Schinazi R.B. From Calculus to Analysis

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200 6 Riemann Integration

4. (a) Give an example for which

M(f +g) < M(f ) +M(g).

(b) Give an example for which

M(f +g) =M(f) +M(g).

5. Assume that a<c<band that f is integrable on [a,c] and on [c, b]. Prove

that f is integrable on [a,b].

6. Give an example of a function f which is positive and integrable on [a,b],

which is not identically 0, but whose integral is 0. Does this contradict Appli-

cation 6.1?

7. Show that S3 holds even when c<aor c>b.

8. Suppose that f is integrable on [a,b]. Show that



f, [a,b]



(b −a) ≤



f ≤M



f, [a,b]



(b −a).

9. Give an example of function f for which









|f |.

10. Assume that f and g are integrable on [a,b].

(a) Show that

(f +g)

−(f −g)

=4fg.

(b) Show that fg is integrable.

11. Assume that f is integrable on [a,b] and that |f(x)| <K for all x in [a,b].

Show that for all x and y in [a,b],wehave







≤K|x −y|.

12. Suppose that f and g are differentiable and that f



and g



are continuous on

[a,b]. Prove the integration-by-parts formula



(fg



) =





g) +f(b)g(b)−f (a)g(a).

13. Assume that f is continuous on [a,b] and that for all x in [a,b],



f =0.

Show that f is identically 0 on [a,b].

14. Let f and g be continuous on [a,b] and assume that g ≥0on[a,b].

(a) Show that there are x

and x

in [a,b] such that for all x in [a,b],

f(x

) ≤f(x)≤f(x

6.3 Uniform Continuity 201

(b) Assume that



g>0. Show that

f(x

) ≤



(fg)



≤f(x



(fg) =f(c)



(d) Show that (c) holds when



g =0 as well. (Use Application 6.1.)

15. We deﬁne a distance between continuous functions f and g by setting

d(f,g)=



|f −g|.

(a) Show that if f and g are continuous on [a,b], then |f − g| is continuous

and therefore integrable.

(b) Show that d(f,g) ≥0 for all continuous functions f and g.

f =g on [a,b]. (Use Application 6.1.)

(d) Show that

d(f,g)=d(g,f)

for all continuous functions f and g.

(e) Show that

d(f,g)≤d(f,h)+d(h,g)

for all continuous functions f , g, and h.

16. Assume that f and g are integrable on [a,b].

(a) Show that



(f −g)

≥0.

(b) Use (a) to show that



(fg) ≤







6.3 Uniform Continuity

We used the notion of uniform continuity in Sect. 6.1 to prove the Riemann integra-

bility of continuous functions. In this section we discuss this notion more in depth.

We start with the deﬁnition.

202 6 Riemann Integration

Uniformly continuous functions

A function f is said to be uniformly continuous on the set D if for every

>0, there is a δ>0 such that if x and y are in D and if |x −y| <δ, then

|f(x)−f(y)|<.

As noted in Sect. 6.1, uniform continuity concerns the behavior of a function on

a whole set. Continuity at a point a, on the other-hand, depends on the behavior of

the function near a. It is clear that if f is uniformly continuous on the set D, then f

is continuous at every point a of D. However, as we will show below, the converse

is not true.

Example 6.7 The function f deﬁned on R by f(x)= x is uniformly continuous

on R.

This is very easy: for any >0, set δ =.If|x −y|<δ, then



f(x)−f(y)



=|x −y|<δ=,

and we are done.

We now state an useful criterion to prove that a function is NOT uniformly con-

tinuous.

Nonuniform continuity criterion

The function f is not uniformly continuous on the set D if and only if there

are two sequences x

and y

in D such that x

−y

converges to 0 and such

that f(x

) −f(y

) is bounded away from 0. That is, there is a>0 such that

for all n ≥1,



f(x

) −f(y

)



≥a.

