functional analysis lecture notes chapter 3. banach...

FUNCTIONAL ANALYSIS LECTURE NOTES

CHAPTER 3. BANACH SPACES

CHRISTOPHER HEIL

1. Elementary Properties and Examples

Notation 1.1. Throughout, F will denote either the real line R or the complex plane C.All vector spaces are assumed to be over the field F.

Definition 1.2. Let X be a vector space over the field F. Then a semi-norm on X is afunction ‖ · ‖ : X → R such that

(a) ‖x‖ ≥ 0 for all x ∈ X,

(b) ‖αx‖ = |α| ‖x‖ for all x ∈ X and α ∈ F,

(c) Triangle Inequality: ‖x + y‖ ≤ ‖x‖ + ‖y‖ for all x, y ∈ X.

A norm on X is a semi-norm which also satisfies:

(d) ‖x‖ = 0 =⇒ x = 0.

A vector space X together with a norm ‖ · ‖ is called a normed linear space, a normed vector

space, or simply a normed space.

Definition 1.3. Let I be a finite or countable index set (for example, I = 1, . . . , N iffinite, or I = N or Z if infinite). Let w : I → [0,∞). Given a sequence of scalars x = (xi)i∈I ,set

‖x‖p,w =

(

∑

i∈I

|xi|pw(i)p

)1/p

, 0 < p <∞,

supi∈I

|xi|w(i), p = ∞,

where these quantities could be infinite. Then we set

`pw(I) =

x = (xi)i∈I : ‖x‖p <∞

.

We call `pw(I) a weighted `p space, and often denote it just by `pw (especially if I = N). Ifw(i) = 1 for all i, then we simply call this space `p(I) or `p and write ‖ · ‖p instead of ‖ · ‖p,w.

Date: April 11, 2006.

These notes closely follow and expand on the text by John B. Conway, “A Course in Functional Analysis,”

Second Edition, Springer, 1990.

1

2 CHRISTOPHER HEIL

Exercise 1.4. Show that if 1 ≤ p ≤ ∞ then ‖ · ‖p,w defines a semi-norm on `pw, and it is anorm if w(i) > 0 for all i.

In particular, if I = 1, . . . , n then `pw = Fn, and each choice of p and w gives a semi-normor norm on Fn.

Hints: The Triangle Inequality on `p is often called Minkowski’s Inequality. It is easy toprove if p = 1 or p = ∞. There are several ways to prove it for other p. One ways is to beginwith

‖x+ y‖pp =

∑

i∈I

|xi + yi|p =

∑

i∈I

|xi + yi|p−1 |xi + yi|

≤∑

i∈I

|xi + yi|p−1 |xi| +

∑

i∈I

|xi + yi|p−1 |yi|.

Then apply Holder’s Inequality to each sum using the exponent p′ on the first factor and pfor the second (recall that p′ = p/(p− 1)). Then divide both sides by ‖x+ y‖p−1

p .

Definition 1.5. Let (X,Ω, µ) be a measure space. Given a measurable f : X → [−∞,∞](if F = R) or f : X → C (if F = C), set

‖f‖p =

(∫

X

|f(x)|p dµ(x)

)1/p

, 0 < p <∞,

ess supx∈X

|f(x)|, p = ∞,

where these quantities could be infinite. Define

Lp(X) =

f : X → [−∞,∞] or C : ‖f‖p <∞

.

Other notations for Lp(X) are Lp(µ), Lp(X, µ), Lp(dµ), Lp(X, dµ), etc.When we write Lp(Rn), it will be assumed that µ is Lebesgue measure on Rn, unless

specifically stated otherwise. In this case we will write dx instead of dµ(x).The space `pw(I) is a special case of Lp(X), where X = I and µ is a weighted counting

measure on I.

Exercise 1.6. Show that if 1 ≤ p ≤ ∞ then ‖ ·‖p is a semi-norm on Lp(X), and it is a normif we identify functions that are equal almost everywhere.

The Triangle Inequality on Lp is often called Minkowski’s Inequality, and its proof is similarto the proof of Minkowski’s Inequality for `p.

Exercise 1.7. Show that every subspace of a normed space is itself a normed space (usingthe same norm).

Definition 1.8 (Distance). Let ‖ · ‖ be a norm on X. Then the distance from x to y in Xis d(x, y) = ‖x− y‖.

CHAPTER 3. BANACH SPACES 3

Exercise 1.9. Show that d(·, ·) defines a metric on X (see Appendix A).

Since X has a metric and hence has an associated topology, all the standard topologicalnotions (open/closed sets, convergence, etc.) apply to X. For convenience, we give someexplicit definitions and facts relating to these topics for the setting of normed spaces.

Definition 1.10 (Convergence). Let X be a normed linear space (such as an inner productspace), and let fnn∈N be a sequence of elements of X.

(a) We say that fnn∈N converges to f ∈ X, and write fn → f , if

limn→∞

‖f − fn‖ = 0,

i.e.,∀ ε > 0, ∃N > 0 such that n > N =⇒ ‖f − fn‖ < ε.

(b) We say that fnn∈N is Cauchy if

∀ ε > 0, ∃N > 0 such that m,n > N =⇒ ‖fm − fn‖ < ε.

Exercise 1.11. Let X be a normed linear space. Prove the following.

(a) Reverse Triangle Inequality:∣

∣‖f‖ − ‖g‖∣

∣ ≤ ‖f − g‖.

(b) Continuity of the norm: fn → f =⇒ ‖fn‖ → ‖f‖.

(c) Continuity of vector addition: fn → f and gn → g =⇒ fn + gn → f + g.

(d) Continuity of scalar multiplication: fn → f and αn → α =⇒ αnfn → αf .

(e) All convergent sequences are bounded, and the limit of a convergent sequence is unique.

(f) Every Cauchy sequence is bounded.

(g) Every convergent sequence is Cauchy.

Exercise 1.12. Let fnn∈N be a sequence of vectors in a normed space X. Show that if‖fn − fn+1‖ < 2−n for every n, then fnn∈N is Cauchy.

Exercise 1.13. Let `pw(I) be the weighted `p space defined in Exercise 1.3, where we assumew(i) > 0 for all i ∈ I. Let xnn∈N be a sequence of vectors in `pw(I), and let x be a vector in`pw(I). Write the components of xn and x as xn = (xn(1), xn(2), . . . ) and x = (x(1), x(2), . . . ).

(a) Prove that if xn → x (i.e., ‖x − xn‖p,w → 0), then xn converges componentwise to x,i.e., for each fixed k we have limn→∞ xn(k) = x(k).

(b) Prove that if I is finite then the converse is also true, i.e., componentwise convergenceimplies convergence with respect to the norm ‖ · ‖p,w.

(c) Prove that if I is infinite then componentwise convergence does not imply convergencein the norm of `wp (I).

4 CHRISTOPHER HEIL

It is not true in an arbitrary normed space that every Cauchy sequence must converge.Normed spaces which do have the property that all Cauchy sequences converge are given aspecial name.

Definition 1.14 (Banach Space). A normed space X is called a Banach space if it iscomplete, i.e., if every Cauchy sequence is convergent. That is,

fnn∈N is Cauchy in X =⇒ ∃ f ∈ X such that fn → f.

Exercise 1.15. Show that the weighted `p space `pw(I) defined in Exercise 1.3 is a Banachspace if w(i) > 0 for all i ∈ I.

Hints: Consider the case I = N. Suppose that xnn∈N is a Cauchy sequence in `pw. Eachxn is a sequence of scalars. Write out the components of xn as

xn = (xn(1), xn(2), . . . ).

Prove that for a fixed component k we have

|xm(k) − xn(k)| ≤ C ‖xm − xn‖p,w,

where C is a fixed constant (determined by the weight and by k but independent of m andn). Conclude that with k fixed, xn(k)n∈N is a Cauchy sequence of scalars, and hence con-verges. Define x(k) = limn→∞ xn(k) and set x = (x(1), x(2), . . . ). Then we have constructeda candidate limit for the sequence xnn∈N. However, so far we only have that each indi-

vidual component of xn converges to the corresponding component of x, i.e., xn convergescomponentwise to x. This is not enough: to complete the proof you must show that xn → xin the norm of `pw. Use the fact that xnn∈N is Cauchy together with the componentwiseconvergence to show that ‖x− xn‖`p

w→ 0 as n→ ∞.

Compare this proof to the proof of Theorem 2.18 given below.

Exercise 1.16. Show that the space Lp(X) defined in Example 1.5 is a Banach space ifwe identify functions that are equal almost everywhere. This is called the Riesz–Fisher

Theorem.Hint: The argument is similar in spirit but more subtle than the one used to prove that

`pw(I) is a Banach space. First find a candidate limit and then show that the sequenceconverges in norm to this limit.

The next two exercises will be useful to us later.

Exercise 1.17. Let X be a normed linear space. If fnn∈N is a Cauchy sequence in X andthere exists a subsequence fnk

k∈N that converges to f ∈ X, then fn → f .

SolutionChoose any ε > 0. Since fnn∈N is Cauchy, there is an N such that ‖fm − fn‖ <

ε2

for m,n > N . Also, there is a k such that nk > N and ‖f − fnk

‖ < ε2. Hence for n > N we have

‖f − fn‖ ≤ ‖f − fnk‖ + ‖fnk

− fn‖ <ε

2+ε

2= ε.


Thus fn → f .

Exercise 1.18. Let fnn∈N be a Cauchy sequence in a normed space X. Show that thereexists a subsequence fnk

k∈N such that

∀ k ∈ N, ‖fnk+1− fnk

‖ < 2−k.

SolutionSince fnn∈N is Cauchy, we can find an n1 such that

m,n ≥ n1 =⇒ ‖fm − fn‖ < 2−1.

Then we can find an n2 > n1 such that

m,n ≥ n2 =⇒ ‖fm − fn‖ < 2−2.

