A Short Introduction to Metric, Banach, and Hilbert Spaces

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Christopher Heil

A Short Introduction to Metric, Banach, and Hilbert Spaces

November 21, 2014

Springer

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Chapter 1 Metric Spaces

We will introduce and quickly study the basics of metrics, norms, and inner products in this short manuscript. This manuscript is essentially self- contained, although the presentation is somewhat compressed. Most major proofs are included, while certain other proofs are assigned as problems, and references are provided for proofs that are omitted. Explicit proofs of most of the statements made in this appendix, with extensive discussion and mo- tivation, will be given in the forthcoming volume [Heil15], but many of these can also be found in the volume [Heil11], which has already appeared.

We assume that the reader is familiar with vector spaces (which are also called linear spaces). The scalar field associated with the vector spaces in this manuscript will always be either the real line R or the complex plane C.

When we specifically deal with the vector space R^d, we assume that the scalar field is R, and likewise we always assume that the scalar field for C^d is C.

For simplicity of presentation, when dealing with a vector space we often implicitly assume that the scalar field is C, but usually only minor (and obvious) changes are required if the scalar field is R. The elements of the scalar field are often called scalars, so for us a scalar will always mean a real or complex number.

1.1 Metrics and Convergence

A metric provides us with a notion of the distance between points in a set.

Definition 1.1 (Metric Space). Let X be a nonempty set. A metric on X is a function d : X × X → R such that for all x, y, z ∈ X we have:

(a) Nonnegativity: 0 ≤ d(x, y) < ∞,

(b) Uniqueness: d(x, y) = 0 if and only if x = y, (c) Symmetry: d(x, y) = d(y, x), and

c

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(d) The Triangle Inequality: d(x, z) ≤ d(x, y) + d(y, z).

If these conditions are satisfied, then X is a called a metric space. The number d(x, y) is called the distance from x to y. ♦

A metric space X does not have to be a vector space, although most of the metric spaces that we will encounter in this manuscript will be vector spaces (indeed, most are actually normed spaces). If X is a generic metric space, then we often refer to the elements of X as “points,” but if we know that X is a vector space, then we usually refer to the elements of X as

“vectors.” We will mostly use letters such as x, y, z to denote elements of a metric or normed space. However, in many cases we know that our space is a space of functions. In such a context we usually use letters such as f, g, h to denote elements of the space, and we may refer to those elements as either

“functions” or “vectors.”

Here is an example of a metric on an infinite-dimensional vector space.

Example 1.2. Given a sequence of real or complex scalars x = (xk)k∈N = (x1, x2, . . . ), we define the ℓ¹-norm of x to be

kxk1 = k(xk)k∈Nk1 =

∞

X

k=1

|xk|. (1.1)

We say that the sequence x is absolutely summable if kxk1< ∞, and we let ℓ¹ denote the space of all absolutely summable sequences, i.e.,

ℓ¹ =

½

x = (xk)k∈N : kxk1 =

∞

X

k=1

|xk| < ∞

¾ .

This set is a vector space; in particular, it is closed under the operations of addition of sequences and multiplication of a sequence by a scalar. Further, a straightforward calculation shows that the function d defined by

d(x, y) = kx − yk1 =

∞

X

k=1

|xk− yk|, x, y ∈ ℓ¹, (1.2)

is a metric on ℓ¹, so ℓ¹is both a vector space and a metric space. ♦ The following particular elements of ℓ¹ appear so often that we introduce a name for them.

Notation 1.3 (The Standard Basis Vectors). Given an integer n ∈ N, we let δn denote the sequence

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1.2 Convergence and Completeness 3

δn = (δnk)k∈N = (0, . . . , 0, 1, 0, 0, . . . ).

That is, the nth component of the sequence δnis 1, while all other components are zero. We call δnthe nth standard basis vector, and we refer to the family {δn}n∈N as the sequence of standard basis vectors, or simply the standard basis. ♦

1.2 Convergence and Completeness

Since a metric space has a notion of distance, we can define a corresponding notion of convergence in the space, as follows.

Definition 1.4 (Convergent Sequence). Let X be a metric space. A sequence of points {xn}n∈N in X converges to the point x ∈ X if

n→∞lim d(xn, x) = 0.

That is, for every ε > 0 there must exist some integer N > 0 such that n ≥ N =⇒ d(xn, x) < ε.

In this case, we write xn→ x. ♦

Convergence implicitly depends on the choice of metric for X, so if we want to emphasize that we are using a particular metric, we may write xn → x with respect to the metric d.

Closely related to convergence is the idea of a Cauchy sequence, which is defined as follows.

Definition 1.5 (Cauchy Sequence). Let X be a metric space. A sequence of points {xn}n∈N in X is a Cauchy sequence if for every ε > 0 there exists an integer N > 0 such that

m, n ≥ N =⇒ d(xm, xn) < ε. ♦

By applying the Triangle Inequality, we immediately obtain the following relation between convergent and Cauchy sequences.

Lemma 1.6 (Convergent Implies Cauchy). If {xn}n∈N is a convergent sequence in a metric space X, then {xn}n∈N is a Cauchy sequence in X.

Proof. Assume that xn → x as n → ∞. Given ε > 0, there exists an N > 0 such that d(x, xn) < ε for all n > N. Consequently, if m, n > N then the Triangle Inequality implies that

d(xm, xn) ≤ d(xm, x) + d(x, xn) < 2ε.

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Therefore {xn}n∈Nis Cauchy. ⊓⊔

However, the converse of Lemma 1.6 does not hold in general.

Example 1.7. The “standard metric” on the real line R or the complex plane Cis

d(x, y) = |x − y|.

Each of R and C is a complete metric space with respect to this metric. That is, any Cauchy sequence of real scalars must converge to a real scalar, and any Cauchy sequence of complex scalars must converge to a complex scalar (for a proof, see [Rud76, Thm. 3.11]).

Now consider the set of rational numbers Q. This is also a metric space with respect to the metric d(x, y) = |x − y|. However, Q is not complete—

there exist Cauchy sequences of rational numbers that do not converge to a rational number. For example, let xnbe the following rational numbers based on the decimal expansion of π:

x1= 3, x2= 3.1, x3= 3.14, . . . .

Then {xn}n∈N is a Cauchy sequence in both Q and R. Moreover, it is a convergent sequence in R because xn → π ∈ R. However, {xn}n∈N is not a convergent sequence in Q, because there is no rational point r ∈ Q such that xn→ r. ♦

The metric space Q in Example 1.7 is not a vector space over the real field. Here is an example of an infinite-dimensional vector space that is not complete.

Example 1.8. Let c00be the space of all sequences of real or complex numbers that have only finitely many nonzero components:

c00 = n

x = (x1, . . . , xN, 0, 0, . . . ) : N > 0, x1, . . . , xN ∈ Co . Since c00 is a subset of ℓ¹, it is a metric space with respect to the metric d defined in equation (1.2).

