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 a 

function · : 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} if 

finite, or I = N or Z if infinite). Let w : I → [0, ∞). Given a sequence of scalars x = (xi)i∈I, 

set 

xp,w = 

⎧ 

 

⎪⎨ |xi| 

i∈I 

⎪⎩ 

p w(i) p 

1/p , 

sup |xi| w(i), 

i∈I 

0 

p = ∞, 

where these quantities could be infinite. Then we set 

 

 

x = (xi)i∈I : xp < ∞ . 

ℓ p w (I) = 

We call ℓ p w(I) a weighted ℓ p space, and often denote it just by ℓ p w (especially if I = N). If 

w(i) = 1 for all i, then we simply call this space ℓ p (I) or ℓ 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 ℓp w, and it is a 

norm if w(i) > 0 for all i. 

In particular, if I = {1, . . . , n} then ℓp w = Fn , and each choice of p and w gives a semi-norm 

or norm on Fn . 

Hints: The Triangle Inequality on ℓp is often called Minkowski’s Inequality. It is easy to 

prove if p = 1 or p = ∞. There are several ways to prove it for other p. One ways is to begin 

with 

 

= |xi + yi| p = 

|xi + yi| p−1 |xi + yi| 

x + y p p 

i∈I 

i∈I 

≤ 

|xi + yi| p−1 |xi| + 

|xi + yi| p−1 |yi|. 

i∈I 

Then apply Hölder’s Inequality to each sum using the exponent p ′ on the first factor and p 

for 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(x)| 

fp = X 

⎪⎩ 

p 1/p dµ(x) , 0 

ess sup |f(x)|, 

x∈X 

p = ∞, 

where these quantities could be infinite. Define 

L p 

 

(X) = f : X → [−∞, ∞] or C : fp < ∞ . 

Other notations for L p (X) are L p (µ), L p (X, µ), L p (dµ), L p (X, dµ), etc. 

When we write L p (R n ), it will be assumed that µ is Lebesgue measure on R n , unless 

specifically stated otherwise. In this case we will write dx instead of dµ(x). 

The space ℓ p w (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 L p (X), and it is a norm 

if we identify functions that are equal almost everywhere. 

The Triangle Inequality on L p is often called Minkowski’s Inequality, and its proof is similar 

to the proof of Minkowski’s Inequality for ℓ p . 

Exercise 1.7. Show that every subspace of a normed space is itself a normed space (using 

the same norm). 

Definition 1.8 (Distance). Let · be a norm on X. Then the distance from x to y in X 

is d(x, y) = x − y. 

i∈I

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 topological 

notions (open/closed sets, convergence, etc.) apply to X. For convenience, we give some 

explicit 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 product 

space), and let {fn}n∈N be a sequence of elements of X. 

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

i.e., 

lim 

n→∞ f − fn = 0, 

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

(b) We say that {fn}n∈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 {fn}n∈N be a sequence of vectors in a normed space X. Show that if 

fn − fn+1 < 2 −n for every n, then {fn}n∈N is Cauchy. 

Exercise 1.13. Let ℓ p w (I) be the weighted ℓp space defined in Exercise 1.3, where we assume 

w(i) > 0 for all i ∈ I. Let {xn}n∈N be a sequence of vectors in ℓ p w(I), and let x be a vector in 

ℓ p w(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 − xnp,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 convergence 

implies convergence with respect to the norm · p,w. 

(c) Prove that if I is infinite then componentwise convergence does not imply convergence 

in the norm of ℓ w p (I).


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 a 

special name. 

Definition 1.14 (Banach Space). A normed space X is called a Banach space if it is 

complete, i.e., if every Cauchy sequence is convergent. That is, 

{fn}n∈N is Cauchy in X =⇒ ∃ f ∈ X such that fn → f. 

Exercise 1.15. Show that the weighted ℓp space ℓp w (I) defined in Exercise 1.3 is a Banach 

space if w(i) > 0 for all i ∈ I. 

Hints: Consider the case I = N. Suppose that {xn}n∈N is a Cauchy sequence in ℓp w 

xn is a sequence of scalars. Write out the components of xn as 

Prove that for a fixed component k we have 

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

|xm(k) − xn(k)| ≤ C xm − xnp,w, 

. Each 

where C is a fixed constant (determined by the weight and by k but independent of m and 

n). Conclude that with k fixed, {xn(k)}n∈N is a Cauchy sequence of scalars, and hence converges. 