Continuing in this way, we inductively construct n1 < n2 < · · · such that for each k,

m,n ≥ nk =⇒ ‖fm − fn‖ < 2−k.

In particular, since nk+1 > nk, we have ‖fnk+1− fnk

‖ < 2−k.

Definition 1.19 (Convergent Series). Let fnn∈N be a sequence of elements of a normedlinear space X. Then the series

∑∞n=1 fn converges and equals f ∈ X if the partial sums

sN =∑N

n=1 fn converge to f , i.e., if

‖f − sN‖ =∥

∥

∥f −

N∑

n=1

fn

∥

∥

∥→ 0 as N → ∞.

Definition 1.20 (Absolutely Convergent Series). Let X be a normed space and let fnn∈N

be a sequence of elements of X. If∞

∑

n=1

‖fn‖ < ∞,

then we say that the series∑

n fn is absolutely convergent in X.

Note that the definition of absolute convergence does not by itself tell us that the series∑

n fn actually converges. That is always true if X is a Banach space, but need not be trueif X is not complete.

Exercise 1.21. Let X be a Banach space and let fnn∈N be a sequence of elements of X.Prove that if

∑

n ‖fn‖ <∞ then the series∑

n fn does converge in X.Hint: You must show that the sequence of partial sums sNN∈N converges. Since X is a

Banach space, you just have to show that this sequence is Cauchy.

The converse of this exercise is also true, and is often a useful method for proving that agiven normed space is a Banach space.

6 CHRISTOPHER HEIL

Proposition 1.22. Let X be a normed space. Prove that X is a Banach space if and onlyif every absolutely convergent series in X converges in X.

Proof. ⇒. This is Exercise 1.21.

⇐. Suppose that every absolutely convergent series is convergent. Let fnn∈N be aCauchy sequence in X. By Exercise 1.18, there exists a subsequence fnk

k∈N such that‖fnk+1

− fnk‖ < 2−k for every k. The series

∑

k(fnk+1− fnk

) is absolutely convergent,because

∞∑

k=1

‖fnk+1− fnk

‖ ≤∞

∑

k=1

2−k = 1 < ∞.

By hypothesis, we conclude that∑

k(fnk+1− fnk

) converges in X, say to f . In terms of thepartial sums, this says that

fnk+1− fn1

=

k∑

j=1

(fnj+1− fnj

) → f as k → ∞,

or fnk→ g = f+fn1

as k → ∞. Thus fnn∈N is a Cauchy sequence which has a subsequencethat converges to g. It therefore follows from Exercise 1.17 that fn → g. Therefore X iscomplete.

Example 1.23 (The Harmonic Series). To illustrate the difference between convergence andabsolute convergence, consider the one-dimensional case, i.e., X = F, the field of scalars. Letxn = 1

n. Then

∑

n xn =∑

n1n

is the harmonic series, and the partial sums sN =∑N

n=11n

ofthis series are unbounded. Thus the harmonic series does not converge. Since sN → ∞, weusually say that

∑

n1n

diverges to infinity, and write∑

n1n

= ∞.

On the other hand, consider the alternating series∑

n(−1)n 1n. Since the terms alternate

signs and since 1n→ 0, it follows that this series does converge to a finite scalar (in fact, it

converges to − ln 2). However, it does not converge absolutely.

Definition 1.24 (Unconditionally Convergent Series). Let X be a Banach space and letfnn∈N be a sequence of elements of X. The series f =

∑∞n=1 fn is said to converge uncon-

ditionally if every rearrangement of the series converges. That is, f =∑∞

n=1 fn convergesunconditionally if for each bijection σ : N → N the series

∞∑

n=1

fσ(n)

converges.

Remark 1.25. It is not obvious, but it can be shown that if∑∞

n=1 fσ(n) is unconditionallyconvergent, then every rearrangement of the series must converge to the same sum, i.e., thereis a single f such that f =

∑∞n=1 fσ(n) for every permutation σ.


Exercise 1.26. Let X be a Banach space. Prove that if a series f =∑∞

n=1 fn convergesabsolutely, then it converges unconditionally.

Remark 1.27. In finite dimensions the converse to Exercise 1.26 is true, i.e., if X is finite-dimensional and a series f =

∑∞n=1 fn converges unconditionally, then it converges absolutely.

However, this fails in infinite dimensions.

Example 1.28 (The Harmonic Series Revisited). To illustrate the importance of uncondi-tional convergence, again consider X = F and the alternating series

∑

n(−1)n 1n. We know

that this series converges, but does not converge absolutely.Now consider what happens if we change the order of summation. Let pn = 1

2nand

qn = 12n+1

, i.e., the pn are the positive terms from the alternative series and the qn are theabsolute values of the negative terms. Each series

∑

n pn and∑

n qn diverges. Hence theremust exist an m1 > 0 such that

p1 + · · ·+ pm1> 1.

Then, there must exist an m2 > m1 such that

p1 + · · ·+ pm1− q1 + pm1+1 + · · ·+ pm2

> 2.

Continuing in this way, we see that

p1 + · · ·+ pm1− q1 + pm1+1 + · · ·+ pm2

− q2 + · · ·

is a rearrangement of∑

n(−1)n 1n

which diverges to +∞.In the same way, we can construct a rearrangement which diverges to −∞, which converges

to any given real number r, or which simply oscillates without ever converging. Moreover,the same can be done for any series of real scalars which converges conditionally.

The following is an equivalent formulation of unconditional convergence.

Proposition 1.29. Let X be a Banach space and let fnn∈N be a sequence of elementsof X. Then the series f =

∑∞n=1 fn converges unconditionally if and only if it converges with

respect to the net of finite subsets of N, i.e., if

∀ ε > 0, ∃ finite F0 ⊆ N such that ∀ finite F ⊇ F0,∥

∥

∥f −

∑

n∈F

fn

∥

∥

∥< ε.

Definition 1.30 (Topology). Let X be a normed linear space.

(a) The open ball in X centered at x ∈ X with radius r > 0 is

Br(x) = B(x, r) = y ∈ X : ‖x− y‖ < r.

(b) A subset U ⊆ X is open if

∀ x ∈ U, ∃ r > 0 such that Br(x) ⊆ U.

(c) A subset F ⊆ X is closed if X \ F is open.

8 CHRISTOPHER HEIL

Definition 1.31 (Limit Points, Closure, Density). Let X be a normed linear space and letA ⊆ X.

(a) A point f ∈ A is called a limit point of A if there exist fn ∈ A with fn 6= f such thatfn → f .

(b) The closure of A is the smallest closed set A such that A ⊆ A. Specifically,

A =⋂

F ⊆ X : F is closed and F ⊇ A.

(c) We say that A is dense in X if A = X.

Exercise 1.32. (a) The closure of A equals the union of A and all limit points of A:

A = A ∪ x ∈ X : x is a limit point of A = z ∈ X : ∃ yn ∈ A such that yn → z.

(b) If X is a normed linear space, then the closure of an open ball Br(x) is the closed ball

Br(x) = y ∈ X : ‖x− y‖ ≤ r.

(c) Prove that A is dense if and only if

∀ x ∈ X, ∀ ε > 0, ∃ y ∈ A such that ‖x− y‖ < ε.

Example 1.33. The set of rationals Q is dense in the real line R.

Exercise 1.34. Let X be a normed linear space and let F ⊆ X. Then

F is closed ⇐⇒ F contains all its limit points.

Solution⇒. Suppose that F is closed but that there exists a limit point f that does not belong to

F . By definition, there must exist fn ∈ F such that fn → f . However, f ∈ X \ F , which isopen, so there exists some r > 0 such that B(f, r) ⊆ X \ F . Yet there must exist some fn

with ‖f − fn‖ < r, so this fn will belong to X \ F , which is a contradiction.

⇐. Exercise.

Remark 1.35. Some authors make the restriction that, when dealing with normed spaces,the terminology “subspace” is used only for closed subspaces. Other authors use the termi-nology “linear manifold” to denote a subspace that need not be closed. To avoid ambiguity,we will use the following terminology.

Definition 1.36. (a) A subset Y of a vector space X is a subspace of X if it is closed underboth vector addition and scalar multiplication, i.e., if for all u, v ∈ Y and α, β ∈ F we haveαu+ βv ∈ Y .

(b) A subset Y of a normed linear space X is a closed subspace of X if it is a subspaceand it is closed with respect to the norm of X.


Exercise 1.37. In finite dimensions, all subspaces are closed sets (this will be proved inProposition 3.5). This exercise demonstrates that, in infinite dimensions, subspaces neednot be closed sets.

(a) Fix 1 ≤ p ≤ ∞. Prove that

c00 = x = (x1, . . . , xN , 0, 0, . . . ) : N > 0, x1, . . . , xN ∈ F

is a subspace of `p(N) that is not closed (with respect to the `p-norm). Prove that c00 isdense in `p(N) if p <∞, but that it is not dense in `∞(N).

(b) Define

c0 = x = (xk)∞k=1 : lim

k→∞xk = 0.

Prove c0 is a closed subspace of `∞(N) (in `∞-norm). Prove that c0 is the closure of c00(under the `∞-norm).

(c) Fix 1 ≤ p ≤ ∞. Prove that Cc(Rn), the space of continuous, compactly supported

functions on Rn, is a subspace of Lp(Rn) that is not closed. Prove that Cc(Rn) is dense in

Lp(Rn) if p <∞, but not if p = ∞.Hints: For a continuous function we have ‖f‖∞ = sup |f(x)|. Hence, if fnn∈N is a

sequence of continuous functions in L∞(Rn) that converges in L∞-norm, then it convergesuniformly. From undergraduate real analysis, we know that the limit of a uniformly conver-gent sequence of continuous functions is continuous.