For each n ∈ N, let xn be the sequence

xn = (2⁻¹, . . . , 2⁻ⁿ, 0, 0, 0, . . . ).

and consider the sequence of vectors {xn}n∈N, which is contained in both c00

and ℓ¹. If m < n, then

kxn− xmk1 =

n

X

k=m+1

2^−k < 2^−m,

and it follows from this that {xn}n∈N is a Cauchy sequence. If we consider this sequence to be a sequence in the space ℓ¹, then it does converge; in fact

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1.2 Convergence and Completeness 5

xn→ x where

x = (2⁻¹, 2⁻², . . . ) = (2^−k)k∈N.

However, this vector x does not belong to c00, and there is no sequence y ∈ c00

such that xn→ y. Therefore c00 is not a complete metric space. ⊓⊔

The set of rationals, Q, is not a vector space over the real field or the complex field (which are the only fields we are considering in this manuscript).

In contrast, c00is a vector space with respect to the real field (if we are using real scalars), or with respect to the complex field (if we are using complex scalars). In fact, c00is the finite linear span of the set of standard basis vectors E = {δk}k∈N:

c00 = span(E) =

½ ^N X

k=1

xkδk : N > 0, x1, . . . , xN ∈ C

¾ .

However, at least when we use the metric d defined by equation (1.2), there are Cauchy sequences in c00 that do not converge to an element of c00. This fact should be compared to the next theorem, which shows that every Cauchy sequence in ℓ¹converges to an element of ℓ¹.

Theorem 1.9. If {xn}n∈N is a Cauchy sequence in ℓ¹, then there exists a vector x ∈ ℓ¹ such that xn → x.

Proof. Assume that {xn}n∈N is a Cauchy sequence in ℓ¹. Each xnis a vector in ℓ¹, and for this proof we will write the components of xn as

xn = ¡xn(1), xn(2), . . .¢

= ¡xn(k)¢

k∈N.

By the definition of a Cauchy sequence, if we fix ε > 0, then there is an integer N > 0 such that d(xm, xn) = kxm− xnk1 < ε for all m, n > N.

Therefore, if we fix a particular index k ∈ N then for all m, n > N we have

|xm(k) − xn(k)| ≤

∞

X

j=1

|xm(j) − xn(j)| = kxm− xnk1 < ε.

Thus, with k fixed,¡xn(k)¢

n∈Nis a Cauchy sequence of scalars and therefore must converge. Define

x(k) = lim

n→∞xn(k), (1.3)

and set x =¡x(1), x(2), . . . ¢. For each fixed k, we have by construction that the kth component of xn converges to the kth component of x as n → ∞.

We therefore say that xnconverges componentwise to x. However, this is not enough. We need to show that x ∈ ℓ¹, and that xn converges to x in ℓ¹-norm.

To do this, again fix an ε > 0. Since {xn}n∈Nis Cauchy, there is an N > 0 such that kxm− xnk1 < ε for all m, n > N. Choose any particular n > N, and fix an integer M > 0. Then, since M is finite,

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M

X

k=1

|x(k) − xn(k)| =

M

X

k=1

m→∞lim |xm(k) − xn(k)|

= lim

m→∞

M

X

k=1

|xm(k) − xn(k)|

≤ lim

m→∞kxm− xnk1

≤ ε.

As this is true for every M, we conclude that

d(xn, x) = kx − xnk1 =

∞

X

k=1

|x(k) − xn(k)|

= lim

M →∞

M

X

k=1

|x(k) − xn(k)|

≤ ε. (1.4)

Even though we do not know yet that x ∈ ℓ¹, this tells us that the vector x − xn has finite ℓ¹-norm and therefore belongs to ℓ¹. Since ℓ¹ is closed under addition, it follows that x = (x − xn) + xn ∈ ℓ¹. Therefore our “candi- date sequence” x does belong to ℓ¹. Further, equation (1.4) establishes that d(xn, x) ≤ ε for all n > N, so we have shown that xn does indeed converge to x as n → ∞. Hence ℓ¹ is complete. ⊓⊔

In summary, some metric spaces have the property that every Cauchy sequence in the space does converge to an element of the space. Since we can test for Cauchyness without having the limit vector x in hand, this is often very useful. We give such spaces the following name.

Definition 1.10 (Complete Metric Space). Let X be a metric space. If every Cauchy sequence in X converges to an element of X, then we say that X is complete. ♦

Theorem 1.9 shows that the space ℓ¹ is a complete metric space. More precisely, it is complete with respect to the metric d(x, y) = kx − yk1; if we impose a different metric on ℓ¹, then it might not be the case that ℓ¹ is complete with respect to that new metric. In particular, Problem 3.30 defines another metric on ℓ¹ and shows that ℓ¹is not complete with respect to that metric.

Remark 1.11. The reader should be aware that the term “complete” is heavily overused and has a number of distinct mathematical meanings. In particular, the notion of a complete space as given in Definition 1.10 is quite different from the notion of a complete sequence that will be introduced in Definition 2.9. ♦

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1.3 Topology in Metric Spaces 7

1.3 Topology in Metric Spaces

Given a metric space X, we make the following definitions.

• Given x ∈ X and r > 0, the open ball in X centered at x with radius r is the set

Br(x) = ©y ∈ X : d(x, y) < rª.

• A set U ⊆ X is open if for each x ∈ U there exists an r > 0 such that Br(x) ⊆ U. Equivalently, U is open if and only if U can be written as a union of open balls.

• The topology of X is the collection of all open subsets of X.

• The interior of a set E ⊆ X is the largest open set E^◦ that is contained in E. Explicitly, E^◦=^S©U : U is open and U ⊆ Eª.

• A set E ⊆ X is closed if its complement X\E is open.

• A set E ⊆ X is bounded if it is contained in some open ball, i.e., there exists some x ∈ X and r > 0 such that E ⊆ Br(x).

• A point x ∈ X is an accumulation point of a set E ⊆ X if there exist xn∈ E with all xn 6= x such that xn → x.

• A point x ∈ X is a boundary point of a set E ⊆ X if for every r > 0 we have both Br(x) ∩ E 6= ∅ and Br(x) ∩ E^C6= ∅. The set of all boundary points of E is called the boundary of E, and it is denoted by ∂E.

• The closure of a set E ⊆ X is the smallest closed set E that contains E.

Explicitly, E =^T©F : F is closed and E ⊆ F ª.

• A set E ⊆ X is dense in X if E = X.

• We say that X is separable if there exists a countable subset E that is dense in X.

The following lemma gives useful equivalent reformulations of some of the notions defined above.

Lemma 1.12. If E is a subset of a metric space X, then the following statements hold.