Define x(k) = limn→∞ xn(k) and set x = (x(1), x(2), . . . ). Then we have constructed 

a candidate limit for the sequence {xn}n∈N. However, so far we only have that each individual 

component of xn converges to the corresponding component of x, i.e., xn converges 

componentwise to x. This is not enough: to complete the proof you must show that xn → x 

in the norm of ℓ p w . Use the fact that {xn}n∈N is Cauchy together with the componentwise 

convergence to show that x − xn ℓ 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 if 

we 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 

ℓp w (I) is a Banach space. First find a candidate limit and then show that the sequence 

converges 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 {fn}n∈N is a Cauchy sequence in X and 

there exists a subsequence {fnk }k∈N that converges to f ∈ X, then fn → f. 

Solution 

Choose any ε > 0. Since {fn}n∈N is Cauchy, there is an N such that fm − fn < ε for m, 

2 

ε 

n > N. Also, there is a k such that nk > N and f − fnk < . Hence for n > N we have 

2 

f − fn ≤ f − fnk + fnk − fn < ε ε 

+ = ε. 

2 2


Thus fn → f. 

Exercise 1.18. Let {fn}n∈N be a Cauchy sequence in a normed space X. Show that there 

exists a subsequence {fnk }k∈N such that 

∀ k ∈ N, fnk+1 − fnk < 2−k . 

Solution 

Since {fn}n∈N is Cauchy, we can find an n1 such that 

Then we can find an n2 > n1 such that 

m, n ≥ n1 =⇒ fm − fn < 2 −1 . 

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 {fn}n∈N be a sequence of elements of a normed 

linear 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 = 

N 

 

f − 

 

 

→ 0 as N → ∞. 

n=1 

fn 

Definition 1.20 (Absolutely Convergent Series). Let X be a normed space and let {fn}n∈N 

be a sequence of elements of X. If 

∞ 

fn < ∞, 

then we say that the series 

n fn is absolutely convergent in X. 

n=1 

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 true 

if X is not complete. 

Exercise 1.21. Let X be a Banach space and let {fn}n∈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 {sN}N∈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 a 

given normed space is a Banach space.


Proposition 1.22. Let X be a normed space. Prove that X is a Banach space if and only 

if every absolutely convergent series in X converges in X. 

Proof. ⇒. This is Exercise 1.21. 

⇐. Suppose that every absolutely convergent series is convergent. Let {fn}n∈N be a 

Cauchy 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 

(fnk+1 k − fnk ) is absolutely convergent, 

because 

∞ 

− fnk ≤ 

∞ 

2 −k = 1 < ∞. 

fnk+1 

k=1 

k=1 

By hypothesis, we conclude that 

(fnk+1 k − fnk ) converges in X, say to f. In terms of the 

partial sums, this says that 

fnk+1 − fn1 = 

k 

j=1 

(fnj+1 

− fnj ) → f as k → ∞, 

or fnk → g = f +fn1 as k → ∞. Thus {fn}n∈N is a Cauchy sequence which has a subsequence 

that converges to g. It therefore follows from Exercise 1.17 that fn → g. Therefore X is 

complete. 

Example 1.23 (The Harmonic Series). To illustrate the difference between convergence and 

absolute convergence, consider the one-dimensional case, i.e., X = F, the field of scalars. Let 

xn = 1 

. Then n n xn = 1 

n n is the harmonic series, and the partial sums sN = N n=1 

this series are unbounded. Thus the harmonic series does not converge. Since sN → ∞, we 

usually say that 

 

1 

1 

n diverges to infinity, and write n n = ∞. n 

On the other hand, consider the alternating series 1 

n (−1)n . Since the terms alternate 

n 

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

n 

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

Definition 1.24 (Unconditionally Convergent Series). Let X be a Banach space and let 

{fn}n∈N be a sequence of elements of X. The series f = ∞ n=1 fn is said to converge unconditionally 

if every rearrangement of the series converges. That is, f = ∞ n=1 fn converges 

unconditionally if for each bijection σ : N → N the series 

∞ 

converges. 

n=1 

fσ(n) 

Remark 1.25. It is not obvious, but it can be shown that if ∞ n=1 fσ(n) is unconditionally 

convergent, then every rearrangement of the series must converge to the same sum, i.e., there 

is a single f such that f = ∞ n=1 fσ(n) for every permutation σ. 