(d) Let C0(Rn) be the set of all continuous functions f : Rn → F such that

lim|x|→∞

f(x) = 0. (1.1)

More precisely, (1.1) means that for every ε > 0 there exists a compact set K such that|f(x)| < ε for all x /∈ K. Prove that C0(R

n) is a closed subspace of L∞(Rn) (closed inL∞-norm). Prove that C0(R

n) is the closure of Cc(Rn) (under the L∞-norm).

(e) Let Cb(Rn) be the set of all bounded, continuous functions f : Rn → F. Prove that

Cb(Rn) is a closed subspace of L∞(Rn).

(f) Fix 1 ≤ p ≤ ∞, and let E be a (Lebesgue) measurable subset of Rn. Let M = f ∈Lp(Rn) : supp(f) ⊆ E. Prove that M is a closed subspace of Lp(Rn).

Remark 1.38. (a) We took the domain in the preceding exercise to be Rn just for conve-nience; the definitions of the spaces Cc, C0, Cb can be extended to domains that are moregeneral topological spaces.

(b) Beware that some authors use the notation C0 for the space that we are calling Cc!

Exercise 1.39. Let X be a Banach space and let M be a subspace of X. Then M is itselfa Banach space (using the norm from X) if and only if M is closed.

10 CHRISTOPHER HEIL

Solution⇒. Suppose that M is a Banach space. Let f be any limit point of M , i.e., suppose

fn ∈ M and fn → f . Then fn is a convergent sequence in X, and hence is Cauchy inX. Since each fn belongs to M , it is also Cauchy in M . Since M is a Banach space, fnmust therefore converge in M , i.e., fn → g for some g ∈ M . However, limits are unique, sof = g ∈M . Therefore M contains all its limit points and hence is closed.

⇐. Exercise.

Notation 1.40 (Notation for Closed Subspaces). Since we will often deal with closed sub-spaces of a Banach space, we declare that the notation

M ≤ X

means that M is a closed subspace of the Banach space X.

Exercise 1.41. Find examples of normed spaces that are not Banach spaces.Hint: Look for examples of normed spaces Y which are subspaces of a larger Banach

space X.

Remark 1.42. Is every normed space Y a subspace of a larger Banach space? The answeris yes, given a normed space Y it is always possible to construct a Banach space X ⊇ Y suchthat the norm on X, when restricted to Y , is the same as the norm on Y , and Y is dense inX with respect to that norm. This space is called the completion of Y .

Definition 1.43. Suppose that X is a normed linear space with respect to a norm ‖ · ‖a

and also with respect to another norm ‖ · ‖b. Then we say that these norms are equivalent

if there exist constants C1, C2 > 0 such that

∀ f ∈ X, C1 ‖f‖a ≤ ‖f‖b ≤ C2 ‖f‖a. (1.2)

Observe that equation (1.2) can be rearranged to read 1C2

‖f‖b ≤ ‖f‖a ≤ 1C1

‖f‖b.

Exercise 1.44. Let X be a vector space. If ‖ · ‖a and ‖ · ‖b are two norms on X, define‖ · ‖a ∼ ‖ · ‖b if ‖ · ‖a and ‖ · ‖b are equivalent. Prove that ∼ is an equivalence relation onthe class of norms on X.

The next result will show that equivalent norms define the same topology and the sameconvergence criterion.

Proposition 1.45. Let ‖·‖a and ‖·‖b be two norms on a vector space X. Then the followingstatements are equivalent.

(a) ‖ · ‖a and ‖ · ‖b are equivalent norms.

(b) ‖ · ‖a and ‖ · ‖b define the same topologies on X. That is, if U ⊆ X, then U is openwith respect to ‖ · ‖a if and only if it is open with respect to ‖ · ‖b.


(c) ‖·‖a and ‖·‖b define the same convergence criterion. That is, if fnn∈N is a sequencein X and f ∈ X, then

limn→∞

‖f − fn‖a = 0 ⇐⇒ limn→∞

‖f − fn‖b = 0.

Proof. (b) ⇒ (a). Assume that statement (b) holds. Let Bar (f) and Bb

r(f) denote the openballs of radius r centered at f ∈ X with respect to ‖ · ‖a and ‖ · ‖b. Since Ba

1(0) is openwith respect to ‖ · ‖a, the hypothesis that statement (b) holds implies that Ba

1 (0) is openwith respect to ‖ · ‖b. Therefore, since 0 ∈ Ba

1 (0), there must exist some r > 0 such thatBb

r(0) ⊂ Ba1 (0).

Now choose any f ∈ X and any ε > 0. Then

(r − ε) f

‖f‖b

∈ Bbr(0) ⊆ Ba

1 (0),

so∥

∥

∥

(r − ε) f

‖f‖b

∥

∥

∥

a< 1.

Rearranging, this implies (r − ε) ‖f‖a < ‖f‖b. Since this is true for every ε, we concludethat r ‖f‖a ≤ ‖f‖b.

A symmetric argument, interchanging the roles of the two norms, shows that there existsan s > such that ‖f‖b ≤ s ‖f‖a for every f ∈ X. Hence the two norms are equivalent

The remaining implications are exercises.

Hints on (c) ⇒ (a): Suppose that statement (c) holds. Show that this implies that thefunction ν : X → R given by ν(f) = ‖f‖b is continuous with respect to the norm ‖ ·‖a. Since(−1, 1) is an open subset of R, it follows that ν−1(−1, 1) is an open subset of X (with respectto ‖ · ‖a). Since 0 ∈ ν−1(−1, 1), there must exist an r > 0 such that Bb

r(0) ⊂ ν−1(−1, 1).But ν−1(−1, 1) = Ba

1 (0), so the remainder of the proof proceeds exactly like the proof of (b)⇒ (a).

2. Linear Operators on Normed Spaces

Definition 2.1 (Notation for Operators). Let X, Y be vector spaces. Let T : X → Y be afunction (= operator = transformation) mapping X into Y . We write either T (f) or Tf todenote the image of an element f ∈ X.

(a) T is linear if T (αf + βg) = αT (f) + βT (g) for every f , g ∈ X and α, β ∈ F.

(b) T is injective if T (f) = T (g) implies f = g.

(c) The kernel or nullspace of T is ker(T ) = f ∈ X : T (f) = 0.

(c) The range of T is range(T ) = T (f) : f ∈ X.

(d) The rank of T is the dimension of its range, i.e., rank(T ) = dim(range(T )). Inparticular, T is finite-rank if range(T ) is finite-dimensional.

(d) T is surjective if range(T ) = Y .

12 CHRISTOPHER HEIL

(e) T is a bijection if it is both injective and surjective.

(f) We use the notation I or IX to denote the identity map of a space X onto itself.

Definition 2.2 (Continuous and Bounded Operators). Let X, Y be normed linear spaces,and let L : X → Y be a linear operator.

(a) L is continuous at a point f ∈ X if fn → f in X implies Lfn → Lf in Y .

(b) L is continuous if it is continuous at every point, i.e., if fn → f inX implies Lfn → Lfin Y for every f .

(c) L is bounded if there exists a finite K ≥ 0 such that

∀ f ∈ X, ‖Lf‖ ≤ K ‖f‖.

Note that ‖Lf‖ is the norm of Lf in Y , while ‖f‖ is the norm of f in X.

(d) The operator norm of L is

‖L‖ = sup‖f‖=1

‖Lf‖.

(e) We let B(X, Y ) denote the set of all bounded linear operators mapping X into Y ,i.e.,

B(X, Y ) = L : X → Y : L is bounded and linear.

If X = Y then we write B(X) = B(X,X).

(f) If Y = F then we say that L is a functional. The set of all bounded linear functionalson X is the dual space of X, and is denoted

X ′ = B(X,F) = L : X → F : L is bounded and linear.

Another common notation for the dual space is X∗.

Note that since the norm on F is just absolutely value, the operator norm of a linearfunctional L ∈ X ′ = B(X,F) is ‖L‖ = sup

‖f‖=1

|Lf |.

Exercise 2.3. Show that if T : X → Y is linear and continuous, then ker(T ) is a closedsubspace of X and that range(T ) is a subspace of Y . Must range(T ) be a closed subspace?

Exercise 2.4. Let X, Y be normed linear spaces. Let L : X → Y be a linear operator.

(a) L is injective if and only if kerL = 0.

(b) If L is a bijection then L−1 : Y → X is also a linear bijection.

(c) L is bounded if and only if ‖L‖ <∞.

(d) If L is bounded then ‖Lf‖ ≤ ‖L‖ ‖f‖ for every f ∈ X (note that three differentmeanings of the symbol ‖ · ‖ appear in this statement!).


(e) If L is bounded then ‖L‖ is the smallest value of K such that ‖Lf‖ ≤ K‖f‖ holdsfor all f ∈ X.

(f) ‖L‖ = sup‖f‖≤1

‖Lf‖ = supf 6=0

‖Lf‖

‖f‖.

Exercise 2.5. Show that B(X, Y ) is a subspace of the vector space V consisting of ALLfunctions A : X → Y . Moreover, show that the operator norm is a norm on the spaceB(X, Y ), i.e.,

(a) 0 ≤ ‖L‖ <∞ for all L ∈ B(X, Y ),

(b) ‖L‖ = 0 if and only if L = 0 (the zero operator that sends every element of X to thezero vector in Y ),

(c) ‖αL‖ = |α| ‖L‖ for every L ∈ B(X, Y ) and every α ∈ F,

(d) ‖L+K‖ ≤ ‖L‖ + ‖K‖ for every L, K ∈ B(X, Y ).

The preceding exercise shows that B(X, Y ) is a normed linear space, and we will showin Theorem 2.18 that it is a Banach space if Y is a Banach space. However, in addition tovector addition and scalar multiplication operations, there is a third operation that we canperform with functions: composition.

