Davidson K.R., Donsig A.P. Real Analysis with Real Applications

7.1 Deﬁnition and Examples 183

Here we will establish the triangle inequality for p = 1. If f and g are in

C[a, b], compute

kf + gk

|f(x) + g(x)|dx

≤

|f(x)| + |g(x)|dx

|f(x)|dx +

|g(x)|dx = kfk

+ kgk

Exercises for Section 7.1

A. Show that k(x, y, z)k = |x| + 2

+ z

is a norm on R

. Sketch the unit ball.

B. Is k(x, y)k =

|x|

1/2

+ |y|

1/2

a norm on R

C. For f in C

[a, b], deﬁne ρ(f ) = kf

∞

. Show that ρ is nonnegative, homogeneous,

and satisﬁes the triangle inequality. Why is it not a norm?

D. If (V, k·k) is a normed vector space, show that

kxk−kyk

≤ kx−yk for all x, y ∈ V .

E. Show that the unit ball of a normed vector space, (V, k · k), is convex, meaning that if

kxk ≤ 1 and kyk ≤ 1, then every point on the line segment between x and y has norm

at most 1.

HINT: Describe the line segment algebraically in terms of x and y and a parameter t.

F. Let K be a compact subset of R

, and let C(K, R

) denote the vector space of all

continuous functions from K into R

. Show that for f in C(K, R

), the quantity

kfk

∞

= sup

x∈K

kf(x)k

is ﬁnite and k · k

∞

is a norm on C(K, R

G. Deﬁne C

[a, b] for p ∈ N and verify that the C

[a, b] norm is indeed a norm. Is there

a reasonable deﬁnition for C

[a, b]?

H. (a) Show that if k · k and ||| · ||| are both norms on V , then kvk

:= max{kvk, |||v|||} is

also a norm on V .

(b) Take V = R

and k(x, y)k =

+ y

and |||(x, y)||| =

|x| + |y|. Then deﬁne

k(x, y)k

as in part (a). Draw a sketch of the unit balls for these three norms.

I. Let S be any subset of R

. Let C

(S) denote the vector space of all bounded continu-

ous functions on S. For f ∈ C(S), deﬁne kfk

∞

= sup

x∈S

|f(x)|.

(a) Show that this is a norm on C

(S).

(b) When is this a norm on the vector space of all continuous functions on S?

J. (a) Let x

, . . . , x

n+1

be distinct points in [a, b]. Show that there are polynomials p

degree n so that p

) = δ

for 1 ≤ i, j ≤ n + 1.

(b) Hence show that the subspace of C[a, b] of all polynomials of degree at most n is

(n + 1)-dimensional.

184 Normed Vector Spaces

K. (a) Let a = x

< ··· < x

= b be distinct points in a compact subset K of R. Let

(x) =

(

x−x

a−x

a≤x ≤x

0 x

≤x≤b

(x) =

(

0 a≤x ≤x

n−1

x−x

n−1

b−x

n−1

≤x≤b

and for 1 ≤ k ≤ n − 1, let

(x) =











0 a ≤ x ≤ x

k−1

and x

k+1

≤ x ≤ b

x−x

k−1

−x

k−1

≤ x ≤ x

k+1

−x

k+1

−x

≤ x ≤ x

k+1

Describe the linear span of {h

, h

, . . . , h

} as a subspace of C(K).

(b) Hence show that C(K) is inﬁnite dimensional if K is an inﬁnite set.

7.2. Topology in Normed Spaces

The point of this section is to show how the notions of convergence and topol-

ogy, which we developed in R and in R

, can be generalized to any normed vector

space. We give the deﬁnitions and some simple properties where everything is al-

most identical to the treatment of R

. Other properties, such as the Heine–Borel

Theorem, do not hold in general. For such properties, we will have to study speciﬁc

kinds of normed vector spaces individually using their special properties.

The notion of convergence is the most fundamental and is exactly the same as

our deﬁnition for R

7.2.1. DEFINITION. In a normed vector space (V, k·k), we say that a sequence

)

∞

n=1

converges to v ∈ V if lim

n→∞

− vk = 0. Equivalently, for every ε > 0,

there is an integer N > 0 so that kv

− vk < ε for all n ≥ N . This is written

lim

n→∞

= v.

As in R

, we can decide if a sequence is attempting to converge without men-

tioning a limit point.

7.2.2. DEFINITION. Call (v

)

∞

n=1

a Cauchy sequence if for every ε > 0, there

is an integer N > 0 so that kv

− v

k < ε for all n, m ≥ N.