(a) E is closed if and only if

xn∈ E and xn→ x ∈ X =⇒ x ∈ E.

(b) The closure of E satisfies

E = E ^S{y ∈ X : y is an accumulation point of E}

= ©y ∈ X : there exist xn∈ E such that xn→ yª. (1.5)

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(c) E is dense in X if and only if for every point x ∈ X there exist xn∈ E such that xn→ x. ♦

Proof. (a) This statement can be deduced from part (b), or it can be proved directly. We assign the proof of statement (a) as Problem 1.31.

(b) Let F be the union of E and the accumulation points of E, i.e., F = E^S{y ∈ X : y is an accumulation point of E}, and let G be the set of all limits of points of E, i.e.,

G = ©y ∈ X : there exist x_n ∈ E such that xn → yª.

We must prove that F = E = G.

To show that E ⊆ F, fix any point x ∈ F^C = X \F. Then x /∈ E and x is not an accumulation point of E. If every open ball Br(x) contained a point y ∈ E, then x would be an accumulation point of E (because y cannot equal x). Therefore there must exist some ball Br(x) that contains no points of E. We claim Br(x) contains no accumulation points of E either. To see this, suppose that y ∈ Br(x) was an accumulation point of E. Then there would exist points xn ∈ E such that xn → y. By taking n large enough it follows that xn ∈ Br(y), which is a contradiction. Thus Br(x) contains no points of E and no accumulation points of E, so Br(x) ⊆ F^C. This shows that F^Cis open, and therefore F is closed. Since E ⊆ F and E is the smallest closed set that contains E, we conclude that E ⊆ F.

Next, we will prove that F ⊆ E by showing that E^C⊆ F^C. Since E^C is open, if x belongs to this set then there exists some r > 0 such that

Br(x) ⊆ E^C.

Hence Br(x) contains no points of E, so x cannot be an accumulation point of E. We also have x /∈ E since x /∈ E, so it follows that x /∈ F.

The above work shows that E = F. Now choose any point y ∈ G. If y ∈ E, then y ∈ F since E ⊆ F. So, suppose that y ∈ G\E. Then, by definition of G, there exist points xn∈ E such that xn→ y. But xn6= y for any n since xn ∈ E and y /∈ G, so y is an accumulation point of E. Therefore y ∈ F.

Thus G ⊆ F.

Finally, suppose that y ∈ F. If y ∈ E, then y is the limit of the sequence {y, y, y, . . . }, so y ∈ G. On the other hand, if y ∈ F \E, then y must be an accumulation point of E. Therefore there exist points xn ∈ E, with xn 6= y for every n, such that xn→ y. This shows that y is a limit of points from E, so y ∈ G. Therefore F ⊆ G.

(c) This follows from equation (1.5) and the fact that E is dense in X if and only if X = E. ⊓⊔

Restating parts of Lemma 1.12, we see that:

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1.4 Compact Sets in Metric Spaces 9

• a set E is closed if and only if it contains every limit of points from E,

• the closure of a set E is the set of all limits of points from E, and

• a set E is dense in X if and only if every point in X is a limit of points from E.

For example, the set of rationals Q is not a closed subset of R because a limit of rational points need not be rational. The closure of Q is R because every point in R can be written as a limit of rational points. Similarly, Q is a dense subset of R because every real number can be written as a limit of rational numbers. Since Q is both countable and dense, this also shows that R is a separable space. Problem 2.26 shows that the space ℓ¹ defined in Example 1.2 is separable.

1.4 Compact Sets in Metric Spaces

Now we define compact sets and discuss their properties.

Definition 1.13 (Compact Set). A subset K of a metric space X is compact if every covering of K by open sets has a finite subcovering. Stated precisely, K is compact if it is the case that whenever

K ⊆ S

i∈I

Ui,

where {Ui}i∈I is any collection of open subsets of X, there exist finitely many i1, . . . , iN ∈ I such that

K ⊆

N

S

k=1

Uik. ♦

First we show that every compact subset of a metric space is both closed and bounded.

Lemma 1.14. If K is a compact subset of a metric space X, then K is closed and there exists some open ball Br(x) such that K ⊆ Br(x).

Proof. Suppose that K is compact, and fix any particular point x ∈ X.

Then the union of the open balls Bn(x) over n ∈ N is all of X (why?), so {Bn(x)}n∈N an open cover of K. This cover must have a a finite subcover, and since the balls are nested it follows that K ⊆ Bn(x) for some single n.

It remains to show that K is closed. If K = X then we are done, so assume that K 6= X. Fix any point y in K^C= X\K. Then given any x ∈ K we have x 6= y, so by the Hausdorff property stated in Problem 1.28, there exist disjoint open sets Ux, Vxsuch that x ∈ Uxand y ∈ Vx. The collection {Ux}x∈K is an open cover of K, so it must contain some finite subcover, say

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K ⊆ Ux1 ∪ · · · ∪ UxN. (1.6) Each Vxj is disjoint from Uxj, so it follows from equation (1.6) that the set

V = Vx1 ∩ · · · ∩ VxN

is entirely contained in the complement of K. Thus, V is an open set that satisfies

y ∈ V ⊆ K^C.

This shows that K^C is open, so we conclude that K is closed. ⊓⊔

If X is a finite-dimensional normed space, then a subset of X is compact if and only if it closed and bounded (for a proof, see [Kre78, Thm. 2.5-3]).

However, every infinite-dimensional normed space contains a set that is closed and bounded but not compact. In fact, if X is an infinite-dimensional normed space then the closed unit ball in X is closed and bounded but not compact.

Problem 2.25 asks for a proof of this fact for the space ℓ¹; for a proof in the general setting see [Heil11, Exercise 1.44].

We will give several equivalent reformulations of compactness for subsets of metric spaces in terms of the following concepts.

Definition 1.15. Let E be a subset of a metric space X.

(a) E is sequentially compact if every sequence {xn}n∈N of points from E contains a convergent subsequence {xnk}k∈N whose limit belongs to E.

(b) E is totally bounded if given any r > 0 we can cover E by finitely many open balls of radius r. That is, for each r > 0 there must exist finitely many points x1, . . . , xN ∈ X such that

E ⊆

N

S

k=1

Br(xk). ♦ We will need the following lemma.

Lemma 1.16. Let E be a sequentially compact subset of a metric space X.

If {Ui}i∈I is an open cover of E, then there exists a number δ > 0 such that if B is an open ball of radius δ that intersects E, then there is an i ∈ I such that B ⊆ Ui.

Proof. Let {Ui}i∈I be an open cover of E. We want to prove that there is a δ > 0 such that

Br(x) ∩ E 6= ∅ =⇒ Br(x) ⊆ Ui for some i ∈ I.