Exercise 2.6. Prove that the operator norm is submultiplicative, i.e., prove if A ∈ B(X, Y )and B ∈ B(Y, Z), then BA ∈ B(X,Z) and

‖BA‖ ≤ ‖B‖ ‖A‖. (2.1)

In particular, when X = Y = Z, we see that B(X) is closed under compositions. Thespace B(X) is an example of an algebra.

Exercise 2.7. Let Fn be n-dimensional Euclidean space over F, under the Euclidean norm,and let Y be any normed linear space. Prove that if L : Fn → Y is linear, then L is bounded.

Hint: If x = (x1, . . . , xn) ∈ Fn then x = x1e1+· · ·+xnen where e1, . . . , en is the standardbasis for Fn. Use the Triangle and the Cauchy-Schwarz Inequalities.

SolutionGiven x ∈ Fn, we have

‖Lx‖ =

∥

∥

∥

∥

n∑

k=1

L(xkek)

∥

∥

∥

∥

≤n

∑

k=1

|xk| ‖Lek‖ ≤

( n∑

k=1

|xk|2

)1/2 ( n∑

k=1

‖Lek‖2

)1/2

= C ‖x‖,

where C =(∑n

k=1 ‖Lek‖2)1/2

. Hence L is bounded.

Remark 2.8. We will prove later that if X is any finite-dimensional vector space and ‖ ·‖ isany norm on X, then any linear function L : X → Y is bounded. To do this we will use thefact (that we will prove later) that all norms on a finite-dimensional space are equivalent.

14 CHRISTOPHER HEIL

The next lemma is a standard fact about continuous functions. L−1(U) denotes the inverseimage of U ⊆ Y , i.e., L−1(U) = f ∈ X : Lf ∈ U.

Exercise 2.9. Let X, Y be normed linear spaces. Let L : X → Y be linear. Then

L is continuous ⇐⇒(

U open in Y =⇒ L−1(U) open in X)

.

Solution⇒. Suppose that L is continuous and that U is an open subset of Y . We will show that

X \ L−1(U) is closed by showing that it contains all its limit points.Suppose that f is a limit point of X \L−1(U). Then there exist fn ∈ X \L−1(U) such that

fn → f . Since L is continuous, this implies Lfn → Lf . However, fn /∈ L−1(U), so Lfn /∈ U ,i.e., Lfn ∈ Y \ U , which is a closed set. Therefore Lf ∈ Y \ U , and hence f ∈ X \ L−1(U).Thus X \ L−1(U) is closed, so L−1(U) is open.

⇐. Exercise.

Theorem 2.10 (Equivalence of Bounded and Continuous Linear Operators). Let X, Y benormed linear spaces, and let L : X → Y be a linear mapping. Then the following statementsare equivalent.

(a) L is continuous at some f ∈ X.

(b) L is continuous at f = 0.

(c) L is continuous.

(d) L is bounded.

Proof. (c) ⇒ (d). Suppose that L is continuous but unbounded. Then ‖L‖ = ∞, sothere must exist fn ∈ X with ‖fn‖ = 1 such that ‖Lfn‖ ≥ n. Set gn = fn/n. Then‖gn − 0‖ = ‖gn‖ = ‖fn‖/n → 0, so gn → 0. Since L is continuous and linear, this impliesLgn → L0 = 0, and therefore ‖Lgn‖ → ‖0‖ = 0. But

‖Lgn‖ =1

n‖Lfn‖ ≥

1

n· n = 1

for all n, which is a contradiction. Hence L must be bounded.

(d) ⇒ (c). Suppose that L is bounded, so ‖L‖ < ∞. Suppose that f ∈ X and thatfn → f . Then ‖fn − f‖ → 0, so

‖Lfn − Lf‖ = ‖L(fn − f)‖ ≤ ‖L‖ ‖fn − f‖ → 0,

i.e., Lfn → Lf . Thus L is continuous.

The remaining implications are exercises.

Definition 2.11 (Isometries and Isometric Isomorphisms). Let X, Y be normed linearspaces and let L : X → Y be linear.


(a) If ‖Lf‖ = ‖f‖ for all f ∈ X then L is called an isometry or is said to be norm-

preserving.

(b) An isometry L : X → Y that is a bijection is called an isometric isomorphism. Inthis case we say that X and Y are isometrically isomorphic.

Exercise 2.12. (a) Suppose that L : X → Y is an isometry. Prove that L is injective andfind ‖L‖.

(b) Find an example of an isometry that is not surjective. Contrast this with the fact thatif A : Cn → Cn is linear, then A is injective if and only if it is surjective.

(c) Prove that if L : X → Y is an isometric isomorphism, then L−1 : Y → X is also anisometric isomorphism.

Exercise 2.13 (Unilateral Shift Operators). Fix 1 ≤ p ≤ ∞.

(a) Define L : `p(N) → `p(N) by L(x) = (x2, x3, . . . ) for x = (x1, x2, . . . ) ∈ `p(N). Provethat this left-shift operator is bounded, linear, surjective, not injective, and is not an isometry.Find ‖L‖.

(b) Define R : `p(N) → `p(N) by R(x) = (0, x1, x2, x3, . . . ) for x = (x1, x2, . . . ) ∈ `p(N).Prove that this right-shift operator is bounded, linear, injective, not surjective, and is anisometry. Find ‖R‖.

(c) Compute LR and RL. Contrast this computation with the fact that in finite dimen-sions, if A, B : Cn → Cn are linear maps (hence correspond to multiplication by n × nmatrices), then AB = I implies BA = I and conversely.

Definition 2.14 (Topological Isomorphisms). Let X, Y be normed linear spaces. If L : X →Y is a linear bijection such that both L and L−1 are bounded, then L is called a topological

isomorphism. In this case we say that X and Y are topologically isomorphic.

Remark 2.15. We will see later that if X and Y are Banach spaces and L : X → Yis a bounded bijection, then L−1 is automatically bounded and hence L is a topologicalisomorphism. Thus, when X and Y are Banach spaces, every continuous invertible map is atopological isomorphism. Sometimes the abbreviation isomorphism or invertible map is usedto mean a topological isomorphism, but it should be noted that these terms are ambiguous.

Exercise 2.16. Prove that if L : X → Y is a topological isomorphism, then

∀ f ∈ X,1

‖L−1‖‖f‖ ≤ ‖Lf‖ ≤ ‖L‖ ‖f‖.

Exercise 2.17. Let X be a Banach space and Y a normed linear space. Suppose thatL : X → Y is bounded and linear. Prove that if there exists c > 0 such that ‖Lf‖ ≥ c‖f‖for all f ∈ X, then L is injective and range(L) is closed.

16 CHRISTOPHER HEIL

The next theorem shows that B(X, Y ) is a Banach space whenever Y is a Banach space.

Theorem 2.18. If X is a normed space and Y is a Banach space, then B(X, Y ) is a Banachspace.

Proof. Assume that Ann∈N is a sequence of operators An ∈ B(X, Y ) that is Cauchy inoperator norm. For any given f ∈ X, we have

‖Amf − Anf‖ ≤ ‖Am − An‖ ‖f‖,

so we conclude that Anfn∈N is a Cauchy sequence of vectors in Y . Since Y is complete,this sequence must converge, say Afn → g ∈ Y . Define Af = g. This gives us a candidatelimit operator A.

Exercise: Show that A defined in this way is a linear operator.To show that A is bounded, first recall that all Cauchy sequences are bounded. Hence we

must have C = sup ‖An‖ <∞. If f ∈ X, then since Anf → Af we have

‖Af‖ = limn→∞

‖Anf‖ ≤ supn∈N

‖Anf‖ ≤ supn∈N

‖An‖ ‖f‖ = C ‖f‖.

Hence A is bounded, and ‖A‖ ≤ C.Finally, we must show that An → A in operator norm. Fix any ε > 0. Then there exists

an N such thatm,n > N =⇒ ‖Am − An‖ <

ε

2.

Choose any f ∈ X with ‖f‖ = 1. Then since Amf → Af , there exists an m > N such that

‖Af − Amf‖ <ε

2.

Hence for any n > N we have

‖Af−Anf‖ ≤ ‖Af−Amf‖+‖Amf−Anf‖ ≤ ‖Af−Amf‖+‖Am−An‖ ‖f‖ <ε

2+ε

2= ε.

Taking the supremum over all unit vectors, we conclude that ‖A − An‖ ≤ ε for all n > N .Thus An → A.

Corollary 2.19. If X is a normed space, then its dual space X ′ = B(X,F) is a Banachspace.

The next exercise deals with the problem of extending an operator defined only a densesubspace to the entire space.

Exercise 2.20 (Extension of Bounded Operators). Let Y be a dense subspace of a normedspace X, and let Z be a Banach space.

(a) Suppose that L : Y → Z is a bounded linear operator. Show that there exists aunique bounded linear operator L : X → Z whose restriction to Y is L. Prove that‖L‖ = ‖L‖.

(b) Prove that if L : Y → range(L) is a topological isomorphism, then L : X → range(L)is also.


(c) Prove that if L : Y → range(L) is an isometry, L : X → range(L) is also.

Solution(a) Fix any f ∈ X. Since Y is dense in X, there exist gn ∈ Y such that gn → f . Since

L is bounded, we have ‖Lgm − Lgn‖ ≤ ‖L‖ ‖gm − gn‖. But gnn∈N is Cauchy in X, so thisimplies that Lgnn∈N is Cauchy in Z. Since Z is a Banach space, we conclude that there

exists an h ∈ Z such that Lgn → h. Define Lf = h.To see that L is well-defined, suppose that we also had g′n → f for some g′n ∈ Y . Then

‖Lg′n−Lgn‖ ≤ ‖L‖ ‖g′n−gn‖ → 0. Since Lgn → h, it follows that Lg′n = Lgn+(Lg′n−Lgn) →h+ 0 = h. Thus L is well-defined.