1 

n of


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

n=1 fn converges 

absolutely, then it converges unconditionally. 

Remark 1.27. In finite dimensions the converse to Exercise 1.26 is true, i.e., if X is finitedimensional 

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 unconditional 

convergence, again consider X = F and the alternating series 1 

n (−1)n . We know 

n 

that this series converges, but does not converge absolutely. 

Now consider what happens if we change the order of summation. Let pn = 1 

2n and 

qn = 1 

2n+1 , i.e., the pn are the positive terms from the alternative series and the qn are the 

absolute values of the negative terms. Each series 

n qn diverges. Hence there 

must exist an m1 > 0 such that 

p1 + · · · + pm1 > 1. 

Then, there must exist an m2 > m1 such that 

Continuing in this way, we see that 

n pn and 

p1 + · · · + pm1 − q1 + pm1+1 + · · · + pm2 > 2. 

p1 + · · · + pm1 − q1 + pm1+1 + · · · + pm2 − q2 + · · · 

is a rearrangement of 1 

n (−1)n which diverges to +∞. 

n 

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 {fn}n∈N be a sequence of elements 

of 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 − 

 

< 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 

(b) A subset U ⊆ X is open if 

Br(x) = B(x, r) = {y ∈ X : x − y < r}. 

∀ x ∈ U, ∃ r > 0 such that Br(x) ⊆ U. 

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

n∈F 

fn


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

A ⊆ X. 

(a) A point f ∈ A is called a limit point of A if there exist fn ∈ A with fn = f such that 

fn → f. 

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

Ā = {F ⊆ X : F is closed and F ⊇ A}. 

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

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

Ā = 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 is 

open, 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 terminology 

“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 under 

both vector addition and scalar multiplication, i.e., if for all u, v ∈ Y and α, β ∈ F we have 

αu + βv ∈ Y . 

(b) A subset Y of a normed linear space X is a closed subspace of X if it is a subspace 

and 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 in 

Proposition 3.5). This exercise demonstrates that, in infinite dimensions, subspaces need 

not be closed sets. 

(a) Fix 1 ≤ p ≤ ∞. Prove that 

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

is a subspace of ℓ p (N) that is not closed (with respect to the ℓ p -norm). Prove that c00 is 

dense in ℓ p (N) if p < ∞, but that it is not dense in ℓ ∞ (N). 

(b) Define 

c0 = {x = (xk) ∞ k=1 : lim 

k→∞ xk = 0}. 

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

(under the ℓ ∞ -norm). 

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

functions on R n , is a subspace of L p (R n ) that is not closed. Prove that Cc(R n ) is dense in 

L p (R n ) if p < ∞, but not if p = ∞. 

Hints: For a continuous function we have f∞ = sup |f(x)|. Hence, if {fn}n∈N is a 

sequence of continuous functions in L ∞ (R n ) that converges in L ∞ -norm, then it converges 

uniformly. From undergraduate real analysis, we know that the limit of a uniformly convergent 

sequence of continuous functions is continuous. 

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

lim f(x) = 0. (1.1) 

|x|→∞ 

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 ∞ (R n ) (closed in 

L ∞ -norm). Prove that C0(R n ) is the closure of Cc(R n ) (under the L ∞ -norm). 

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

Cb(R n ) is a closed subspace of L ∞ (R n ). 

(f) Fix 1 ≤ p ≤ ∞, and let E be a (Lebesgue) measurable subset of R n . Let M = {f ∈ 

L p (R n ) : supp(f) ⊆ E}. Prove that M is a closed subspace of L p (R n ). 

Remark 1.38. (a) We took the domain in the preceding exercise to be R n just for convenience; 

the definitions of the spaces Cc, C0, Cb can be extended to domains that are more 

general 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 itself 

a Banach space (using the norm from X) if and only if M is closed.


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 in 

X. Since each fn belongs to M, it is also Cauchy in M. Since M is a Banach space, {fn} 

must therefore converge in M, i.e., fn → g for some g ∈ M. However, limits are unique, so 

f = g ∈ M. Therefore M contains all its limit points and hence is closed. 


Notation 1.40 (Notation for Closed Subspaces). Since we will often deal with closed subspaces 

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 answer 

is yes, given a normed space Y it is always possible to construct a Banach space X ⊇ Y such 

that the norm on X, when restricted to Y , is the same as the norm on Y , and Y is dense in 

X 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 fa ≤ fb ≤ C2 fa. (1.2) 

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

C2 fb ≤ fa ≤ 1 

C1 fb. 