Suppose that no δ > 0 has this property. Then for each positive integer n, there must exist some open ball with radius ¹_n that intersects E but is not contained in any set Ui. Call this open ball Gn. For each n, choose a point

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1.4 Compact Sets in Metric Spaces 11

xn ∈ Gn∩ E. Then since {xn}n∈N is a sequence in E and E is sequentially compact, there must be a subsequence {xnk}k∈N that converges to a point x ∈ E. Since {Ui}i∈I is a cover of E, we must have x ∈ Ui for some i ∈ I, and since Ui is open there must exist some r > 0 such that Br(x) ⊆ Ui. Now choose k large enough that we have both

1 nk

< r

3 and d¡x, xnk¢ < r 3.

Keeping in mind the facts that Gnk contains xnk, Gnk is an open ball with radius 1/nk, the distance from x to xnk is less than r/3, and Br(x) has radius r, it follows that Gnk ⊆ Br(x) ⊆ Ui, which is a contradiction. ⊓⊔

Now we prove some reformulations of compactness that hold for subsets of metric spaces.

Theorem 1.17. If K is a subset of a complete metric space X, then the following statements are equivalent.

(a) K is compact.

(b) K is sequentially compact.

(c) K is totally bounded and every Cauchy sequence of points from K converges to a point in K.

Proof. (a) ⇒ (b). Suppose that E is not sequentially compact. Then there exists a sequence {xn}n∈N in E that has no subsequence that converges to an element of E. Choose any point x ∈ E. If every open ball centered at x contains infinitely many of the points xn, then by considering radii r = _k¹ we can construct a subsequence {xnk}k∈N that converges to x. This is a contradiction, so there must exist some open ball centered at x that contains only finitely many xn. If we call this ball Bx, then {Bx}x∈Eis an open cover of E that contains no finite subcover. Consequently E is not compact.

(b) ⇒ (c). Suppose that E is sequentially compact, and suppose that {xn}n∈N is a Cauchy sequence in E. Then since E is sequentially compact, there is a subsequence {xn_k}k∈N that converges to some point x ∈ E. Hence {xn}n∈Nis Cauchy and has a convergent subsequence. Appealing to Problem 1.29, this implies that xn→ x. Therefore E is complete.

Suppose that E is not totally bounded. Then there is a radius r > 0 such that E cannot be covered by finitely many open balls of radius r centered at points of X. Choose any point x1 ∈ E. Since E cannot be covered by a single r-ball, E cannot be a subset of Br(x1). Hence there exists a point x2∈ E \Br(x1). In particular, d(x2, x1) ≥ r. But E cannot be covered by two r-balls, so there must exist a point x3that belongs to E \¡Br(x1) ∪ Br(x2)¢.

In particular, we have both d(x3, x1) ≥ r and d(x3, x2) ≥ r. Continuing in this way we obtain a sequence of points {xn}n∈Nin E that has no convergent subsequence, which is a contradiction.

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(c) ⇒ (b). Assume that E is complete and totally bounded, and let {xn}n∈N be any sequence of points in E. Since E can be covered by finitely many open balls of radius ¹₂, at least one of those balls must contain infinitely many of the points xn. That is, there is some infinite subsequence {x⁽¹⁾n }n∈N

of {xn}n∈N that is contained in an open ball of radius ¹₂. The Triangle In- equality therefore implies that

∀ m, n ∈ N, d¡x⁽¹⁾_m, x⁽¹⁾_n ¢

< 1.

Similarly, since E can be covered by finitely many open balls of radius ¹₄, there is some subsequence {x⁽²⁾n }n∈N of {x⁽¹⁾n }n∈N such that

∀ m, n ∈ N, d¡x⁽²⁾_m, x⁽²⁾_n ¢

< 1 2.

Continuing by induction, for each k > 1 we find a subsequence {x^(k)n }n∈N of {x^(k−1)n }n∈Nsuch that d¡x^(k)m , x^(k)n ¢ <_k¹ for all m, n ∈ N.

Now consider the diagonal subsequence {x^(k)_k }k∈N. Given ε > 0, let N be large enough that _N¹ < ε. If j ≥ k > N, then x^(j)_j is one element of the sequence {x^(k)n }n∈N, say x^(j)_j = x^(k)n . Hence

d¡x^(j)_j , x^(k)_k ¢

= d¡x^(k)_n , x^(k)_k ¢

< 1 k < 1

N < ε.

Thus {x^(k)_k }k∈N is a Cauchy subsequence of the original sequence {xn}n∈N. Since E is complete, this subsequence must converge to some element of E.

Hence E is sequentially compact.

(b) ⇒ (a). Assume that E is sequentially compact. Since we have already proved that statement (b) implies statement (c), we know that E is complete and totally bounded.

Suppose that {Ui}i∈Jis any open cover of E. By Lemma 1.16, there exists a δ > 0 such that if B is an open ball of radius δ that intersects E, then there is an i ∈ J such that B ⊆ Ui. However, E is totally bounded, so we can cover E by finitely many open balls of radius δ. Each of these balls is contained in some Ui, so E is covered by finitely many Ui. ⊓⊔

1.5 Continuity for Functions on Metric Spaces

Here is the abstract definition of continuity for functions on metric spaces.

Definition 1.18 (Continuous Function). Let X and Y be metric spaces.

We say that a function f : X → Y is continuous if given any open set V ⊆ Y, its inverse image f⁻¹(V ) is an open subset of X. ♦

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1.5 Continuity for Functions on Metric Spaces 13

By applying the definition of continuity, the following lemma shows that the direct image of a compact set under a continuous function is compact.

Lemma 1.19. Let X and Y be metric spaces. If f : X → Y is continuous and K is a compact subset of X, then f (K) is a compact subset of Y.

Proof. Let {Vi}i∈Jbe any open cover of f (K). Each set Ui= f⁻¹(Vi) is open, and {Ui}i∈J is an open cover of K. Since K is compact, this cover must have a finite subcover {Ui1, . . . , UiN}. But then {Vi1, . . . , ViN} is a finite subcover of f (K), so we conclude that f (K) is compact. ⊓⊔

The next lemma gives a useful reformulation of continuity for functions on metric spaces in terms of preservation of limits.

Lemma 1.20. Let X be a metric space with metric dX, and let Y be a metric space with metric dY. If f : X → Y, then the following three statements are equivalent.

(a) f is continuous.

(b) If x is any point in X, then for every ε > 0 there exists a δ > 0 such that dX(x, y) < δ =⇒ dY¡f(x), f(y)¢ < ε.

(c) Given any point x ∈ X and any sequence {xn}n∈N in X, xn→ x in X =⇒ f (xn) → f (x) in Y. ♦

Proof. For this proof we let B_r^X(x) and B_s^Y(y) denote open balls in X and Y, respectively.