This leads to the notion of completeness in this context. Completeness is the

fundamental property that distinguishes the real numbers from the rational num-

bers. We have already seen several important theorems that depend on this notion,

such as the Least Upper Bound Principle (2.5.3) and the completeness both of R

(Theorem 2.7.4) and of R

(Theorem 4.2.5). So it should not surprise the reader

to ﬁnd out that this is also a fundamental property of bigger normed spaces such as

C(K), k·k

∞

. That C(K) is complete will be established in the next chapter; see

Theorem 8.2.2.

7.2 Topology in Normed Spaces 185

7.2.3. DEFINITION. Say that (V, k·k) is complete if every Cauchy sequence in

V converges to some vector v ∈ V . A complete normed space is called a Banach

space.

Now we can reformulate convergence in terms of open and closed sets, again

exactly as for R

7.2.4. DEFINITION. For a normed vector space (V, k · k), we deﬁne the open

ball with centre a ∈ V and radius r > 0 to be B

(a) = {v ∈ V : kv − ak < r}.

A subset U of V is open if for every a ∈ U, there is some r > 0 so that

(a) ⊂ U .

A subset C of V is closed if it contains all of its limit points. That is, whenever

) is a convergent sequence of points in C with limit x = lim

n→∞

, then x belongs

to C.

Proposition 4.3.8 works just as well for any normed space. So the open sets are

precisely the complements of closed sets. Here is a sample result in showing the

relationship between convergence and topology.

7.2.5. PROPOSITION. A sequence x

in a normed vector space V converges

to a vector x if and only if for each open set U containing x, there is an integer N

so that x

∈ U for all n ≥ N.

PROOF. Suppose that x = lim

n→∞

and U is an open set containing x. Then there

is an r > 0 so that B

(x) is contained in U. From the deﬁnition of limit, there is

an integer N so that

kx − x

k < r for all n ≥ N.

This just says that x

∈ B

(x) ⊂ U for all n ≥ N.

Conversely, suppose that the latter condition holds. In order to establish that

x = lim

n→∞

, let r > 0 be given. Take the open set U = B

(x). By hypothesis,

there is an integer N so that x

∈ U for all n ≥ N. As before, this just means that

kx − x

k < r for all n ≥ N.

Hence lim

n→∞

= x. ¥

There is one more fundamental property that we can deﬁne in this general

context, compactness. However, the reader should be warned that the main the-

orem about compactness in R

, the Heine–Borel Theorem, is not valid in inﬁnite-

dimensional spaces. The correct characterization of compact sets is given by the

Borel–Lebesgue Theorem (Theorem 9.2.3), which we will prove in the context of

metric spaces in Chapter 9.

186 Normed Vector Spaces

7.2.6. DEFINITION. A subset K of a normed vector space V is compact if

every sequence (x

) of points in K has a subsequence (x

) which converges to a

point in K.

Exercises for Section 7.2

A. If x = lim

n→∞

and y = lim

n→∞

in a normed space V and α = lim

n→∞

, show that

x + y = lim

n→∞

+ y

and αx = lim

n→∞

B. Show that every convergent sequence is any normed space is a Cauchy sequence.

C. If A is a subset of (V, k · k), let

A denote its closure. Show that if x ∈ V and α ∈ R,

then

x + A = x +

A and α

A = αA.

D. Show that if A is an arbitrary subset of a normed space V and U is an open subset,

then A + U = {a + u : a ∈ A, u ∈ U} is open.

E. Prove that if two norms on V have the same unit ball, then the norms are equal.

F. Which of the following sets are open in C

[0, 1]? Explain.

(a) A = {f ∈ C

[0, 1] : f(x) > 0, kf

∞

< 1, |f

(0)| > 2}

(b) B = {f ∈ C

[0, 1] : f(1) < 0, f

(1) = 0, f

(1) > 0}

[0, 1] : f(x)f

(x) > 0 for 0 ≤ x ≤ 1}.

HINT: Extreme Value Theorem and Intermediate Value Theorem

(d) D = {f ∈ C

[0, 1] : f(x)f

(x) > 0 for 0 < x < 1}.

HINT: Why is this different from the previous example?

G. Prove that a compact subset of a normed vector space is closed and bounded.

H. (a) Prove that a compact subset of a normed vector space is complete.

(b) Prove that a closed subset of a complete normed vector space is complete.

I. Consider the functions in C[−1, 1] given by f

(x) =











0 −1 ≤ x ≤ 0

nx 0 ≤ x ≤

≤ x ≤ 1

(a) Show that kf

− f

∞

≥

if m ≥ 2n.

(b) Hence show that no subsequence of (f

)

∞

n=1

converges.

(d) Show that the unit ball of C[−1, 1] is closed and bounded and complete.