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 on 

the class of norms on X. 

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

convergence criterion. 

Proposition 1.45. Let ·a and ·b be two norms on a vector space X. Then the following 

statements 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 open 

with 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 {fn}n∈N is a sequence 

in X and f ∈ X, then 

lim 

n→∞ f − fna = 0 ⇐⇒ lim f − fnb = 0. 

n→∞ 

Proof. (b) ⇒ (a). Assume that statement (b) holds. Let Ba r (f) and Bb r(f) denote the open 

balls of radius r centered at f ∈ X with respect to · a and · b. Since Ba 1 (0) is open 

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

with respect to · b. Therefore, since 0 ∈ Ba 1 (0), there must exist some r > 0 such that 

Bb r(0) ⊂ Ba 1(0). 

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

(r − ε) f 

∈ B b r (0) ⊆ Ba 1 (0), 

fb 

so 

 

(r 

− ε) f 

 

 

< 1. 

fb a 

Rearranging, this implies (r − ε) fa < fb. Since this is true for every ε, we conclude 

that r fa ≤ fb. 

A symmetric argument, interchanging the roles of the two norms, shows that there exists 

an s > such that fb ≤ s fa 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 the 

function ν : X → R given by ν(f) = fb 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 respect 

to · a). Since 0 ∈ ν −1 (−1, 1), there must exist an r > 0 such that B b r(0) ⊂ ν −1 (−1, 1). 

But ν −1 (−1, 1) = B a 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 a 

function (= operator = transformation) mapping X into Y . We write either T (f) or T f to 

denote 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 )). In 

particular, T is finite-rank if range(T ) is finite-dimensional. 

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


(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 in X implies Lfn → Lf 

in 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 Lf. 

f=1 

(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 functionals 

on 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 linear 

|Lf|. 

functional L ∈ X ′ = B(X, F) is L = sup 

f=1 

Exercise 2.3. Show that if T : X → Y is linear and continuous, then ker(T ) is a closed 

subspace 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 ker L = {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 different 

meanings of the symbol · appear in this statement!).


(e) If L is bounded then L is the smallest value of K such that Lf ≤ Kf holds 

for all f ∈ X. 

Lf 

(f) L = sup Lf = sup 

f≤1 

f=0 f . 

Exercise 2.5. Show that B(X, Y ) is a subspace of the vector space V consisting of ALL 

functions A: X → Y . Moreover, show that the operator norm is a norm on the space 

B(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 the 

zero 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 show 

in Theorem 2.18 that it is a Banach space if Y is a Banach space. However, in addition to 

vector addition and scalar multiplication operations, there is a third operation that we can 

perform 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. The 

space B(X) is an example of an algebra. 

Exercise 2.7. Let F n be n-dimensional Euclidean space over F, under the Euclidean norm, 

and let Y be any normed linear space. Prove that if L: F n → Y is linear, then L is bounded. 

Hint: If x = (x1, . . . , xn) ∈ F n then x = x1e1+· · ·+xnen where {e1, . . . , en} is the standard 

basis for F n . Use the Triangle and the Cauchy-Schwarz Inequalities. 

Solution 

Given x ∈ Fn , we have 

 

 

n 

 

 

Lx = 

L(xkek) 

≤ 

k=1 

n 

|xk| Lek ≤ 

k=1 

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 · is 

any norm on X, then any linear function L: X → Y is bounded. To do this we will use the 

fact (that we will prove later) that all norms on a finite-dimensional space are equivalent.


The next lemma is a standard fact about continuous functions. L −1 (U) denotes the inverse 

image 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. 


Theorem 2.10 (Equivalence of Bounded and Continuous Linear Operators). Let X, Y be 

normed linear spaces, and let L: X → Y be a linear mapping. Then the following statements 

are 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 = ∞, so 

there 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 implies 

Lgn → L0 = 0, and therefore Lgn → 0 = 0. But 

Lgn = 1 

n Lfn ≥ 1 

· n = 1 

n 

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 that 

fn → f. Then fn − f → 0, so 

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

Lfn − Lf = L(fn − f) ≤ L fn − f → 0, 

The remaining implications are exercises. 