(a) ⇒ (b). Suppose that f is continuous, and choose any point x ∈ X and any ε > 0. Then the ball V = B^Y_ε¡f(x)¢ is an open subset of Y, so U = f⁻¹(V ) must be an open subset of X. As x ∈ U, there exists some δ > 0 such that B^X_δ (x) ⊆ U. If y ∈ X is any point that satisfies dX(x, y) < δ, then we have y ∈ B_δ^X(x) ⊆ U, and therefore

f (y) ∈ f (U ) ⊆ V = B^Y_ε¡f(x)¢.

Consequently dY¡f(x), f(y)¢ < ε.

(b) ⇒ (c). Assume that statement (b) holds, choose any point x ∈ X, and suppose that xn ∈ X are such that xn→ x. Fix any ε > 0, and let δ > 0 be the number whose existence is given by statement (b). Since xn → x, there must exist some N > 0 such that dX(x, xn) < δ for all n > N. Statement (b) therefore implies that dY¡f(x), f(xn)¢ < ε for all n > N, so we conclude that f (xn) → f (x) in Y.

(c) ⇒ (a). Suppose that statement (c) holds, and let V be any open subset of Y. Suppose that f⁻¹(V ) was not open. Then f⁻¹(V ) cannot be the empty

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set, so there must exist some point x ∈ f⁻¹(V ). By definition, this means that y = f (x) ∈ V. Since V is open, there exists some ε > 0 such that B_ε^Y(y) ⊆ V. But f⁻¹(V ) is not open, so for no n ∈ N can the ball B_1/n^X (x) be entirely contained in f⁻¹(V ). Hence for every n there is some xn ∈ B_1/n^X (x) such that xn ∈ f/ ⁻¹(V ). Since dX(x, xn) < 1/n, we have xn → x. Applying statement (c), it follows that f (xn) → f (x). Therefore there must exist some N > 0 such that dY¡f(x), f(xn)¢ < ε for all n > N. Hence for every n > N we have f (xn) ∈ B_ε^Y(y) ⊆ V, and therefore xn ∈ f⁻¹(V ). This contradicts the definition of xn, so we conclude that f⁻¹(V ) must be open. Therefore f is continuous. ⊓⊔

The number δ that appears in statement (b) of Lemma 1.20 depends both on the point x and the number ε. We say that a function f is uniformly continuous if δ can be chosen independently of x. Here is the precise definition.

Definition 1.21 (Uniform Continuity). Let X be a metric space with metric dX, and let Y be a metric space with metric dY. If E ⊆ X, then we say that a function f : X → Y is uniformly continuous on E if for every ε > 0 there exists a δ > 0 such that

∀ x, y ∈ E, dX(x, y) < δ =⇒ dY¡f(x), f(y)¢ < ε. ♦ The next lemma shows that a continuous function whose domain is a compact set is uniformly continuous on that set.

Lemma 1.22. Let X and Y be metric spaces. If K is a compact subset of X and f : K → Y is continuous, then f is uniformly continuous on K.

Proof. For this proof, let dX and dY denote the metrics on X and Y, and let B_r^X(x) and B_s^Y(y) denote open balls in X and Y, respectively.

Suppose that f : K → Y is continuous, and fix ε > 0. For each point z ∈ Y, let Uz= f⁻¹(B_ε^Y(z)). Then {Uz}z∈Y is an open cover of K. Since K is compact, it is sequentially compact. Therefore Lemma 1.16 implies that there exists a number δ > 0 such that if B is any open ball of radius δ in X that intersects K, then B ⊆ Uz for some z ∈ Y.

Now choose any points x, y ∈ K with dX(x, y) < δ. Then x, y ∈ B_δ^X(x), and B^X_δ (x) must be contained in some set Uz, so

f (x), f (y) ∈ f (B_δ^X(x)) ⊆ f (Uz) ⊆ B_ε^Y(z).

Therefore dY(f (x), f (y)) < 2ε, so f is uniformly continuous. ⊓⊔

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1.6 Urysohn’s Lemma and the Tietze Extension Theorem 15

1.6 Urysohn’s Lemma and the Tietze Extension Theorem

Urysohn’s Lemma and the Tietze Extension Theorem are two basic results about the existence of continuous functions. We will state these results for functions on metric spaces, although both results can be generalized to functions on normal topological spaces (for precise statements and proofs, see texts such as [Mun75] or [ST76]).

To prove Urysohn’s Lemma for functions on a metric space, we will use the following lemma, which states that the distance from a point x to a subset E of a metric space is a continuous function of x.

Lemma 1.23. Let X be a metric space. If E is a nonempty subset of X, then the function f : X → R defined by

f (x) = dist(x, E) = inf©d(x, z) : z ∈ Eª, x ∈ X, is uniformly continuous on X.

Proof. Fix ε > 0, and set δ = ε/2. Choose any points x, y ∈ X such that d(x, y) < δ. By definition of the distance function, there exist points a, b ∈ E such that

d(x, a) < dist(x, E) + δ and d(y, b) < dist(y, E) + δ.

Consequently,

f (y) = dist(y, E) ≤ d(y, a)

≤ d(y, x) + d(x, a)

< δ +¡dist(x, E) + δ¢

= f (x) + ε.

Interchanging the roles of x and y, we similarly obtain f (x) < f (y) + ε.

Therefore |f (x) − f (y)| < ε whenever d(x, y) < δ. This tells us that f is uniformly continuous on X. ⊓⊔

Now we prove Urysohn’s Lemma, which states that if E and F are disjoint closed sets, then we can find a continuous function that is identically 0 on E and identically 1 on F. In this sense, E and F can be “separated” by a continuous function.

Theorem 1.24 (Urysohn’s Lemma). If E, F are disjoint closed subsets of a metric space X, then there exists a continuous function θ : X → R such that

(a) 0 ≤ θ ≤ 1 on X,

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(b) θ = 0 on E, and (c) θ = 1 on F.

Proof. Since E is closed, if x /∈ E then dist(x, E) > 0. Lemma 1.23 tells us dist(x, E) and dist(x, F ) are each continuous functions of x. It follows that the function

θ(x) = dist(x, E) dist(x, E) + dist(x, F ) has the required properties. ⊓⊔

The Tietze Extension Theorem is a generalization of Urysohn’s Lemma. It states that if F is a closed set and f : F → R is continuous, then there exists a continuous function g : X → R that equals f on the set F. The first step in the proof is given in the following lemma, which constructs a continuous function on X that is related to f on F in a certain way.

Lemma 1.25. Let F be a closed subset of a metric space X. If f : F → R is continuous and |f (x)| ≤ M for all x ∈ F, then there exists a continuous function g : X → R such that

(a) |f (x) − g(x)| ≤ ²₃M for x ∈ F, (b) |g(x)| ≤ ¹₃M for x ∈ F, and

(c) |g(x)| <¹₃M for x /∈ F.