J. Prove that the following are equivalent for a normed vector space (V, k · k).

(1) (V, k · k) is complete.

(2) Every decreasing sequence of closed balls has a nonempty intersection.

Note that the balls need not be concentric.

(3) Everydecreasing sequence of closed balls with radii going to zero has nonempty

intersection.

HINT: (1) =⇒ (2) show that the centres of the balls form a Cauchy sequence.

K. (a) Show that if A is a closed subset of a normed vector space V and C is a compact

subset, then A + C = {a + c : a ∈ A, c ∈ C} is closed.

(b) Is it enough for C to also be closed, or is compactness necessary?

7.3 Inner Product Spaces 187

L. Let X

= {f ∈ C[0, 1] : f(0) = 0, kfk

∞

≤ 1, and f(x) ≥

for x ≥

(a) Show that X

is a closed bounded subset of C[0, 1].

(b) Show that X

n+1

is a proper subset of X

for n ≥ 1, and compute

n≥1

the theorem fail in this context?

M. Let c

be the vector space of all sequences x = (x

)

∞

n=1

such that lim

n→∞

= 0.

Deﬁne a norm on c

by kxk

∞

= sup

n≥1

|. Prove that c

is complete.

HINT: Let x

= (x

k,n

)

∞

n=1

be a Cauchy sequence in c

(a) Show that (x

k,n

)

∞

k=1

is Cauchy for each n ≥ 1. Hence deﬁne x = (x

) by

= lim

k→∞

k,n

(b) Show that kx

∞

is Cauchy and kxk ≤ lim

k→∞

∞

− x

k,n

| ≤ ε for all n ≥ 1 and all k ≥ K.

(d) Conclude that x belongs to c

and that lim

k→∞

= x.

N. Consider the sequence f

in C[−1, 1] from Exercise 7.2.I, but use the L

[−1, 1] norm.

(a) Show that f

is Cauchy in the L

norm.

(b) Show that f

converges to

(0,1]

, the characteristic function of (0, 1], in the L

norm.

(0,1]

− hk

> 0 for every h in C[−1, 1].

(d) Conclude that C[−1, 1] is not complete in the L

norm.

7.3. Inner Product Spaces

In studying R

, we constructed the Euclidean norm using the dot product. An

inner product on a vector space is a generalization of the dot product. It is one of

the most important sources of norms, and the norms obtained from inner products

are particularly tractable. For example, the L

norm on C[a, b] arises this way.

7.3.1. DEFINITION. An inner product on a vector space V is a function hx, yi

on pairs (x, y) of vectors in V × V taking values in R satisfying the following

properties:

(1) (positive deﬁniteness) hx, xi ≥ 0 for all x ∈ V and

hx, xi = 0 only if x = 0.

(2) (symmetry) hx, yi = hy, xi for all x, y ∈ V .

(3) (bilinearity) For all x, y, z ∈ V and scalars α, β ∈ R,

hαx + βy, zi = αhx, zi + βhy, zi.

Given an inner product space, it is easy to check that the following deﬁnition gives

us a norm:

kxk = hx, xi

1/2

188 Normed Vector Spaces

The dot product on R

is an inner product and the norm obtained from it is the

usual Euclidean norm.

The bilinearity condition just given is linearity in the ﬁrst variable. But as

the term suggests, it really means a twofold linearity because combining it with

symmetry yields linearity in the second variable as well. For x, y in V and scalars

α, β in R,

hz, αx + βyi = hαx + βy, zi = αhx, zi + βhy, zi = αhz, xi + βhz, yi.

7.3.2. EXAMPLE. The space C[a, b] can be given an inner product

hf, gi =

f(x)g(x) dx.

This gives rise to the L

norm, which we deﬁned in Example 7.1.5. Positive def-

initeness of this norm was established in Example 7.1.5 for arbitrary p, including

p = 2. The other two properties follow from the linearity of the integral and are

left as an exercise for the reader.

7.3.3. EXAMPLE. The space R

can be given other inner product structures by

weighting the vectors by a matrix A =

hx, yi

= hAx, yi =

i=1

j=1

This is easily seen to be bilinear because A is linear and the standard inner product

is bilinear. To be symmetric, we must require that a

= a

, so A is a symmetric

matrix. Moreover, we need an additional condition to ensure that the inner product

is positive deﬁnite. It turns out that the necessary condition is that the eigenvalues

of A are all strictly positive. The proof of this is an important result from linear

algebra known as the Spectral Theorem for Symmetric Matrices or as the Principal

Axis Theorem.

For the purposes of this example, consider the 2 × 2 matrix A =

3 1

1 2

. We

noted that since A is symmetric, the inner product h·, ·i

is symmetric and bilinear.