Definition 2.11 (Isometries and Isometric Isomorphisms). Let X, Y be normed linear 

spaces 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 normpreserving. 

(b) An isometry L: X → Y that is a bijection is called an isometric isomorphism. In 

this 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 and 

find L. 

(b) Find an example of an isometry that is not surjective. Contrast this with the fact that 

if A: C n → C n 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 an 

isometric isomorphism. 

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

(a) Define L: ℓ p (N) → ℓ p (N) by L(x) = (x2, x3, . . . ) for x = (x1, x2, . . . ) ∈ ℓ p (N). Prove 

that this left-shift operator is bounded, linear, surjective, not injective, and is not an isometry. 

Find L. 

(b) Define R: ℓ p (N) → ℓ p (N) by R(x) = (0, x1, x2, x3, . . . ) for x = (x1, x2, . . . ) ∈ ℓ p (N). 

Prove that this right-shift operator is bounded, linear, injective, not surjective, and is an 

isometry. Find R. 

(c) Compute LR and RL. Contrast this computation with the fact that in finite dimensions, 

if A, B : C n → C n are linear maps (hence correspond to multiplication by n × n 

matrices), 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 → Y 

is a bounded bijection, then L −1 is automatically bounded and hence L is a topological 

isomorphism. Thus, when X and Y are Banach spaces, every continuous invertible map is a 

topological isomorphism. Sometimes the abbreviation isomorphism or invertible map is used 

to 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 

1 

∀ f ∈ X, 

L−1 f ≤ Lf ≤ L f. 

 

Exercise 2.17. Let X be a Banach space and Y a normed linear space. Suppose that 

L: X → Y is bounded and linear. Prove that if there exists c > 0 such that Lf ≥ cf 

for all f ∈ X, then L is injective and range(L) is closed.


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 Banach 

space. 

Proof. Assume that {An}n∈N is a sequence of operators An ∈ B(X, Y ) that is Cauchy in 

operator norm. For any given f ∈ X, we have 

Amf − Anf ≤ Am − An f, 

so we conclude that {Anf}n∈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 candidate 

limit 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 = lim Anf ≤ sup 

n→∞ 

n∈N 

Anf ≤ sup An f = C f. 

n∈N 

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 that 

m, 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 ε 

+ = 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 Banach 

space. 

The next exercise deals with the problem of extending an operator defined only a dense 

subspace to the entire space. 

Exercise 2.20 (Extension of Bounded Operators). Let Y be a dense subspace of a normed 

space X, and let Z be a Banach space. 

(a) Suppose that L: Y → Z is a bounded linear operator. Show that there exists a 

unique 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. 

2 

2


(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 {gn}n∈N is Cauchy in X, so this 

implies that {Lgn}n∈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 → c ˜ Lf + ˜ Lg, 

by definition we have that ˜ L(cf + g) = c ˜ Lf + ˜ Lg. 

To see that ˜ L is an extension of L, suppose that g ∈ Y is fixed. If we set gn = g, then 

gn → g and Lgn → Lg, so by definition we have ˜ Lg = Lg. Hence the restriction of ˜ L to Y 

is L. Consequently, 

˜ L = sup 

f∈X, f=1 

˜ Lf ≥ sup 

f∈Y, f=1 

˜ Lf = sup Lf = L. 

f∈Y, f=1 

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

so 

˜ Lf = lim 

n→∞ Lgn ≤ lim 

n→∞ 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 that 

˜L: 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 a 

topological isomorphism, we have by Exercise 2.16 that gn ≤ L −1 Lgn. Hence 

˜ Lf = lim 

n→∞ Lgn ≥ lim 

n→∞ 

gn 

L −1 

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

= f 

L −1 . 

˜ L −1 h ≤ L −1 ˜ L( ˜ L −1 h) = 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).


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). Hence 

f = ˜ 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 onto 

F n for some n. 

Example 3.1 (Coordinates). Let X be a finite-dimensional vector space over F. Then X 

has a finite basis, say B = {e1, . . . , en}. Every element of X can be written uniquely in this 

basis, 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 → F n given by x ↦→ [x]B is, by definition of basis, a linear bijection 

of X onto F n . 

Since we already know how to construct many norms on F n , by transferring these to X 

we obtain a multitude of norms for X. 