Assume ﬁrst that f is not uniformly continuous on D. Then, there is an >0

such that for every δ>0, there are x and y in D such that |x −y|<δand |f(x)−

f(y)|≥. For every n ≥1, we may pick δ = 1/n, and the corresponding x and y

can be denoted by x

and y

. We construct like this two sequences x

and y

in D

such that for all n ≥1,

−y

|< 1/n and



f(x

) −f(y

)



≥.

Hence, x

−y

converges to 0, and f(x

) − f(y

) is bounded away from 0. This

proves the direct implication.

For the converse, we do a proof by contradiction. Assume that f is uniformly

continuous and that there are two sequences x

and y

in D such that x

− y

converges to 0 and that there is a>0 such that for all n ≥1,



f(x

) −f(y

)



≥a.

6.3 Uniform Continuity 203

For  =a>0, there must be a δ>0 such that if |x −y|<δ, then |f(x)−f(y)|<a.

Since x

−y

converges to 0, there is a natural N such that if n ≥N , then

−y

|<δ.

Hence, for n ≥N,



f(x

) −f(y

)



<a.

But we know that |f(x

) −f(y

)|≥a. We have a contradiction, and thus f is not

uniformly continuous on D. This completes the proof.

We apply the nonuniform continuity criterion in the next example.

Example 6.8 Let h be deﬁned by h(x) = 1/x for x in (0, 1]. Show that h is not

uniformly continuous on (0, 1].

By looking at the graph of h one can guess that this function is not uniformly

continuous because of its behavior near 0. As we approach 0, the function grows

faster and faster to inﬁnity. This suggests that we pick sequences x

and y

that

approach 0. Let x

= 1/n and y

= 2/n. Therefore, x

− y

=−1/n converges

to 0. On the other hand,



h(x

) −h(y

)



=n/2 ≥1/2

for all n ≥1. Hence, we have found two sequences x

and y

such that x

−y

con-

verges to 0 and f(x

) −f(y

) bounded away from 0. By the nonuniform continuity

criterion, h is not uniformly continuous on (0, 1].

We now state the most important result of this section.

Continuity on a closed and bounded interval

Assume that f is continuous at every point of a closed and bounded interval

[a,b]. Then f is uniformly continuous on [a,b].

We do a proof by contradiction. Assume that f is continuous but not uniformly

continuous on [a,b]. By the nonuniform continuity criterion, there are two se-

quences x

and y

on [a,b] such that x

−y

converges to 0 and such that there is

an >0forwhich



f(x

) −f(y

)



≥ for all n ≥1.

Since x

is a bounded sequence (why?), by the Bolzano–Weierstrass theorem, it has

a convergent subsequence x

. The limit  must be in [a,b] (why?). On the other

hand, x

−y

is a subsequence of x

−y

and therefore must converge to 0. Hence,

−x

converges to 0 + =.

By the continuity of f at  we have that

f(x

) and f(y

) both converge to .

204 6 Riemann Integration

Hence, f(x

) −f(y

) converges to 0. Therefore, there is a natural K such that if

k ≥K, then



f(x

) −f(y

)



</2.

This contradicts |f(x

) −f(y

)|≥ for every n. The function f must be uniformly

continuous on [a,b].

Note that Example 6.8 shows that it is crucial for the set to be closed for the

theorem to hold. The function 1/x is continuous on (0, 1] but not uniformly contin-

uous. It is also crucial for the set to be bounded. In the Exercises, one shows that the

function x

is continuous on R but not uniformly continuous.

Lipschitz functions

Assume that there is K>0 such that



f(x)−f(y)



≤K|x −y|

for all x and y in D. Then, f is said to be a Lipschitz function. If f is a

Lipschitz function on D, then it is uniformly continuous.

We now prove that if f is Lipschitz on D, then it is uniformly continuous. Let

>0 and pick δ =/K.If|x −y|<δ, then



f(x)−f(y)



≤K|x −y| <Kδ=.

Hence, the function f is uniformly continuous, and we are done.

We apply the preceding result in the next example.

Example 6.9 Show that the function f deﬁned f(x)=

√

x is uniformly continuous

on [1, ∞).