Proof. Since f is real-valued and continuous on the closed set F, the sets A = ©x ∈ F : f(x) ≤ −¹₃Mª

and B = ©x ∈ F : f(x) ≥¹₃Mª are closed subsets of X (why?). Further, they are disjoint by construction.

The remainder of the proof, we we assign to the reader as Problem 1.33, consists of showing that in each of the following cases the given function g has the required properties.

Case 1 : A = B = ∅, g = 0.

Case 2 : A 6= ∅, B 6= ∅, g(x) = M 3

dist(x, A) − dist(x, B) dist(x, A) + dist(x, B). Case 3 : A = ∅, B 6= ∅, g(x) = M

3

dist(x, B) 1 + dist(x, B). Case 4 : A 6= ∅, B = ∅ (similar to Case 3). ⊓⊔ Now we prove the Tietze Extension Theorem.

Theorem 1.26 (Tietze Extension Theorem). If F is a closed subset of a metric space X and f : F → R is continuous, then the following statements hold.

(a) There exists a continuous function g : X → R such that g = f on F.

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(b) If |f (x)| ≤ M for all x ∈ F, then we can choose the function g in statement (a) so that |g(x)| ≤ M for all x and |g(x)| < M for x /∈ F. ♦ Proof. (b) We will prove this statement first, i.e., we assume that f is continuous on F and |f (x)| ≤ M on F. For integers n = 0, 1, 2, . . . we will inductively define a function gn that is continuous on X and satisfies

sup

x∈F

¯

¯ f (x) −

n

X

k=0

gk(x)

¯

≤ ¡₂

3

¢n

M. (1.7)

For n = 0, equation (1.7) holds if we simply take g0 = 0. Assume that n ≥ 0 and g0, . . . , gm are continuous functions on X that satisfy equation (1.7) for n = 0, . . . , m. Set

hm = f −

m

X

k=0

gk.

This function is continuous on F, and |hm(x)| ≤ (2/3)^mM for all x ∈ F.

Applying Lemma 1.25 to hm, we obtain a continuous function gm+1 that satisfies

|hm(x) − gm+1(x)| ≤ ²₃¡₂

3

¢m

M, x ∈ F, (1.8)

|gm+1(x)| ≤ ¹₃¡₂

3

¢m

M, x ∈ F, (1.9)

|gm+1(x)| < ¹₃¡₂

3

¢m

M, x /∈ F. (1.10)

In particular, equation (1.8) implies that

sup

x∈F

¯

¯ f (x) −

m+1

X

k=0

gk(x)

¯

≤ ¡₂

3

¢m+1

M.

This completes the inductive step in the definition of g0, g1, . . . .

Each function g0, g1, . . . is continuous, and by using the estimate in equation (1.9) we see that the series

g =

∞

X

k=0

gk

converges uniformly on X. Consequently g is continuous. Using equation (1.9), for each x ∈ F we compute that

|g(x)| ≤

∞

X

k=0

|gk(x)| ≤

∞

X

k=1 1 3

¡₂

3

¢k−1

M = M.

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If x /∈ F then a similar calculation based on equation (1.10) shows that

|g(x)| < M. Finally, taking the limit as n → ∞ in equation (1.7), we obtain g(x) = f (x) for all x ∈ F.

(a) Now let f be an arbitrary continuous function on F. Set h(x) = arctan x and consider the composition h ◦ f. This function is bounded and continuous on F, so by applying statement (b) to h ◦ f we conclude that there is a continuous function G that coincides with h ◦ f on F. But then the function g(x) = (h⁻¹◦ G)(x) = tan(G(x)) is continuous on X and coincides with f on F. ⊓⊔

We can sometimes improve on Urysohn’s Lemma or the Tietze Extension Theorem when dealing with particular concrete domains. For example, the following result says that if we take X = R and fix a compact set K ⊆ R, then we can find an infinitely differentiable, compactly supported function that is identically 1 on K.

Theorem 1.27. If K ⊆ R is compact and U ⊇ K is open, then there exists a function f ∈ C_c^∞(R) such that

(a) 0 ≤ f ≤ 1 everywhere, (b) f = 1 on K, and

(c) supp(f ) ⊆ U. ♦

One way to prove Theorem 1.27 is by combining Urysohn’s Lemma with the operation of convolution. For details on convolution we refer to texts such as [DM72], [Ben97], or [Kat04]. For comparison, Problem 1.34 presents a less ambitious result. Specifically, it shows that there exists an infinitely differentiable function g that is identically zero outside of the finite interval [0, 1] and nonzero on the interior of this interval. Problems 1.35 and 1.36 give some related constructions.

Problems

1.28. Prove that every metric space X is Hausdorff, i.e., if x 6= y are two distinct elements of X, then there exist disjoint open sets U, V such that x ∈ U and y ∈ V.

1.29. Suppose that {xn}n∈N is a Cauchy sequence in a metric space X, and suppose there exists a subsequence {xn_k}k∈N that converges to x ∈ X, i.e., xn_k → x as k → ∞. Prove that xn→ x as n → ∞.

1.30. Let A be a subset of a metric space X.

(a) Prove that A is open if and only if A = A^◦. (b) Prove that A is closed if and only if A = A.

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1.31. Prove part (a) of Lemma 1.12.

1.32. Let X be a complete metric space, and let A be a subset of X. Prove that the following two statements are equivalent.

(a) A is closed.

(b) A is complete, i.e., every Cauchy sequence of points in A converges to a point of A.

1.33. Complete the proof of Lemma 1.25.

1.34. Define

γ(x) = e^−1/xχ_[0,∞)(x) =

(e^−1/x, x > 0,

0, x ≤ 0,

and

β(x) = γ¡1 − x²¢

=

(e^−1/(x²⁻¹⁾, −1 < x < 1,

0, |x| ≥ 1.

Prove the following statements.

(a) γ(x) = 0 for all x ≤ 0, and γ(x) > 0 for all x > 0.

(b) For each n ∈ N, there exists a polynomial pn of degree n − 1 such that γ⁽ⁿ⁾(x) = pn(x)

x²ⁿ γ(x).

(c) γ ∈ C^∞(R) and γ⁽ⁿ⁾(0) = 0 for every n ≥ 0.

(d) β ∈ C_c^∞(R), β(x) > 0 for |x| < 1, and β(x) = 0 for |x| ≥ 1.

1.35. Let f be the piecewise linear function pictured in Figure 1.1. Show that the function

g(x) = Z x

−3

f (t) dt, x ∈ R, has the following properties.

(a) 0 ≤ g(x) ≤ 1 for every x ∈ R.

(b) g(x) = 1 for |x| ≤ 1.