Exercise 3.2 (ℓ p w Norms on X). Let X be a finite-dimensional vector space over F and let 

B = {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 

xp,w = ⎧ 

⎪⎨ 

n 

 

|ck(x)| 

[x]B 

= p,w k=1 

⎪⎩ 

p w(k) p 

1/p , 1 ≤ p < ∞, 

max |ck(x)| w(k), p = ∞. 

k 

Note that while we use the same symbol · p,w to denote a function on X and on F n , by 

context it has different meanings depending on whether it is being applied to an element of 

X or to an element of F n . 

Prove the following. 

(a) · p,w is a norm on X. 

(b) x ↦→ [x]B is a isometric isomorphism of X onto F n (using the norm · p,w on X and 

the norm · p,w on F n ).


(c) Let {xn}n∈N be a sequence of vectors in X and let x ∈ X. Prove that xn → x with 

respect to the norm · p,w on X if and only if the coordinate vectors [xn]B converge 

componentwise 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 X 


Proof. Let B = {e1, . . . , en} be any basis for X, and let · ∞ be the norm on X defined in 

Exercise 3.2. Since equivalence of norms is an equivalence relation, it suffices to show that 

an 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, 

n 

x ≤ |ck(x)| ek ≤ 

k=1 

n 

k=1 

 

max 

ek |ck(x)| 

k 

= 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, let 

D = {x ∈ X : x∞ = 1} 

be the ℓ∞-unit circle in X. 

Exercise: Show that D is compact (with respect to the norm · ∞). Hints: Suppose 

that {xn}n∈N is a sequence of vectors in D. Then for each n, we have |ck(xn)| = 1 for some 

k ∈ {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}, we 

can select a subsequence whose first coordinates converge. Repeat for each coordinate, and 

remember that the kth coordinate is always 1. Hence we can construct a subsequence that 

converges to x ∈ D with respect to the ℓ ∞ -norm. 

Our next goal is to show that D is also compact with respect to the norm · . Let 

{xn}n∈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 {xnk }k∈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 defined 

by · . A real-valued continuous function on a compact set must achieve a maximum and 

minimum on that set. Hence, there must exist constants m and M such that m ≤ x ≤ M 

for 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 then 

x = 0, which implies x∞ = 0, contradicting the fact that x ∈ D. Hence m > 0, so we can 

take C1 = m.


Consequently, from now on we need not specify the norm on a finite-dimensional vector 

space X—we can take any norm that we like whenever we need it. 

Exercise 3.4. Let F m×n be the space of all m × n matrices with entries in F. F m×n is 

naturally isomorphic to F mn . 

Prove that if · a is any norm on F m×n , · b is any norm on F n×k , and · c is any norm 

on F m×k , then there exists a constant C > 0 such that 

∀ A ∈ F m×n , ∀ B ∈ F n×k , ABc ≤ C Aa Bb. 

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

closed. 

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

M1 = span{M, y} = {m + cy : m ∈ M, c ∈ F}. 

Then M1 is finite-dimensional subspace of X. Moreover, every element z ∈ M1 can be 

written uniquely as z = m(z) + c(z)y where m(z) ∈ M and c(z) is a scalar. For z ∈ M1 

define 

zM1 = m(z) + |c(z)|. 

Exercise: Show that · M1 is 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 zM1 ≤ C z for all z ∈ M1. Since 

c(xn) = 0 for every n, we therefore have 

|c(y)| = |c(y) − c(xn)| = |c(y − xn)| 

≤ m(y − xn) + |c(y − xn)| 

= y − xnM1 

≤ 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 proof 

identical to the one used in the preceding proposition, except using the given norm · on 

X, 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 linear 

space. 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) = inf g − f > 1 − ε. 

f∈M 

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

Fix δ > 0 small enough that a 

a+δ 

that a ≤ u − v < a + δ. Set 

a = dist(u, M) = inf u − f > 0. 

f∈M 

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

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 − h 

u − v = > 

u − v 

a 

> 1 − ε. 

a + δ 

 

Exercise 3.9. Let X be a normed linear space. Let B = {x ∈ X : x ≤ 1} be the closed 

unit 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, by 

Lemma 3.8 there exists e2 ∈ X \span{e1} with e2 ≤ 1 such that e2 −e1 > 1. 

Continue in 

2 

this way to construct vectors ek such that {e1, . . . , en} are independent for any n. Conclude 

that X is infinite-dimensional. 

