∫ ∞a
ft(t, y)dy
)dt
=
∫ ∞a
(∫ xb
ft(t, y)dt
)dy by Theorem 6.3
=
∫ ∞a
[f(x, y)− f(b, y)]dy
= F (x)− F (b).
Now, G(t) is continuous, so that
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d
dx
∫ xb
G(t)dt = G(x),
and from this it follows that F (x) is differentiable and
F ′(x) = G(x) =
∫ ∞a
fx(x, y)dy,
and the theorem is proved.We now turn to two examples. First we look at the onesided Laplace transform
F (s) =
∫ ∞0
f(t)e−stdt.
If f(t) satisfies the condition that f(t) ≤ Aeat for some constants A > 0, a (we thensay that f(t) is of exponential order) and that f(t) is continuous for t > 0, then F (s)converges uniformly for all s > a, since we then have f(t)e−st ≤ Ae−(s−a)t and then,according to the Theorem 6.1, F (s) converges uniformly for s > a. Note that tn for anyn ∈ N is of exponential order since we have tne−t → 0 as t→∞, so that for t ≥ 0 thereis for a given n ∈ N a constant A > 0 with tne−t ≤ A so that tn ≤ Aet. Hence itfollows that tnf(t) is of exponential order if f(t) is. Then it follows that if F (s) convergesuniformly for s > a, we have by Theorem 6.4 that F (s) is differentiable and
F ′(s) =
∫ ∞0
(−t)f(t)e−stdt.
The second application is to integrals of the form
F (ω) =
∫ ∞−∞
f(t)e−iωtdt.
Here we define
F (ω) = limR→∞
FR(ω),
with
FR(ω) =
∫ R−R
f(t)e−iωtdt
for each R > 0. To ensure convergence we require that f(t) satisfy the condition that∫ ∞−∞f(t)dt = lim
R→∞
∫ R−Rf(t)dt
exist. The set of such f(t) is denoted by L1(R, dt). Then it follows that, if f ∈ L1(R, dt) iscontinuous, we have that F (ω) converges uniformly by Theorem 6.1, and then by Theorem6.2 we have that F (ω) is continuous as a function of ω. If we have f(t) continuous withf(t) ∈ L1(R, dt) and tf(t) ∈ L1(R, dt) we then find that F (ω) is differentiable and that
F ′(ω) =
∫ ∞−∞
(−it)f(t)e−iωtdt.
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7 APPENDIX.
7.1 Supremum and Infimum: a recapitulation.
Definition 7.1 Let A ⊂ R. Then the supremum of A, denoted by supA, is defined asthe smallest number a ∈ R with the property that x ≤ a for all x ∈ A. In mathematicalshorthand we have
supA = min{a ∈ R : x ≤ a for all x ∈ A}.
Similarly, the infimum of A, denoted by inf A, is defined as the largest numberb ∈ R with the property that x ≥ b for all x ∈ A. In mathematical shorthand we have
inf A = max{b ∈ R : x ≥ b for all x ∈ A}.
Remark 7.1 Note that in these definitions neither the supremum nor the infimum needbelong to the set A.
Example 7.1
(i) A = [−1, 3]. Here we have supA = 3, inf A = −1, and both these belong to A.
(ii) A =]− 1, 3]. Here supA = 3, inf A = −1, but only inf A belongs to A.
(iii) A =]− 1, 3[. Here supA = 3, inf A = −1, and both are not in A.
(iv) A = [−1,∞[. Here inf A = −1 whereas supA does not exist.
Definition 7.2 Let f : R → R be a function. Then the supremum of f(x) over A isdefined as the smallest number a ∈ R with the property that f(x) ≤ a for all x ∈ A.In mathematical shorthand we have
supx∈A
f(x) = min{a ∈ R : f(x) ≤ a for all x ∈ A}.
Similarly, the infimum of f(x) over A is defined as the largest number b ∈ Rwith the property that f(x) ≥ b for all x ∈ A. In mathematical shorthand we have
infx∈A
= max{b ∈ R : f(x) ≥ b for all x ∈ A}.
Example 7.2 (i) f(x) = x3, A = [−1, 3] Then, since f(x) is a strictly increasingfunction, we have
supx∈A
f(x) = 27, infx∈A
f(x) = −1.
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(ii) f(x) = x2, A = [−1, 3] Then note that f(x) = x2 is not strictly increasing on thisinterval: it is decreasing on [−1, 0] and then strictly increasing on [0, 3]. So we have
supx∈[−1,0]
f(x) = 1, infx∈[−1,0]
f(x) = 0
and
supx∈[0,3]
f(x) = 9, infx∈[0,3]
f(x) = 0.
Combining these two observations, we find that
supx∈[−1,3]
f(x) = 9, infx∈[−1,3]
= 0.
(iii) f(x) = arctan x, A = R. Here we have a strictly increasing function, and we have
supx∈R
f(x) =π
2, inf
x∈Rf(x) = −π
2.