(c) g(x) = 0 for |x| ≥ 3.

(d) g ∈ C_c¹(R), but g /∈ C_c²(R).

1.36. Prove that there exists an infinitely differentiable function θ that satisfies:

(a) 0 ≤ θ(x) ≤ 1 for every x ∈ R, (b) θ(x) = 1 for all |x| ≤ 1, (c) θ(x) = 0 for all |x| ≥ 3. ♦

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f

-4 -2 2 4

-1 -0.5 0.5 1

Fig. 1.1 The function f from Problem 1.35.

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Chapter 2 Norms and Banach Spaces

While a metric provides us with a notion of the distance between points in a space, a norm gives us a notion of the length of an individual vector. A norm can only be defined on a vector space, while a metric can be defined on arbitrary sets.

2.1 The Definition of a Norm

Here is the definition of a norm, and the slightly weaker notion of a seminorm.

Definition 2.1 (Seminorms and Norms). Let X be a vector space over the real field R or the complex field C. A seminorm on X is a function k · k : X → R such that for all vectors x, y ∈ X and all scalars c we have:

(a) Nonnegativity: kxk ≥ 0,

(b) Homogeneity: kcxk = |c| kxk, and

(c) The Triangle Inequality: kx + yk ≤ kxk + kyk.

A seminorm is a norm if we also have:

(d) Uniqueness: kxk = 0 if and only if x = 0.

A vector space X together with a norm k · k is called a normed vector space, a normed linear space, or simply a normed space. ♦

We refer to the number kxk as the length of a vector x, and we say that kx − yk is the distance between the vectors x and y. A vector x that has length 1 is called a unit vector, or is said to be normalized.

We usually use k · k to denote a norm or seminorm, sometimes with sub- scripts to distinguish among different norms (for example, k · ka and k · kb).

Other common symbols for norms or seminorms are | · |, ||| · |||, or ρ(·). The absolute value function |x| is a norm on the real line R and on the complex plane C. Likewise, the Euclidean norm is a norm on R^d and C^d.

c

°2014 Christopher Heil

21

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Example 2.2. The space ℓ¹introduced in Example 1.2 is a vector space, and it is straightforward to check that the “ℓ¹-norm”

kxk1 =

∞

X

k=1

|xk|, x = (xk)k∈N∈ ℓ¹,

defined in equation (1.1) is indeed a norm on ℓ¹. Therefore ℓ¹ is a normed space. ♦

2.2 The Induced Norm

If X is a normed space, then it follows directly from the definition of a norm that

d(x, y) = kx − yk, x, y ∈ X,

defines a metric on X. This is called the metric on X induced from k · k, or simply the induced metric on X.

Consequently, whenever we are given a normed space X, we have a metric on X as well as a norm, and this metric is the induced metric. Therefore all definitions made for metric spaces apply to normed spaces, using the induced norm d(x, y) = kx−yk. For example, convergence in a normed space is defined by

xn → x ⇐⇒ lim

n→∞kx − xnk = 0.

It may be possible to place a metric on X other than the induced metric, but unless we explicitly state otherwise, all metric-related statements on a normed space are taken with respect to the induced metric.

2.3 Properties of Norms

Here are some useful properties of norms (we assign the proof as Problem 2.19).

Lemma 2.3. If X is a normed space, then the following statements hold.

(a) Reverse Triangle Inequality:¯

¯kxk − kyk¯

¯≤ kx − yk.

(b) Convergent implies Cauchy: If xn → x, then {xn}n∈N is Cauchy.

(c) Boundedness of Cauchy sequences: If {xn}n∈Nis a Cauchy sequence, then sup kxnk < ∞.

(d) Continuity of the norm: If xn→ x, then kxnk → kxk.

(e) Continuity of vector addition: If xn→ x and yn → y, then xn+yn→ x+y.

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2.5 Banach Spaces 23

(f) Continuity of scalar multiplication: If xn → x and cn → c, then cnxn→ cx. ♦

2.4 Convexity

If x and y are vectors in a vector space X, then the line segment joining x to y is the set of all points of the form tx + (1 − t)y where 0 ≤ t ≤ 1.

A subset K of a vector space is convex if given any two points x, y ∈ K, the line segment joining x to y is entirely contained within K. That is, K is convex if

x, y ∈ K, 0 ≤ t ≤ 1 =⇒ tx + (1 − t)y ∈ K.

If X is a normed space, then every open ball Br(x) in X is convex (this is Problem 2.20). However, if X is a metric space that is a vector space, it need not be true that open balls in X need be convex.

2.5 Banach Spaces

Every convergent sequence in a normed vector space must be Cauchy, but the converse does not hold in general. In some normed spaces it is true that every Cauchy sequence in the space is convergent. We give spaces that have this property the following name.

Definition 2.4 (Banach Space). A normed space X is a Banach space if it is complete, i.e., if every Cauchy sequence in X converges to an element of X. ♦

The terms “Banach space” and “complete normed space” are interchange- able, and we will use whichever is more convenient in a given context.

The set of real numbers R is complete with respect to absolute value (using real scalars), and likewise the complex plane C is complete with respect to absolute value (using complex scalars). More generally, R^dand C^dare Banach spaces with respect to the Euclidean norm. In fact, every finite-dimensional vector space V is complete with respect to any norm that we place on V (for a proof, see [Con90, Sec. III.3].

We proved in Theorem 1.9 that the space ℓ¹ is complete. Therefore ℓ¹ is an example of an infinite-dimensional Banach space. One example of an infinite-dimensional normed space that is not a Banach space is the space c00

studied in Example 1.8. A space that is complete with respect to one norm may be incomplete if replace that norm with another norm. For example, ℓ¹ is no longer complete if we replace the norm k · k1 with the “ℓ²-norm” (see Problem 3.30).

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2.6 Infinite Series in Normed Spaces

An infinite series is a limit of the partial sums of the series. Hence, the definition of an infinite series involves both addition and limits, each of which are defined in normed spaces. The precise definition of an infinite series in a normed space is as follows.

Definition 2.5 (Convergent Series). Let {xn}n∈N be a sequence in a normed space X. We say that the seriesP∞

n=1xnconverges and equals x ∈ X if the partial sums sN =PN

n=1xn converge to x, i.e., if

N →∞lim kx − sNk = lim

N →∞

°

° x −

N

X

n=1

xn

°

= 0.

In this case, we write x =P∞

n=1xn, and we also use the shorthands x =P xn

or x =P

nxn. ♦

In order for an infinite series to converge in X, the norm of the difference between f and the partial sum sN must converge to zero. If we wish to emphasize which norm we are referring to, we may write that x = P xn

converges with respect to k · k, or we may say that x =P xn converges in X.

Here is another type of convergence notion for series.