Definition 3.10. We say that a normed linear space X is locally compact if for each f ∈ X 

there 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 that 

is contained in a compact subset of X. For example, F n 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 has 

compact closure.


4. Quotients and Products of Normed Spaces 

Any vector space is an abelian group under the operation of vector addition. So, if you are 

familiar with the basic notions of abstract algebra, the concept of a coset will be familiar to 

you. However, even if you have not studied abstract algebra, the idea of a coset in a vector 

space is very natural. 

Example 4.1 (Cosets in R 2 ). Consider the vector space X = R 2 . Let M be any onedimensional 

subspace of R 2 , i.e., M is a line in R 2 through the origin. A coset of M is 

simply a rigid translate of M by a vector in R 2 . For concreteness, let us specifically consider 

the case where M is the x1-axis in R 2 , i.e., M = {(x1, 0) : x1 ∈ R}. Then given a vector 

y = (y1, y2) ∈ R 2 , 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 R 2 , but it is a rigid 

translate of the x1-axis. Note that there are infinitely many different choices of y that give 

the 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 R 2 . 

(c) The set of distinct cosets is a partition of R 2 . 

The preceding example is entirely typical. 

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

are 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 quotient 

space 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 simply 

the set of all the distinct cosets of M. 

Example 4.5. Again let M = {(x1, 0) : x1 ∈ R} be the x1-axis in R 2 . Then, by Example 4.1, 

we have that 

R 2 /M = {y + M : y ∈ R 2 } = {(x1, 0) + M : x1 ∈ R}, 

i.e., R 2 /M is the set of all horizontal lines in R 2 . Note that R 2 /M is in 1-1 correspondence 

with the set of distinct heights, i.e., there is a natural bijection of R 2 /M onto R. This is a 

special 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 multiplication 

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 addition 

of 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 welldefined. 

After all, there need not be just one f that determines the coset f + M—how do 

we know that if we choose different vectors that determine the same cosets, we will get the 

same result when we compute (f + g) + M? We must show that f1 + M = f2 + M and 

g1 + M = g2 + M then (f1 + g1) + M = (f2 + g2) + M in order to know that Definition 4.6 

makes sense. 

Proposition 4.8. If M is a subspace of a vector space X, then the addition of cosets of M 

given 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 have 

h = 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 operations 

just defined. 

Proposition 4.10. If M is a subspace of a vector space X, then X/M is a vector space 

with respect to the operations given in Definition 4.6.


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 codimension 

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 subspace 

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 + P 

can be thought of as “f modulo the polynomials,” i.e., it is the equivalence class obtained 

by identifying functions which differ by a polynomial. 

In the same way, a coset f + M can be thought of as the equivalence class obtained by 

identifying vectors which differ by an element of M. We can imagine the mapping that takes 

f 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 or 

the 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 + m 

with u ∈ E and m ∈ M. 

1 Conway 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 result 

shows 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) = inf f − m. 

m∈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 inf m∈M f − m = 0. Hence 

there exist vectors gn ∈ M such that f − gn → 0 as n → ∞. But M is closed, so this 

implies f ∈ M. By Exercise 4.3(e), we therefore have f + M = M = 0 + M, which is the 

zero 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 the 

following statements hold. 

(a) π(f) = f + M ≤ f for each f ∈ X. 

(b) Let B X r 

(f) denote the open ball of radius r in X centered at f, and let BX/M r (f + M) 

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

and r > 0 we have 

π B X r (f) = B X/M 

r (f + M). 

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

in 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 = inf f − m ≤ f − 0 = f. 

m∈M 

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

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

(0 + M). 

(0 + M). Then infm∈M g − m = g + M < r. Hence 

g + M = h + M for some h ∈ BX r 

so g + M ∈ B X/M 

r 

Now suppose that g + M ∈ B X/M 

r 

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

g + M = g − m + M = π(g − m) ∈ π B X r (0) .


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 is 

open in X/M. 

⇐. Suppose that W is a subset of X/M such that π −1 (W ) is open in X. We must show 

that W is open in X/M. Choose any point f + M ∈ W . Then f ∈ π −1 (W ), which is open 

in X. Hence, there exists an r > 0 such that B X r (f) ⊂ π−1 (W ). By part (b) we therefore 

have 

B X/M 

r (f + M) = π B X r (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} = 

(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 in 

X/M. 

Exercise 4.17. Let M be a closed subspace of a normed space X, and let π be the canonical 

projection π of X onto X/M. Prove that π = 1. 