Observe that



f(x)−f(y)



√

x −

√



|x −y|

√

x +

√

Since x ≥1, we have

√

x ≥

√

1 =1. Therefore,

√

x +

√

y ≥2,

and since the inverse function is decreasing on the positive reals, we have

√

x +

√

≤

Hence,



f(x)−f(y)



≤

|x −y|.

Therefore, f is Lipschitz (with K =1/2) and so uniformly continuous on [1, ∞).

We will show in the exercises that a function may be uniformly continuous with-

out being Lipschitz.

6.3 Uniform Continuity 205

Exercises

1. Let g be deﬁned by g(x) =x

(a) Show that g is not uniformly continuous on R.

(b) Is there a set on which g is uniformly continuous?

2. Show that the function h deﬁned by h(x) = 1/x is uniformly continuous on

[1, ∞).

3. Let the function f be deﬁned by f(x)=

√

x on [0, 1].

(a) Show that there is a sequence x

in (0, 1] such that f(x

)/x

is unbounded.

(b) Use (a) to show that the function f is not Lipschitz on [0, 1].

[0, 1] but which is uniformly continuous.

4. Assume that f and g are uniformly continuous on D.

(a) Show that f +g is uniformly continuous on D.

(b) Is fg necessarily uniformly continuous on D?

5. Assume that f is uniformly continuous on [0, 1] and on [1, ∞). Show that f is

uniformly continuous on [0, ∞).

6. Show that the function f deﬁned by f(x)=

√

x is uniformly continuous on

[0, ∞).(UseExercise5.)

7. Let f be Riemann integrable on [0, 1]. Deﬁne

F(x)=



f for x ∈[0, 1].

Show that F is a Lipschitz function. (Recall that f is bounded.)

8. Let g(x) =sin(1/x) for x in (0, 1).

(a) Show that g is continuous on (0, 1).

(b) Prove that g is not uniformly continuous on (0, 1).

9. Let f be uniformly continuous on [0, ∞). In this exercise we will show that f

can grow at most linearly.

(a) Show that there is a δ>0 such that if |x −y|<δ, then |f(x)−f(y)|< 1.

Let d =δ/2.

(b) Show that



f(d)



< 1 +



f(0)



)



<n+



f(0)



(d) For any x in [0, ∞), there is a positive integer n(x) such that

n(x) ≤ x/d < n(x) +1.

(For a proof, see the lemma in Sect. 1.3.) Show that



f(x)



< 1 +





n(x)d





< 1 +n(x) +



f(0)



<x/d+



f(0)



+1.

The function on the r.h.s. is linear in x, and this concludes the proof.

Chapter 7

Convergence of Functions

In many situations we have a sequence of functions f

that converges to some func-

tion f and f is not easy to study directly. Can we use the functions f

to get some

information about f ? For instance, if the f

are continuous, is f necessarily con-

tinuous? Another question that often comes up is: can I compute



fdx using



dx? More precisely, is it true that

lim

n→∞



dx =



fdx?

We can rewrite the question as

lim

n→∞



dx =



lim

n→∞

dx?

In other words, can we interchange the limit and the integral? We will give some

partial answers to these questions in this chapter. First, we need to deﬁne what we

mean by the convergence of a sequence of functions. There are many different ways

a sequence of functions can converge. In this chapter we will just consider two of

them. Here is the ﬁrst.

Pointwise convergence

Consider a sequence of functions f

all deﬁned on the same set S ⊂ R.The

sequence f

is said to converge pointwise to f on S if for every x,inS we

have that

lim

n→∞

(x) =f(x).

Observe that we are using the deﬁnition of numerical sequences f

(x) to deﬁne

the convergence of a sequence of functions f

. More precisely, if f

converges

pointwise to f on S, it means that for every (ﬁxed) x in S and every >0, there

exists N (that depends on  and x) such that for n ≥ N,wehave|f

(x)−f(x)|<.