To see that L is linear, suppose that f , g ∈ X are given and c ∈ F. Then there exist fn,gn ∈ Y such that fn → f and gn → g. Since cfn + gn → cf + g and

L(cfn + gn) = cLfn + Lgn → cLf + Lg,

by definition we have that L(cf + g) = cLf + Lg.

To see that L is an extension of L, suppose that g ∈ Y is fixed. If we set gn = g, thengn → g and Lgn → Lg, so by definition we have Lg = Lg. Hence the restriction of L to Yis L. Consequently,

‖L‖ = supf∈X, ‖f‖=1

‖Lf‖ ≥ supf∈Y, ‖f‖=1

‖Lf‖ = supf∈Y, ‖f‖=1

‖Lf‖ = ‖L‖.

Finally, suppose that f ∈ X. Then there exist gn ∈ Y such that gn → f and Lgn → Lf ,so

‖Lf‖ = limn→∞

‖Lgn‖ ≤ limn→∞

‖L‖ ‖gn‖ = ‖L‖ ‖f‖.

Hence ‖L‖ ≤ ‖L‖. Combining this with the opposite inequality derived above, we conclude

that ‖L‖ = ‖L‖.

(b) Suppose that L : Y → range(Y ) is a topological isomorphism. We already know thatL : X → range(L) is bounded. We need to show that L is injective, that L−1 : range(L) → X

is bounded, and that range(L) = range(L).

Fix any f ∈ X. Then there exist gn ∈ Y such that gn → f and Lgn → Lf . Since L is atopological isomorphism, we have by Exercise 2.16 that ‖gn‖ ≤ ‖L−1‖ ‖Lgn‖. Hence

‖Lf‖ = limn→∞

‖Lgn‖ ≥ limn→∞

‖gn‖

‖L−1‖=

‖f‖

‖L−1‖.

Consequently, L is injective and for any h ∈ range(L) we have

‖L−1h‖ ≤ ‖L−1‖ ‖L(L−1h)‖ = ‖L−1‖ ‖h‖.

Therefore L−1 : range(L) → X is bounded.It remains only to show that the range of L is the closure of the range of L. If f ∈ X, then

by definition there exist gn ∈ Y such that gn → f and Lgn → Lf . Hence Lf ∈ range(L), so

range(L) ⊂ range(L).

18 CHRISTOPHER HEIL

On the other hand, suppose that h ∈ range(L) Then there exist gn ∈ Y such that Lgn → h.

Since L−1 is bounded and L extends L, we conclude that gn = L−1(Lgn) → L−1(h). Hencef = L−1(h), so f ∈ range(L).

3. Finite-Dimensional Normed Spaces

In this section we will prove some basic facts about finite-dimensional spaces.First, recall that a finite-dimensional vector space has a finite basis, which gives us a

natural notion of coordinates of a vector, which in turn yields a linear bijection of X ontoFn for some n.

Example 3.1 (Coordinates). Let X be a finite-dimensional vector space over F. Then Xhas a finite basis, say B = e1, . . . , en. Every element of X can be written uniquely in thisbasis, say,

x = c1(x) e1 + · · ·+ cn(x) en, x ∈ X.

Define the coordinates of x with respect to the basis B to be

[x]B =

c1(x)...

cn(x)

.

Then the mapping T : X → Fn given by x 7→ [x]B is, by definition of basis, a linear bijectionof X onto Fn.

Since we already know how to construct many norms on Fn, by transferring these to Xwe obtain a multitude of norms for X.

Exercise 3.2 (`pw Norms on X). Let X be a finite-dimensional vector space over F and letB = e1, . . . , en be any basis. Fix any 1 ≤ p ≤ ∞ and any weight w : 1, . . . , n → (0,∞).

Using the notation of Example 3.1, given x ∈ X define

‖x‖p,w =∥

∥[x]B∥

∥

p,w=

( n∑

k=1

|ck(x)|p w(k)p

)1/p

, 1 ≤ p <∞,

maxk

|ck(x)|w(k), p = ∞.

Note that while we use the same symbol ‖ · ‖p,w to denote a function on X and on Fn, bycontext it has different meanings depending on whether it is being applied to an element ofX or to an element of Fn.

Prove the following.

(a) ‖ · ‖p,w is a norm on X.

(b) x 7→ [x]B is a isometric isomorphism of X onto Fn (using the norm ‖ · ‖p,w on X andthe norm ‖ · ‖p,w on Fn).


(c) Let xnn∈N be a sequence of vectors in X and let x ∈ X. Prove that xn → x withrespect to the norm ‖ · ‖p,w on X if and only if the coordinate vectors [xn]B convergecomponentwise to the coordinate vector [x]B.

(d) X is complete in the norm ‖ · ‖p,w.

Now we can show that all norms on a finite-dimensional space are equivalent.

Theorem 3.3. If X is a finite-dimensional vector space over F, then any two norms on Xare equivalent.

Proof. Let B = e1, . . . , en be any basis for X, and let ‖ · ‖∞ be the norm on X defined inExercise 3.2. Since equivalence of norms is an equivalence relation, it suffices to show thatan arbitrary norm ‖ · ‖ on X is equivalent to ‖ · ‖∞.

Using the notation of Exercise 3.1, given x ∈ X we can write x uniquely as x = c1(x) e1 +· · ·+ cn(x) en. Therefore,

‖x‖ ≤n

∑

k=1

|ck(x)| ‖ek‖ ≤

( n∑

k=1

‖ek‖

)

(

maxk

|ck(x)|)

= C2 ‖x‖∞,

where C2 =∑n

k=1 ‖ek‖ is a nonzero constant independent of x.It remains to show that there is a constant C1 > 0 such that C1 ‖x‖∞ ≤ ‖x‖ for every x.

First, letD = x ∈ X : ‖x‖∞ = 1

be the `∞-unit circle in X.Exercise: Show that D is compact (with respect to the norm ‖ · ‖∞). Hints: Suppose

that xnn∈N is a sequence of vectors in D. Then for each n, we have |ck(xn)| = 1 for somek ∈ 1, . . . , n. Hence there must be some k such that |ck(xnj

)| = 1 for infinitely many j.Since c1(xnj

)j∈N is an infinite sequence of scalars in the compact set c ∈ F : |c| ≤ 1, wecan select a subsequence whose first coordinates converge. Repeat for each coordinate, andremember that the kth coordinate is always 1. Hence we can construct a subsequence thatconverges to x ∈ D with respect to the `∞-norm.

Our next goal is to show that D is also compact with respect to the norm ‖ · ‖. Letxnn∈N be any sequence of vectors in D. Since D is compact with respect to ‖ · ‖∞,there exists a subsequence xnk

k∈N and an x ∈ D such that ‖x − xnk‖∞ → 0. But then

‖x− xnk‖ ≤ C2 ‖x− xnk

‖∞ → 0, so xnkk∈N is a subsequence that converges to x ∈ X with

respect to ‖ · ‖. Hence D is compact with respect to ‖ · ‖.Now, ‖ · ‖ is a continuous function with respect to the convergence criteria defined by ‖ · ‖

(this is part (b) of Exercise 1.11). The set D is compact with respect to the topology definedby ‖ · ‖. A real-valued continuous function on a compact set must achieve a maximum andminimum on that set. Hence, there must exist constants m and M such that m ≤ ‖x‖ ≤Mfor all x ∈ D. Since x ∈ D if and only if ‖x‖∞ = 1, this implies that

∀ x ∈ X, m ‖x‖∞ ≤ ‖x‖ ≤ M ‖x‖∞.

If we had m = 0, then this would imply that there is an x ∈ D such that ‖x‖ = 0. But thenx = 0, which implies ‖x‖∞ = 0, contradicting the fact that x ∈ D. Hence m > 0, so we cantake C1 = m.

20 CHRISTOPHER HEIL

Consequently, from now on we need not specify the norm on a finite-dimensional vectorspace X—we can take any norm that we like whenever we need it.

Exercise 3.4. Let Fm×n be the space of all m × n matrices with entries in F. Fm×n isnaturally isomorphic to Fmn.

Prove that if ‖ · ‖a is any norm on Fm×n, ‖ · ‖b is any norm on Fn×k, and ‖ · ‖c is any normon Fm×k, then there exists a constant C > 0 such that

∀A ∈ Fm×n, ∀B ∈ Fn×k, ‖AB‖c ≤ C ‖A‖a ‖B‖b.

Proposition 3.5. If M is a subspace of a finite-dimensional vector space X, then M isclosed.

Proof. Let ‖ · ‖ be any norm on X. Suppose that xn ∈M and that xn → y ∈ X. Set

M1 = spanM, y = m+ cy : m ∈M, c ∈ F.

Then M1 is finite-dimensional subspace of X. Moreover, every element z ∈ M1 can bewritten uniquely as z = m(z) + c(z)y where m(z) ∈ M and c(z) is a scalar. For z ∈ M1

define

‖z‖M1= ‖m(z)‖ + |c(z)|.

Exercise: Show that ‖ · ‖M1is a norm on M1.

Since ‖ ·‖ is also a norm on M1 and all norms on a finite-dimensional space are equivalent,we conclude that there is a constant C > 0 such that ‖z‖M1

≤ C ‖z‖ for all z ∈ M1. Sincec(xn) = 0 for every n, we therefore have

|c(y)| = |c(y) − c(xn)| = |c(y − xn)|

≤ ‖m(y − xn)‖ + |c(y − xn)|

= ‖y − xn‖M1

≤ C ‖y − xn‖ → 0.

Therefore c(y) = 0, so y ∈M .

If X is any normed vector space and M is a finite-dimensional subspace of X, then a proofidentical to the one used in the preceding proposition, except using the given norm ‖ · ‖ onX, shows that M is closed.

Exercise 3.6. Let X be a normed linear space. If M is a finite-dimensional subspace of X,then M is closed.