Let us establish directly that it is positive deﬁnite. Take a vector x = (x, y).

hx, xi

= 3x

+ xy + yx + 2y

= 2x

+ (x + y)

+ y

From this identity, it is clear that hx, xi

≥ 0. Moreover, equality requires that x,

y and x + y all be 0, whence x = 0. So it is positive deﬁnite.

It is a fundamental fact that every inner product space satisﬁes the Schwarz

inequality. Our proof for R

was special, using the speciﬁc formula for the dot

product. We now show that it follows just from the basic properties of an inner

product.

7.3 Inner Product Spaces 189

7.3.4. CAUCHY–SCHWARZ INEQUALITY.

For all vectors x, y in an inner product space V ,

|hx, yi| ≤ kxkkyk.

Equality holds if and only if x and y are collinear.

PROOF. If either x or y is 0, both sides of the inequality are 0. Equality holds here,

and these vectors are collinear. So we may assume that x and y are nonzero.

Apply the positive deﬁnite property to the vector x − ty for t ∈ R.

0 ≤ hx − ty, x − tyi

= hx, x − tyi − thy, x − tyi

= hx, xi − thx, yi − thx, yi + t

hy, yi

= kxk

− 2thx, yi + t

kyk

Substitute t = hx, yi/kyk

to obtain

0 ≤ kxk

−

hx, yi

kyk

Hence

hx, yi

≤ kxk

kyk

This establishes the inequality.

For equality to hold, the vector x − ty must have norm 0. By the positive

deﬁnite property, this means that x = ty and so they are collinear. Conversely, if

x = ty, then kxk

= hty, tyi = t

kyk

and

|hx, yi| = |t|hy, yi =

hy, yi

hy, yi = kxkkyk.

7.3.5. COROLLARY. For f, g ∈ C[a, b], we have

f(x)g(x) dx

≤

f(x)

1/2

g(x)

1/2

As for R

, the triangle inequality in an immediate consequence. In particular,

the L

norms on C[a, b] are indeed norms.

7.3.6. COROLLARY. An inner product space V satisﬁes the triangle inequality

kx + yk ≤ kxk + kyk for all x, y ∈ V.

Moreover, if equality occurs, then x and y are collinear.

190 Normed Vector Spaces

PROOF. This proof is identical to the R

case. The key is the use of Cauchy–

Schwarz inequality.

kx + yk

= hx + y, x + yi

= hx, xi + 2hx, yi + hy, yi

≤ kxk

+ 2kxkkyk + kyk

kxk + kyk

This establishes the inequality.

Moreover, equality only occurs if hx, yi = kxkkyk, which by the Cauchy–

Schwarz inequality can only happen when x and y are collinear. ¥

Another fundamental consequence of the Cauchy–Schwarz inequality is that

the inner product is continuous with respect to the induced norm. We leave the

proof as an exercise.

7.3.7. COROLLARY. Let V be an inner product space with induced norm k · k.

Then the inner product is continuous (i.e., if x

converges to x and y

converges

to y, then hx

, y

i converges to hx, yi).

Exercises for Section 7.3

A. Let A =





3 1 2

1 2 1

2 1 4





. Show that the form h·, ·i

is positive deﬁnite on R

B. Minimize the quantity kxk

− 2thx, yi + t

kyk

over t ∈ R. You will see why we

chose t as we did in the proof of the Cauchy–Schwarz inequality.

C. Show that it is possible for x and y to be collinear yet the triangle inequality is still a

strict inequality.

D. Prove Corollary 7.3.7.

E. Let w(x) be a strictly positive continuous function on [a, b]. Deﬁne a form on C[a, b]

by the formula hf, gi

f(x)g(x)w(x) dx for f, g ∈ C[a, b]. Show that this is

an inner product.

F. A normed vector space V is strictly convex if kuk = kvk = k(u + v)/2k = 1 for

vectors u, v ∈ V implies that u = v.

(a) Show that an inner product space is always strictly convex.

(b) Show that R

with the norm k(x, y)k

∞

= max{|x|, |y|} is not strictly convex.

G. Let T be an n×n matrix. Deﬁne a form on R

by hx, yi

= hT x, T yi for x, y ∈ R

Show that this is an inner product if and only if T is invertible.

H. Let T be an invertible n × n matrix. Prove that a sequence of vectors x

in R

converges to a vector x in the usual Euclidean norm if and only if it converges to x in

the norm of Exercise G, namely kxk

:= kT xk.

HINT: Using Exercise G, show

kxk

−1

≤ kT xk ≤ kT k

kxk; see Exercises I and J.