Definition 2.6. Let {xn}n∈N be a sequence in a normed space X. We say that the seriesP∞

n=1xn is absolutely convergent if

∞

X

n=1

kxnk < ∞. ♦

Note that Definition 2.6 does not imply that an absolutely convergent series will converge in the sense of Definition 2.5. We will prove in the next theorem that if X is complete then every absolutely convergent series in X converges, while if X is not complete then there will exist some vectors xn ∈ X such thatP kxnk < ∞ yetP xn does not converge.

Theorem 2.7. If X is a normed space, then the following two statements are equivalent.

(a) X is complete (i.e., X is a Banach space).

(b) Every absolutely convergent series in X converges in X. That is, if {xn}n∈N is a sequence in X and P kxnk < ∞, then the series P xn

converges in X.

Proof. (a) ⇒ (b). Assume that X is complete, and suppose thatP kxnk < ∞.

Set

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2.6 Infinite Series in Normed Spaces 25

sN =

N

X

n=1

xn and tN =

N

X

n=1

kxnk.

Given any N > M we have

ksN − sMk =

°

N

X

n=M +1

xn

°

≤

N

X

n=M +1

kxnk = |tN− tM|.

Since {tN}N ∈N is a Cauchy sequence of scalars, this implies that {sN}N ∈N is a Cauchy sequence of vectors in X, and hence converges. By definition, this means thatP∞

n=1xn converges in X.

(b) ⇒ (a). Suppose that every absolutely convergent series in X is convergent. Let {xn}n∈N be a Cauchy sequence in X. Applying Problem 2.22, there exists a subsequence {xnk}k∈N such that

∀ k ∈ N, kxnk+1− xnkk < 2^−k. The seriesP∞

k=1(xnk+1− xnk) is absolutely convergent, because

∞

X

k=1

kxnk+1− xnkk ≤

∞

X

k=1

2^−k = 1 < ∞.

Therefore the seriesP∞

k=1(xnk+1− xnk) converges in X, so we can set x =

∞

X

k=1

(xnk+1− xnk).

By definition, this means that the partial sums

sM =

M

X

k=1

(xnk+1− xnk) = xnM +1− xn1

converge to x as M → ∞. Setting y = x + xn1, it follows that xnM +1 = sM + xn1 → x + xn1 = y as M → ∞.

Reindexing (replace M + 1 by k), we conclude that xnk→ y as k → ∞.

Thus {xn}n∈N is a Cauchy sequence that has a subsequence {xnk}k∈N

that converges to the vector y. Appealing to Problem 1.29, this implies that xn→ y. Hence every Cauchy sequence in X converges, so X is complete. ⊓⊔ The ordering of the vectors in a series may be important. If we reorder a series, or in other words consider a new seriesP∞

n=1x_σ(n)where σ : N → N is a bijection, there is no guarantee that this reordered series will still converge.

IfP∞

n=1xσ(n)does converge for every bijection σ, then we say that the series

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P∞

n=1xn converges unconditionally. A series that converges but does not converge unconditionally is said to be conditionally convergent.

The following theorem gives some relationships between absolute and un- conditional convergence in Banach spaces. The proof is nontrivial; one proof can be found in [Heil11, Sec. 3.6]. In particular, statement (c) of Theorem 2.8 is a consequence of a important result known as the Dvoretzky–Rogers The- orem (see Theorem 3.33 in [Heil11]).

Theorem 2.8. Assume that X is a Banach space.

(a) IfP xn converges absolutely in X, thenP xn converges unconditionally.

(b) If X is finite dimensional andP xnconverges unconditionally, thenP xn

converges absolutely.

(c) If X is infinite dimensional, then there exist vectors xn ∈ X such that P xn converges unconditionally but not absolutely.

A corollary of Theorem 2.8 is that if cn is a scalar for every n, then the seriesP cn converges absolutely if and only if it converges unconditionally.

2.7 Span and Closed Span

If A is a subset of a vector space X, then the finite linear span of A, denoted by span(A), is the set of all possible finite linear combinations of elements of A. If the scalar field is C, then

span(A) =

½ ^N X

n=1

cnxn : N > 0, xn ∈ A, cn∈ C

¾

, (2.1)

and if the scalar field is R then we need only replace C by R on the preceding line. We often call span(A) the finite span, the linear span, or simply the span of A. We say that A spans X if span(A) = X.

If X is a normed space, then we can take limits of elements, and consider the closure of the span. The closure of span(A) is called the closed linear span or simply the closed span of A. For compactness of notation, usually write the closed span as span(A) instead of span(A).

It follows from part (b) of Lemma 1.12 that the closed span consists of all limits of elements of the span:

span(A) = ©y ∈ X : ∃ yn∈ span(A) such that yn→ yª. (2.2) Suppose that xn ∈ A, cn is a scalar, and x = P cnxn is a convergent infinite series in X. Then the partial sums

(31)

2.8 Hamel Bases and Schauder Bases 27

sN =

N

X

n=1

cnxn

belong to span(A) and sN → x, so it follows from equation (2.2) that x belongs to span(A). As a consequence, we have the inclusion

½ ^∞ X

n=1

cnxn: xn∈ A, cn scalar,

∞

X

n=1

cnxn converges

¾

⊆ span(A). (2.3)

However, equality need not hold in equation (2.3). A specific example of such a situation is presented in Example 2.18.

According to the following definition, if the closed span of a sequence is the entire space X, then we say that sequence is complete.

Definition 2.9 (Complete Sequence). Let {xn}n∈Nbe a sequence of vectors in a normed vector space X. We say that the sequence {xn}n∈Nis complete in X if span{xn}n∈Nis dense in X, i.e., if

span{xn}n∈N = X.

Complete sequences are also known as total or fundamental sequences. ♦ If M is a subspace of a normed space X, then, by definition, M is finite dimensional if it has a finite Hamel basis. Equivalently, a subspace M is finite dimensional if and only if we can write M = span{x1, . . . , xn} for some finitely many vectors x1, . . . , xn ∈ X. The following result states that every finite-dimensional subspace of a normed space is closed. For a proof of this theorem, see [Con90, Prop. III.3.3].

Theorem 2.10. If M is a finite-dimensional subspace of a normed space X, then M is a closed subset of X. ♦

2.8 Hamel Bases and Schauder Bases

A Hamel Basis is simply another name for the usual notion of a basis for a vector space. We formalize this in the following definition.

Definition 2.11 (Hamel Basis). Let V be a vector space. A Hamel basis, vector space basis, or simply a basis for V is a set B ⊆ V such that

(a) B is linearly independent, and (b) span(B) = V. ♦

The span of B was defined in equation (2.1) to be the set of all finite linear combinations of vectors from B. Likewise, independence is defined in terms

A Short Introduction to Metric, Banach, and Hilbert Spaces

Christopher Heil