Exercise 3.7. Let X be a finite-dimensional normed space, and let Y be a normed linearspace. Prove that if L : X → Y is linear, then L is bounded.

The following lemma will be needed for Exercise 3.9.


Lemma 3.8 (F. Riesz’s Lemma). Let M be a proper, closed subspace of a normed space X.Then for each ε > 0, there exists g ∈ X with ‖g‖ = 1 such that

dist(g,M) = inff∈M

‖g − f‖ > 1 − ε.

Proof. Choose any u ∈ X \M . Since M is closed, we have

a = dist(u,M) = inff∈M

‖u− f‖ > 0.

Fix δ > 0 small enough that aa+δ

> 1− ε. By definition of infimum, there exists v ∈M such

that a ≤ ‖u− v‖ < a + δ. Set

g =u− v

‖u− v‖,

and note that ‖g‖ = 1. Given f ∈M we have h = v + ‖u− v‖ f ∈M , so

‖g − f‖ =

∥

∥

∥

∥

u− v − ‖u− v‖ f

‖u− v‖

∥

∥

∥

∥

=‖u− h‖

‖u− v‖>

a

a + δ> 1 − ε.

Exercise 3.9. Let X be a normed linear space. Let B = x ∈ X : ‖x‖ ≤ 1 be the closedunit ball in X. Prove that if B is compact, then X is finite-dimensional.

Hints: Suppose that X is infinite-dimensional. Given any nonzero e1 with ‖e1‖ ≤ 1, byLemma 3.8 there exists e2 ∈ X \spane1 with ‖e2‖ ≤ 1 such that ‖e2−e1‖ >

12. Continue in

this way to construct vectors ek such that e1, . . . , en are independent for any n. Concludethat X is infinite-dimensional.

Definition 3.10. We say that a normed linear space X is locally compact if for each f ∈ Xthere exists a compact K ⊂ X with nonempty interior K such that f ∈ K.

In other words, X is locally compact if for every f ∈ X there is a neighborhood of f thatis contained in a compact subset of X. For example, Fn is locally compact.

With this terminology, we can reword Exercise 3.9 as follows.

Exercise 3.11. Let X be a normed linear space.

(a) Prove that if X is locally compact, then X is finite-dimensional.

(b) Prove that if X is infinite-dimensional, then no nonempty open subset of X hascompact closure.

22 CHRISTOPHER HEIL

4. Quotients and Products of Normed Spaces

Any vector space is an abelian group under the operation of vector addition. So, if you arefamiliar with the basic notions of abstract algebra, the concept of a coset will be familiar toyou. However, even if you have not studied abstract algebra, the idea of a coset in a vectorspace is very natural.

Example 4.1 (Cosets in R2). Consider the vector space X = R2. Let M be any one-dimensional subspace of R2, i.e., M is a line in R2 through the origin. A coset of M issimply a rigid translate of M by a vector in R2. For concreteness, let us specifically considerthe case where M is the x1-axis in R2, i.e., M = (x1, 0) : x1 ∈ R. Then given a vectory = (y1, y2) ∈ R2, the coset y +M is the set

y +M = y +m : m ∈M = (y1 + x1, y2 + 0) : x1 ∈ R = (x1, y2) : x1 ∈ R,

which is the horizontal line at height y2. This is not a subspace of R2, but it is a rigidtranslate of the x1-axis. Note that there are infinitely many different choices of y that givethe same coset. Furthermore, we have the following facts for this particular setting.

(a) Two cosets are either identical or entirely disjoint.

(b) The union of all the cosets is all of R2.

(c) The set of distinct cosets is a partition of R2.

The preceding example is entirely typical.

Definition 4.2 (Cosets). Let M be a subspace of a vector space X. Then the cosets of Mare the sets

f +M = f +m : m ∈M, f ∈M.

Exercise 4.3. Let X be a vector space, and let M be a subspace of X. Given f , g ∈ M ,define f ∼ g if f − g ∈M . Prove the following.

(a) ∼ is an equivalence relation on X.

(b) The equivalence class of f under the relation ∼ is [f ] = f +M .

(c) If f , g ∈M then either f +M = g +M or (f +M) ∩ (g +M) = ∅.

(d) f +M = g +M if and only if f − g ∈M .

(e) f +M = M if and only if f ∈M .

(f) If f ∈ X and m ∈M then f +M = f +m+M .

(g) The set of distinct cosets of M is a partition of X.

Definition 4.4 (Quotient Space). If M is a subspace of a vector space X, then the quotientspace X/M is

X/M = f +M : f ∈ X.


Since two cosets of M are either identical or disjoint, the quotient space X/M is simplythe set of all the distinct cosets of M .

Example 4.5. Again let M = (x1, 0) : x1 ∈ R be the x1-axis in R2. Then, by Example 4.1,we have that

R2/M = y +M : y ∈ R2 = (x1, 0) +M : x1 ∈ R,

i.e., R2/M is the set of all horizontal lines in R2. Note that R2/M is in 1-1 correspondencewith the set of distinct heights, i.e., there is a natural bijection of R2/M onto R. This is aspecial case of a more general fact that we will explore.

Next we define two natural operations on the set of cosets: addition of cosets and multi-plication of a coset by a scalar. These are defined formally as follows.

Definition 4.6. Let M be a subspace of a vector space X. Given f , g ∈ X, define additionof cosets by

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

Given f ∈ X and c ∈ F, define scalar multiplication by

c(f +M) = cf +M.

Remark 4.7. Before proceeding, we must show that these operations are actually well-defined. After all, there need not be just one f that determines the coset f + M—how dowe know that if we choose different vectors that determine the same cosets, we will get thesame result when we compute (f + g) + M? We must show that f1 + M = f2 + M andg1 +M = g2 +M then (f1 + g1) +M = (f2 + g2) +M in order to know that Definition 4.6makes sense.

Proposition 4.8. If M is a subspace of a vector space X, then the addition of cosets of Mgiven in Definition 4.6 is well-defined.

Proof. Suppose that f1 + M = f2 + M and g1 + M = g2 + M . Then by Exercise 4.3(d)we know that f1 − f2 = k ∈ M and g1 − g2 = l ∈ M . If h ∈ (f1 + g1) + M then we haveh = f1 + g1 +m for some m ∈M . Hence

h = (f2 + k) + (g2 + l) +m = (f2 + g2) + (k + l +m) ∈ (f2 + g2) +M.

Thus (f1 + g1) +M ⊂ (f2 + g2) +M , and the converse inclusion is symmetric.

Exercise 4.9. Show that scalar multiplication is likewise well-defined.

Now we can show that the quotient space is actually a vector space under the operationsjust defined.

Proposition 4.10. If M is a subspace of a vector space X, then X/M is a vector spacewith respect to the operations given in Definition 4.6.

24 CHRISTOPHER HEIL

Proof. Addition of cosets is commutative because

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

The zero vector in X/M is the coset 0+M = M , because (f+M)+(0+M) = (f+0)+M =f +M .

Exercise: Show that the remaining axioms of a vector space are satisfied.

Definition 4.11 (Codimension). If M is a subspace of a vector space X, then the codimen-sion of M is the dimension of X/M , i.e.,

codim(M) = dim(X/M).

Example 4.12. Let C(R) be space of continuous functions on R, and let P be the sub-space containing the polynomials. Given f ∈ C(R), the coset f + P is f + P = f + p :p is a polynomial. Further, f +P = g+P if and only if f − g is a polynomial. Thus, f +Pcan be thought of as “f modulo the polynomials,” i.e., it is the equivalence class obtainedby identifying functions which differ by a polynomial.

In the same way, a coset f + M can be thought of as the equivalence class obtained byidentifying vectors which differ by an element of M . We can imagine the mapping that takesf to f +M as “collapsing information modulo M .”1

Definition 4.13. If M is a subspace of a vector space X, then the canonical projection orthe canonical mapping of X onto X/M is π : X → X/M defined by

π(f) = f +M, f ∈ X.

Exercise 4.14. Let M be a subspace of a vector space X.

(a) Prove that the canonical projection π is linear.

(b) Prove that π is surjective and ker(π) = M .

(c) Prove that if E ⊂ X, then the inverse image of π(E) is

π−1(

π(E))

= E +M = u+m : u ∈ E,m ∈M.

Solution(c) Suppose that u ∈ E and m ∈M are given. Then

π(u+m) = u+m +M = u+M = π(u) ∈ π(E).

Hence u+m ∈ π−1(

π(E))

.

Now suppose that v ∈ π−1(

π(E))

. Then, by definition, π(v) ∈ π(E) = u+M : u ∈ E.Hence v + M = π(v) = u + M for some u ∈ E. But then m = v − u ∈ M , so v = u + mwith u ∈ E and m ∈M .

1Conway calls this map Q, but I prefer to call it π for “projection.”


We will mostly be interested in the case where X is a normed space. The following resultshows that X/M is a semi-normed space in general, and is a normed space if M is closed.

Proposition 4.15. Let M be a subspace of a normed linear space X. Given f ∈ X, define

‖f +M‖ = dist(f,M) = infm∈M

‖f −m‖.

Then the following statements hold.

(a) ‖ · ‖ is well-defined.

(b) ‖ · ‖ is a semi-norm on X/M .

(c) If M is closed, then ‖ · ‖ is a norm on X/M .

Proof. (a) Exercise.Hint: Show that if f1 +M = f2 +M then f1 −m : m ∈M = f2 −m : m ∈M.

(b) Exercise.

(c) Suppose that M is closed, and that ‖f +M‖ = 0. Then infm∈M ‖f −m‖ = 0. Hencethere exist vectors gn ∈ M such that ‖f − gn‖ → 0 as n → ∞. But M is closed, so thisimplies f ∈ M . By Exercise 4.3(e), we therefore have f + M = M = 0 + M , which is thezero vector in X/M .