7.4 Orthonormal Sets 191

I. Show that there is an inner product on the space M

of all n × n matrices given by

hA, Bi = Tr(AB

) =

i=1

j=1

where Tr(A) =

i=1

denotes the trace. The

norm kAk

= hA, Ai

1/2

is called the Hilbert–Schmidt norm.

J. Let A =

be an n × n matrix, and let x = (x

, . . . , x

) be a vector in R

. Show

that kAxk ≤ kAk

kxk.

HINT: Compute kAxk

using coordinates, and apply the Cauchy–Schwarz inequality.

K. Given an inner product space V , deﬁne a function on V \ {0} by Rx = x/kxk

. This

R is called

inversion

with respect to the unit sphere {x ∈ V : kxk = 1}.

(a) Prove that kRx − Ryk =

kx − yk

kxkkyk

(b) Hence show that the inversion R is continuous.

kw − ykkx − zk ≤ kw − xkky − zk + kw − zkkx − yk.

HINT: Reduce to the case w = 0, and reinterpret the inequality using inversion.

7.4. Orthonormal Sets

7.4.1. DEFINITION. Two vectors x and y are called orthogonal if hx, yi = 0.

A collection of vectors {e

: n ∈ S} in V is called orthonormal if ke

k = 1 for

all n ∈ S and he

, e

i = 0 for n 6= m ∈ S. This set is called an orthonormal

basis if in addition this set is maximal with respect to being an orthonormal set.

The space R

has many orthonormal bases, including the canonical one given

by e

= (1, 0, . . . , 0), . . . , e

= (0, . . . , 0, 1).

7.4.2. PROPOSITION. An orthonormal set is linearly independent.

An orthonormal basis for a ﬁnite-dimensional inner product space is a basis.

PROOF. Suppose that {e

: n ∈ S} is orthonormal and that there is a relation

n∈S

= 0 with only ﬁnitely many α

6= 0. Then

0 = h0, e

i =

n∈S

hα

, e

i = α

So α

= 0 for all k ∈ S and this set is linearly independent.

If V is a ﬁnite-dimensional inner product space and {e

: n ∈ S} is an or-

thonormal basis, then by the ﬁrst paragraph it follows that {e

: n ∈ S} is linearly

independent. Let V

= span{e

: n ∈ S}. If V

= V , then {e

: n ∈ S} is a

spanning linearly independent set, and hence a basis.

192 Normed Vector Spaces

On the other hand, if V

is a proper subspace, then there is a vector v ∈ V that

is not in V

. Deﬁne

w = v −

n∈S

hv, e

Observe that w 6= 0 because w = 0 would mean that v belonged to the span of

: n ∈ S}. Moreover,

hw, e

i = hv, e

i −

n∈S

hv, e

ihe

, e

i = hv, e

i − hv, e

i = 0.

Thus w is orthogonal to {e

: n ∈ S}. Therefore, {e

: n ∈ S} ∪ {w/kwk} is a

larger orthonormal set. This contradiction establishes that V

= V as desired. ¥

7.4.3 THE GRAM–SCHMIDT PROCESS. The idea used in the proof of the

proposition can be turned into an algorithm for producing orthonormal sets with

nice properties. It provides an orthonormal set with the same span as some initial

set of vectors {x

, . . . , x

}. Set y

= x

and inductively deﬁne

k+1

= x

k+1

−

i=1

k+1

, y

for 1 ≤ k < n,

where we interpret y

/ky

to be 0 if y

= 0. Now delete those terms y

that are

zero, and for the rest deﬁne f

= y

/ky

k. This yields a set with three properties:

(1) span{f

, . . . , f

} = span{x

, . . . , x

} for 1 ≤ k ≤ n,

(2) y

= 0 (and f

is not deﬁned) if and only if x

∈ span{x

: i < k}, and

(3) {f

, . . . , f

} is an orthonormal basis for span{x

, . . . , x

The proof of these facts is left to the exercises. By applying this algorithm

to any basis for a ﬁnite-dimensional inner product space, it follows that an inner

product space of dimension n always has an orthonormal basis consisting of n

vectors.

Let us look at a fundamental example that underlies Fourier series, a subject to

which we devote two chapters later on.

7.4.4. EXAMPLE. We use the natural inner product on C[−π, π] given by

hf, gi =

2π

−π

f(θ)g(θ) dθ.

The factor

2π

is put in to make the constant function 1 have unit length. The norm

is given by

kfk

:= hf, f i

1/2

2π

−π

|f(θ)|

dθ

1/2

This is called the L

norm.

Davidson K.R., Donsig A.P. Real Analysis with Real Applications

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