R.B. Schinazi, From Calculus to Analysis,

207

208 7 Convergence of Functions

Example 7.1 Consider the sequence f

(x) = e

−nx

and take S =[0, ∞). Does the

sequence f

converge?

We have

(x) =e

−nx



−x



where r = e

−x

. There are two cases. If x = 0, then r = 1 and r

= 1. Therefore,

(0) converges to 1. On the other hand, if x>0, then r<1, and r

converges to 0.

That is, f

(x) converges to 0 for all x>0. In summary, we have proved that f

converges pointwise to the function f on [0, ∞) deﬁned by f(0) =1 and f(x)=0

for x>0.

Example 7.1 shows that a sequence of continuous functions f

may converge

pointwise to a limit f which is not continuous!

Example 7.2 Consider the sequence f

deﬁned on [0, 1] by

(0) =0,

(x) =n for x ∈





(x) =0forx ∈



, 1



We ﬁrst show that f

converges pointwise to 0 on [0, 1].Ifx = 0, then f

(0) is

always 0 and therefore converges to 0. If x is in (0, 1], then there exists N such that

(why?) and therefore x>

. Hence, for all n ≥ N,wehavex>

.Bythe

deﬁnition of f

we have f

(x) =0 for all n ≥N. This implies that f

(x) converges

to 0. We have proved that f

converges pointwise to 0 on [0, 1].

We now turn to a different question. Is it true that

lim

n→∞



dx =



lim

n→∞

dx?

We will show that this is not true for this example. First note that each f

is Riemann

integrable on [0, 1] and that



(x) dx =



1/n

ndx =1.

Hence, the sequence



(x) dx is always 1 and therefore converges to 1. On the

other hand, we know that for every x in [0, 1], f

(x) converges to 0, so



lim

n→∞

dx =



0 dx =0.

This is an example for which we cannot interchange the limit and the integral.

Examples 7.1 and 7.2 show that pointwise convergence is not enough for our

purposes. A sequence of continuous functions need not converge to a continuous

7 Convergence of Functions 209

function, and the interchange of limit and integral need not be true. This is why we

turn to a more stringent type of convergence.

Uniform convergence

Consider a sequence of functions f

all deﬁned on the same set S ⊂ R.The

sequence f

is said to converge uniformly to f on S if for every >0, there

exists a natural N such that for all n ≥N ,wehave



(x) −f(x)



< for all x ∈S.

The critical difference between pointwise and uniform convergence is the follow-

ing. In the case of uniform convergence we ask for the same N for all x in S.This

is why it is called uniform convergence. In the case of pointwise convergence, for

each x in S, we have a possibly different N . The following criterion will be useful

in checking for uniform convergence.

Uniform convergence criterion

Consider a sequence of functions f

. For each n ≥1, f

is deﬁned on S.For

a ﬁxed function f deﬁned on S and for n ≥1, let

=sup





(x) −f(x)



:x ∈S



The sequence f

converges uniformly to f if and only if m

converges to 0.

We now prove the criterion. Assume ﬁrst that f

converges uniformly to f on S.

Let >0. There exists a natural N such that for all n ≥N ,wehave



(x) −f(x)



< for all x ∈S.

This shows that for n ≥N ,thesetA

={|f

(x) −f(x)|:x ∈S} is bounded above

by . Hence the supremum (i.e., the least upper bound) m

of this set exists (why?)

and is less than the upper bound . That is, for n ≥N,wehavem

=|m

−0|≤.

This proves that m

converges to 0. The direct implication is proved.

We now prove the converse. Assume that m

converges to 0. Let >0. There is

a natural N such that if n ≥ N, then |m

− 0| <. Since m

is positive, we have

<. That is, the least upper bound m

of the set A

is less than  for n ≥ N.

Hence,  is an upper bound for A

. Therefore, for n ≥N ,wehave



(x) −f(x)



≤

for all x in S. This proves that f

converges uniformly to f on S. The proof of the

criterion is complete.

It is easy to see that uniform convergence implies pointwise convergence. We

will see shortly that the converse is not true.