Now we derive some basic properties of the canonical projection π of X onto X/M .

Proposition 4.16. Let M be a closed subspace of a normed linear space X. Then thefollowing statements hold.

(a) ‖π(f)‖ = ‖f +M‖ ≤ ‖f‖ for each f ∈ X.

(b) Let BXr (f) denote the open ball of radius r in X centered at f , and let B

X/Mr (f +M)

denote the open ball of radius r in X/M centered at f + M . Then for any f ∈ Xand r > 0 we have

π(

BXr (f)

)

= BX/Mr (f +M).

(c) W ⊂ X/M is open in X/M if and only if π−1(W ) = f ∈ X : f +M ∈ W is openin X.

(d) π is an open mapping, i.e., if U is open in X then π(U) is open in X/M .

Proof. (a) Choose any f ∈ X. Since 0 is one of the elements of M , we have

‖π(f)‖ = ‖f +M‖ = infm∈M

‖f −m‖ ≤ ‖f − 0‖ = ‖f‖.

(b) First consider the case f = 0 and r > 0. Suppose that g + M ∈ π(

BXr (0)

)

. Theng +M = h +M for some h ∈ BX

r (0), i.e., ‖h‖ < r. Hence ‖g +M‖ = ‖h +M‖ ≤ ‖h‖ < r,

so g +M ∈ BX/Mr (0 +M).

Now suppose that g +M ∈ BX/Mr (0 +M). Then infm∈M ‖g −m‖ = ‖g +M‖ < r. Hence

there exists m ∈M such that ‖g −m‖ < r. Thus g −m ∈ BXr (0), so

g +M = g −m+M = π(g −m) ∈ π(

BXr (0)

)

.

26 CHRISTOPHER HEIL

Exercise: Show that statement (b) holds for an arbitrary f ∈ X.

(c) ⇒. Part (a) implies that π is continuous. Hence π−1(W ) must be open in X if W isopen in X/M .

⇐. Suppose that W is a subset of X/M such that π−1(W ) is open in X. We must showthat W is open in X/M . Choose any point f +M ∈ W . Then f ∈ π−1(W ), which is openin X. Hence, there exists an r > 0 such that BX

r (f) ⊂ π−1(W ). By part (b) we thereforehave

BX/Mr (f +M) = π

(

BXr (f)

)

⊆ π(

π−1(W ))

= W.

Therefore W is open.

(d) Suppose that U is an open subset of X. Then by Exercise 4.14(c), we have

π−1(

π(U))

= U +M = u+m : u ∈ U,m ∈M =⋃

m∈M

(U +m).

But each set U +m, being the translate of the open set U , is itself open. Hence π−1(

π(U))

is open, since it is a union of open sets. Part (c) therefore implies that π(U) is open inX/M .

Exercise 4.17. Let M be a closed subspace of a normed space X, and let π be the canonicalprojection π of X onto X/M . Prove that ‖π‖ = 1.

Hint: Lemma 3.8.

Now we can prove that ifX is a Banach space, then X/M inherits a Banach space structurefrom X.

Theorem 4.18. If M is a closed subspace of a Banach space X, then X/M is a Banachspace.

Proof. We have already shown that X/M is a normed space, so we must show that it iscomplete in that norm.

Suppose that fn +Mn∈N is a Cauchy sequence in X/M . It would be convenient if thisimplies that fnn∈N is a Cauchy sequence in X, but this need not be the case. For, thevectors fn are not unique in general: if we replace fn by any vector fn + m with m ∈ M ,then we obtain the same coset. We will show that by choosing an appropriate subsequencefnk

k∈N and replacing the fnkby appropriate vectors that determine the same cosets fnk

+M ,we can create a sequence in X that is Cauchy and hence converges, and then use this toshow that the original sequence of cosets fn +Mn∈N converges in X/M .

We begin by applying Exercise 1.18: there exists a subsequence fnk+Mk∈N such that

∀ k ∈ N, ‖(fnk+1− fnk

) +M‖ = ‖(fnk+1+M) − (fnk

+M)‖ < 2−k.

Now we seek to create vectors gk ∈ M so that fnk−gkk∈N will converge in X. Note that

the cosets determined by fnkand by fnk

− gk are identical.Set g1 = 0. Then

infg∈M

‖(fn1− g1) − (fn2

− g)‖ = infg∈M

‖(fn1− fn2

) + g‖ = ‖(fn1− fn2

) +M‖ <1

2.


Therefore, there exists a g2 ∈M such that

‖(fn1− g1) − (fn2

− g2)‖ < 2 ·1

2.

Then, since g2 ∈M ,

infg∈M

‖(fn2− g2) − (fn3

− g)‖ = infg∈M

‖(fn2− fn3

) + g‖ = ‖(fn2− fn3

) +M‖ <1

22.

Therefore, there exists a g3 ∈M such that

‖(fn2− g2) − (fn3

− g3)‖ < 2 ·1

22.

Continuing in this way, by induction we construct hk = fnk− gk such that

‖hk − hk+1‖ <1

2k−1.

Exercise 1.12 therefore implies that hkk∈N is a Cauchy sequence in X. Since X is complete,this sequence converges, say hk → h.

Since

‖(fnk+M) − (h+M)‖ = ‖fnk

− gk − h+M‖ (since gk ∈M)

= ‖hk − h+M‖

≤ ‖hk − h‖ → 0,

we see that fnk+Mk∈N is a convergent subsequence of fn +Mn∈N. Since we know that

fn +Mn∈N is Cauchy, it follows from Exercise 1.17 that fnk+ M → h +M . Thus X/M

is complete.

The following exercise shows that the converse of the preceding theorem is true as well.

Exercise 4.19. Let M be a closed subspace of a normed space X. Prove that if M andX/M are both complete, then X must be complete.

Hints: Suppose that fnn∈N is a Cauchy sequence in X. Show that fn + Mn∈N is aCauchy sequence in X/M , hence converges to some coset f +M . Thus ‖f − fn +M‖ → 0.Does this imply that there exists a g ∈M such that ‖f − fn + g‖ → 0? Or perhaps vectorsgn ∈M such that ‖f − fn + gn‖ → 0?

The quotient space and canonical map will be useful tools for proving many later results.The following proof illustrates their utility.

Proposition 4.20. Let X be a normed linear space. If M is a closed subspace of X and Nis finite-dimensional, then M +N is a closed subspace of X.

Proof. Let π be the canonical projection of X onto X/M . Since N is finite-dimensional, ithas a finite basis, say e1, . . . , en. Then since π is linear,

π(N) = π(

spane1, . . . , en)

= spanπ(e1), . . . , π(en) = spane1 +M, . . . , en +M.

28 CHRISTOPHER HEIL

Thus π(N) is a finite-dimensional subspace of X/M , and therefore is closed by Proposi-tion 3.6. Since π is continuous, it follows that π−1(π(N)) is closed in X. However, byExercise 4.14(c), we have π−1(π(N)) = M +N .

Exercise 4.21. Let M be a closed subspace of a normed linear space X.

(a) Prove that if X is separable, then X/M is separable.

(b) Prove that if X/M and M are both separable, then X is separable.

(c) Give an example of X, M such that X/M is separable, but X is not separable.

Solution(b) Let fn +Mn∈N be a countable dense subset of X/M , and let gnn∈N be a countable

dense subset of M . Then S = fm + gnm,n∈N is a countable subset of X, and we claim thatit is dense.

To see this, fix any f ∈ X. Then there exists an m such that

infh∈M

‖f − fm + h‖ = ‖f − fm +M‖ = ‖(f +M) − (fm +M)‖ <ε

2.

Hence, there exists some h ∈ M such that ‖f − fm + h‖ < ε2. Since h ∈ M , there exists an

n such that ‖h− gn‖ <ε2. Therefore

‖f − (fm + gn)‖ ≤ ‖f − fm − h‖ + ‖h− gn‖ <ε

2+ε

2= ε.

Exercise 4.22. Recall from Exercise 1.37 that c0 is a closed subspace of the Banach space `∞.

(a) Prove that `∞ is not separable.Hint: Consider

S = (x1, x2, . . . ) : xk = 0 or 1 for every k.

What is the distance between two distinct elements of S?

(b) Let x, y be vectors in S Prove that if xk 6= yk for at most finitely many k, thenx + c0 = y + c0. Prove that if xk 6= yk for infinitely many k, then ‖x− y + c0‖ = 1.

(c) Use part (b) to prove directly that `∞/c0 is not separable.

(e) Prove that the standard basis enn∈N is a Schauder basis for c0 (with respect to the`∞-norm).

(f) Use part (e) to show that c0 is separable.Hint: Use one of the exercises about Schauder bases from the Chapter 1 lecture notes.

(g) Use part (e) and Exercise 5.4 to show that `∞/c0 is not separable.

In the Hilbert space case, there is a very close relationship between π and the orthogonalprojection of H onto M⊥.


Exercise 4.23. Let M be a closed subspace of a Hilbert space H, and let π be the canonicalprojection ofH ontoH/M . Prove that the restriction of π toM⊥ is an isometric isomorphismof M⊥ onto H/M .

Remark 4.24. There is no analog of the preceding result for arbitrary Banach spaces. If Xis a Banach space and M is a closed subspace then we say that M is complemented in X ifthere exists another closed subspace N such that M ∩N = 0 and M +N = X.

It is not true that every closed subspace of every Banach space is complemented. Inparticular, c0 is not complemented in `∞. Also, if 1 < p ≤ ∞ and p 6= 2, then `p hasuncomplemented subspaces.

Next we will define the product or direct sum of normed spaces. An analogous definitionholds for the case of a finite collection of spaces.

Definition 4.25. Let Xii∈N be a countable family of Banach spaces, and let ‖ · ‖i denotethe norm on Xi. Define

∞∏

i=1

Xi =

f = (f1, f2, . . . ) : fi ∈ Xi

.