It is tempting to take the largest value of a function on an interval as the supremum,and the least value for the infimum. The last example shows that the neither the supremum nor the infimum need be attainable values of a function. However, we have thefollowing simple but useful result:
Lemma 7.1 Suppose that f(x) is a realvalued continuous function on the closed,bounded interval [a, b]. Then
supx∈[a,b]
f(x) = max{f(x) : x ∈ [a, b]}, infx∈[a,b]
f(x) = min{f(x) : x ∈ [a, b]}.
That is, the supremum of a continuous function over a closed, bounded interval is equalto its largest value over that interval, and the infimum is the least value of the functionover the interval.
Proof: Since f(x) is continuous and the interval is closed, then f(x) has a largest valueand a least value on the interval: there exist x1, x2 ∈ [a, b] so that f(x1) ≤ f(x) ≤ f(x2)for all x ∈ [a, b], and we now see that
supx∈[a,b]
f(x) = f(x2), infx∈[a,b]
f(x) = f(x1),
and the result is proved.
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7.2 Proof of Lemma 6.1
We sketch the proof and refer to any good book on analysis for further details.
Our task is to prove that for each x ∈ [c, d] we have
FR(x+ h) −→ FR(x) as h→ 0.
Then we have
FR(x+ h)− FR(x) ≤∫ Ra
f(x+ h, y)− f(x, y)dy.
Now if f is continuous on [c, d]× [a,R] it can be shown that for any � > 0 there is a δ > 0so that
f(x0, y0)− f(x1, y1) < �
whenever√
(x0 − x1)2 + (y0 − y1)2 < δ. That is, whenever the distance between thepoints (x0, y0) and (x1, y1) is less than δ. This is called uniform continuity. Using thisfact, we choose � > 0 (arbitrarily small) and then for each given R we find a δ > 0 so that
f(x+ h, y)− f(x, y) < �R− a
whenever h < δ. From this it follows that
FR(x+ h)− FR(x) < � whenever h < δ.
Note that we have � > 0 arbitrarily small, and for each such choice there is a correspondingδ. From this it follows that
FR(x+ h)− FR(x) −→ 0 as h→ 0.
Hence FR(x) is continuous on [c, d], for each choice of R > 0.
7.3 Cauchy’s condition for uniform convergence of series
In this section we record, without proof, a result which is of some interest: Cauchy’s criterion for uniform convergence of functional series, and we make some commentson some aspects of uniform convergence.
Theorem 7.1 Suppose that {uk(x)} is a sequence of continuous functions defined on aninterval I. Then the series
∞∑k=0
uk(x)
is uniformly convergent on I if and only if for each choice of � > 0, however small, thereexists a (corresponding) integer N > 0 so that
uk+1(x) + uk+2(x) + · · ·+ um(x) < �
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for all m > k ≥ N and for all x ∈ I. In particular, we then have, on putting m = k + 1,
uk+1(x) < �
for all k ≥ N and all x ∈ I.
The proof of this result requires more mathematical machinery than we have at hand,and can be found in any good textbook on Mathematical Analysis.
As an illustration of the usefulness of this result we look at the series expansion of ex.We have
ex = 1 + x+x2
2!+ · · ·+ x
n
n!+ · · · =
∞∑k=0
xk
k!.
This expansion is true for each x ∈ R, so we have pointwise convergence of the series.However, we do not have uniform convergence on R. To see this, we apply the lastcomment in the theorem: we need to find, for a given � > 0, an N > 0 so that for allx ∈ R we have
uk+1(x) =xk+1
k!< �
whenever k ≥ N . However, if we choose any k we may choose x so that
xk+1
k!
is as large as we like, contradicting the requirement for uniform convergence. Hencewe do not have uniform convergence on the whole of R. However, if we only considerx ∈ [−a, a] for some a > 0 then we can prove uniform convergence of the series on thisclosed, bounded interval. This can be done using Weierstrass’ Majorant Theorem.This phenomenon occurs often, and then we say that the series converges uniformlyon closed, bounded intervals or converges on compact sets. Another example ofthis phenomenon occurs in power series (which are the first kind of functional seriestaught in elementary calculus courses). For instance, for the geometric series
∞∑k=0
xk
we have absolute convergence for x < 1 and divergence for x ≥ 1. For x < 1 we have∞∑k=0
xk =1
1− x= S(x).
The corresponding partial sums are
SN(x) =N∑k=0
xk =1− xN+1
1− x.
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The sequence SN(x) does not converge uniformly to S(x) on the interval ] − 1, 1[ : wehave
SN(x)− S(x) =xN+1
1− xand as x → 1− we see that xN+1 → 1 and 1/(1 − x) → ∞, so that we may makeSN(x) − S(x) as large as we like, and it then follows that ‖SN − S‖ does not exist, soit is impossible for ‖SN − S‖ → 0 as N → ∞. However, if we consider the series on theclosed, bounded interval [−a, a] with a fixed 0 < a < 1, we have
SN(x)− S(x) ≤aN+1
1− afor all x ∈ [−a, a], and therefore
‖SN − S‖ = supx∈[−a,a]
SN(x)− S(x) ≤aN+1
1− a→ 0 as N →∞,
because 0 < a < 1 gives aN+1 → ∞ when N → ∞. So we have uniform convergence ofthe series on compact subsets of ]− 1, 1[.
24