For 1 ≤ p <∞, define

⊕

pXi =

f ∈∞∏

i=1

Xk : ‖f‖p =

( ∞∑

i=1

‖fi‖pi

)1/p

<∞

.

For p = ∞, define

⊕

∞Xi =

f ∈∞∏

i=1

Xk : ‖f‖∞ = supi

‖fi‖i <∞

.

Exercise 4.26. Let Xii∈N be a countable family of Banach spaces and fix 1 ≤ p ≤ ∞.Let X =

⊕

pXi. Prove the following.

(a) X is a normed space.

(b) For each i, the projection Pi : X → Xi given by Pi(f1, f2, . . . ) = fi is continuous, and‖Pi‖ = 1.

(c) X is a Banach space if and only if each Xi is a Banach space.

Exercise 4.27. Let X1, . . . , Xn be finitely many normed spaces. Prove that the spaces ⊕pXi

are equal for 1 ≤ p ≤ ∞, and that all the norms ‖ · ‖p are equivalent. For this reason, weoften denote this space by X1 × · · · ×Xn.

30 CHRISTOPHER HEIL

Exercise 4.28. Let X and Y be normed vector spaces. Define T : B(X, Y ) × X → Y byT (A, f) = Af .

(a) Prove that T is continuous if and only if An → A and fn → f implies Anfn → Af .

(b) Prove that T is continuous. Conclude that T ∈ B(B(X, Y ) ×X, Y ).

5. Linear Functionals

Definition 5.1 (Hyperplane). Let M be a subspace of a vector space X. Then we say thatM is a hyperplane if it has codimension 1, i.e., if codim(M) = dim(X/M) = 1.

Example 5.2. Let H be a Hilbert space, and let L be a bounded linear functional on H.That is, L : H → F is a bounded linear operator, so L∗ : F → H is also a bounded linearoperator. Now, F is one-dimensional, and a linear operator is entirely determined by whatit does to the basis elements. In particular, 1 is a basis for F, so L∗(c) = L∗(c1) = c L∗(1).Hence range(L∗) = spanL∗(1). If L∗(1) = 0 then L∗ is the zero operator, and hence L isthe zero operator as well (why?). Otherwise, range(L∗) must be one-dimensional, and henceis a closed subspace of H.

Therefore,

H = range(L∗) ⊕ range(L∗)⊥ = range(L∗) ⊕ ker(L).

Hence, every vector f ∈ H can be written uniquely as

f = c L∗(1) + k

for some scalar c ∈ range(L∗) and k ∈ ker(L). Hence, if π is the canonical projection of Honto H/ ker(L), then since k ∈ ker(L) we have

f + ker(L) = π(f) = cL∗(1) + k + ker(L) = cL∗(1) + ker(L).

Therefore

H/ ker(L) = f + ker(L) : f ∈ H = cL∗(1) + k + ker(L) : c ∈ F, k ∈ ker(L)

= cL∗(1) + ker(L) : c ∈ F

= spanL∗(1) + ker(L),

which is one-dimensional.Thus, if L is a bounded linear functional on a Hilbert space H, then we conclude that

ker(L) is a hyperplane in H.

We will show that a similar result holds for arbitrary normed spaces. We will need thefollowing tool. In abstract algebra, the group version of the next result is called the FirstIsomorphism Theorem or the Homomorphism Theorem. In the group setting, an isomor-phism is a bijective homomorphism. In the vector space setting, an isomorphism is a linearbijection.


Exercise 5.3 (Isomorphism Theorem). Let X and Y be vector spaces, and let ϕ : X → Ybe a linear surjection. Let M = ker(ϕ). Prove that ψ : X/M → Y given by ψ(f+M) = ϕ(f)is a well-defined linear bijection, and that ϕ = ψ π.

Exercise 5.4. Fix 1 ≤ p ≤ ∞, and set

M = x ∈ `p : x(2k) = 0 for every k.

Prove that M is a closed subspace of `p, and that `p/M is isometrically isomorphic to `p.

Now we can show that every hyperplane is the kernel of some (not necessarily continuous)linear functional.

Proposition 5.5. Let M be a subspace of a normed space X. Then the following statementsare equivalent.

(a) M is a hyperplane.

(b) M = ker(µ) for some nonzero linear functional µ : X → F.

Proof. (b) ⇒ (a). Suppose that M = ker(µ) where µ is a nonzero linear functional. Thenby the Isomorphism Theorem, there is a linear bijection ψ : X/ ker(µ) → F. Since F isone-dimensional, we conclude that X/ ker(µ) is as well.

(a) ⇒ (b). Assume that M is a hyperplane. Then X/M is one-dimensional, so thereexists a linear bijection ψ : X/M → F. Set ϕ = ψ π, then ψ : X → F. If f ∈ M , thenπ(f) = f + M = 0 + M , so ψ(f) = ψ(π(f)) = ψ(0 + M) = 0, so f ∈ ker(ϕ). Conversely,if f ∈ ker(ϕ) then we have ψ(π(f)) = ϕ(0 + M) = 0. But ψ is a bijection, so this impliesf +M = π(f) = 0 +M . Hence f ∈M , so ker(ϕ) ⊂M .

Proposition 5.6. Let X be a normed linear space. If µ, ν : X → F are nonzero linearfunctionals, then

ker(µ) = ker(ν) ⇐⇒ µ = cν for some nonzero scalarc.

Proof. ⇐. Trivial.

⇒. Suppose that ker(µ) = ker(ν). Since µ 6= 0, there exists some f ∈ X such thatµ(f) 6= 0, and by rescaling, we can assume that µ(f) = 1. Since f /∈ ker(µ) = ker(ν), wehave ν(f) 6= 0. Given any g ∈ X, we have

µ(

g − µ(g)f)

= µ(g) − µ(g) · µ(f) = 0,

so g − µ(g)f ∈ ker(µ) = ker(ν). Therefore

ν(g) − µ(g) · ν(f) = ν(

g − µ(g)f)

= 0,

so after rearranging we see that ν = ν(f)µ.

Proposition 5.7. If M is a hyperplane in a normed linear space X, then M is either closedor is dense in X.

32 CHRISTOPHER HEIL

Proof. We are given that X/M is one-dimensional. Let π be the canonical projection of Xonto X/M . The closure M of M is a subspace of X, so since π is linear we know that π(M)is a subspace of X/M . But X/M is one-dimensional, so there only two possibilities.

First, we could have π(M) = 0+M. In this case, M ⊂ ker(π) = M , so we have M = Mand M is closed.

Second, we could have π(M) = X/M . In this case, we have by Exercise 4.14(c) that

X = π−1(X/M) = π−1(π(M)) = M +M = M,

and thus M is dense.

Proposition 5.8. Let µ : X → F be a linear functional on a normed space X. Then:

µ is continuous ⇐⇒ ker(µ) is closed.

Proof. ⇒. Exercise.

⇐. Suppose that ker(µ) is closed. We know by Proposition 5.5 that ker(µ) is a hyperplane,so X/ ker(µ) is one-dimensional. Let π be the canonical projection of X onto X/ ker(µ).Because M is closed, we know that π is continuous. By the Isomorphism Theorem, thereexists a linear bijection ψ : X/ ker(µ) → F. Since X/ ker(µ) is a one-dimensional normedlinear space, and linear map from X/ ker(µ) into another normed space is continuous byExercise 3.7. Therefore ψ is continuous, and hence µ = ψ π is continuous.

Recall now the definition of the dual space of a normed space X:

X∗ = X ′ = B(X,F) = L : X → F : L is bounded and linear.

Since the norm on F is just absolute value, the operator norm of a linear functional L ∈X ′ = B(X,F) is

‖L‖ = sup‖f‖=1

|Lf |.

Since F is a Banach space, the dual space X ′ is a Banach space even if X is not.In order to give some examples of dual spaces, we recall Holder’s Inequality.

Theorem 5.9 (Holder’s Inequality). Let (X,Ω, µ) be a measure space. If 1 ≤ p ≤ ∞ and1p

+ 1p′

= 1, then

∀ f ∈ Lp(X), ∀ g ∈ Lp′

(X), ‖fg‖1 ≤ ‖f‖p ‖g‖p′.

In particular, if f ∈ Lp(X) and g ∈ Lp′

(X), then fg ∈ L1(X).


Appendix A. Appendix: Topological and Metric Spaces

Definition A.1 (Topological Space). A topological space (X, T ) is a nonempty set X to-gether with a family T of subsets of X such that the following statements hold.

(a) ∅, X ∈ T .

(b) Closure under finite intersections: If U , V ∈ T , then U ∩ V ∈ T .

(c) Closure under arbitrary unions: If I is any index set and Ui ∈ T for i ∈ I, then∪i Ui ∈ T .

We call T a topology on X. The elements of T are the open subsets of X.

Definition A.2 (Metric Space). Let X be a nonempty set. A metric on X is a functiond(·, ·) → R such that

(a) d(f, g) ≥ 0 for all f , g ∈ X,

(b) d(f, g) = 0 if and only if f = g,

(c) Triangle Inequality: d(f, h) ≤ d(f, g) + d(g, h) for all f , g, h ∈ X.

A space X together with a metric d(·, ·) is called a metric space.

Definition A.3 (Topology on a Metric Space). Let X be a metric space.

(a) The open ball in X centered at x ∈ X with radius r > 0 is

Br(x) = B(x, r) = y ∈ X : d(x, y) < r.

(b) A subset U ⊆ X is open if

∀ x ∈ U, ∃ r > 0 such that Br(x) ⊆ U.

(c) The topology on X is T = U ⊆ X : U is open.

Exercise A.4. Prove that if X is a metric space, then (X, T ) is a topological space usingthe preceding definition.

functional analysis lecture notes chapter 3. banach...

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