Hint: Lemma 3.8. 

Now we can prove that if X is a Banach space, then X/M inherits a Banach space structure 

from X. 

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

space. 

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

complete in that norm. 

Suppose that {fn + M}n∈N is a Cauchy sequence in X/M. It would be convenient if this 

implies that {fn}n∈N is a Cauchy sequence in X, but this need not be the case. For, the 

vectors 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 subsequence 

{fnk }k∈N and replacing the fnk by 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 to 

show that the original sequence of cosets {fn + M}n∈N converges in X/M. 

We begin by applying Exercise 1.18: there exists a subsequence {fnk + M}k∈N such that 

m∈M 

∀ k ∈ N, (fnk+1 − fnk ) + M = (fnk+1 + M) − (fnk + M) < 2−k . 

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

the cosets determined by fnk and by fnk − gk are identical. 

Set g1 = 0. Then 

inf 

g∈M (fn1 − g1) − (fn2 − g) = inf 

g∈M (fn1 − fn2) + g = (fn1 − fn2) + M < 1 

2 .

Therefore, there exists a g2 ∈ M such that 

Then, since g2 ∈ M, 


(fn1 − g1) − (fn2 − g2) < 2 · 1 

2 . 

inf 

g∈M (fn2 − g2) − (fn3 − g) = inf 

g∈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 {hk}k∈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 + M}k∈N is a convergent subsequence of {fn + M}n∈N. Since we know that 

{fn + M}n∈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 and 

X/M are both complete, then X must be complete. 

Hints: Suppose that {fn}n∈N is a Cauchy sequence in X. Show that {fn + M}n∈N is a 

Cauchy 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 vectors 

gn ∈ 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 N 

is 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, it 

has a finite basis, say {e1, . . . , en}. Then since π is linear, 

π(N) = π span{e1, . . . , en} = span{π(e1), . . . , π(en)} = span{e1 + M, . . . , en + M}.


Thus π(N) is a finite-dimensional subspace of X/M, and therefore is closed by Proposition 

3.6. Since π is continuous, it follows that π −1 (π(N)) is closed in X. However, by 

Exercise 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 + M}n∈N be a countable dense subset of X/M, and let {gn}n∈N be a countable 

dense subset of M. Then S = {fm + gn}m,n∈N is a countable subset of X, and we claim that 

it is dense. 

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

inf 

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

2 . 

Hence, there exists some h ∈ M such that f − fm + h < ε 

. Since h ∈ M, there exists an 

2 

n such that h − gn < ε. 

Therefore 

2 

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

+ = ε. 

2 2 

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

(a) Prove that ℓ ∞ 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 = yk for at most finitely many k, then 

x + c0 = y + c0. Prove that if xk = yk for infinitely many k, then x − y + c0 = 1. 

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

(e) Prove that the standard basis {en}n∈N is a Schauder basis for c0 (with respect to the 

ℓ ∞ -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 ℓ ∞ /c0 is not separable. 

In the Hilbert space case, there is a very close relationship between π and the orthogonal 

projection of H onto M ⊥ .


Exercise 4.23. Let M be a closed subspace of a Hilbert space H, and let π be the canonical 

projection of H onto H/M. Prove that the restriction of π to M ⊥ is an isometric isomorphism 

of M ⊥ onto H/M. 

Remark 4.24. There is no analog of the preceding result for arbitrary Banach spaces. If X 

is a Banach space and M is a closed subspace then we say that M is complemented in X if 

there 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. In 

particular, c0 is not complemented in ℓ ∞ . Also, if 1 

uncomplemented subspaces. 

Next we will define the product or direct sum of normed spaces. An analogous definition 

holds for the case of a finite collection of spaces. 

Definition 4.25. Let {Xi}i∈N be a countable family of Banach spaces, and let · i denote 

the norm on Xi. Define 

For 1 ≤ p < ∞, define 

For p = ∞, define 

 

p Xi = 

∞ 

i=1 

 

∞ Xi = 

Xi = 

f = (f1, f2, . . . ) : fi ∈ Xi . 

 

f ∈ 

∞ 

Xk : fp = 

i=1 

 

f ∈ 

∞ 

i=1 

∞ 

i=1 

fi p 

1/p 

i < ∞ . 

Xk : f∞ = sup fii < ∞ 

i 

 

. 

Exercise 4.26. Let {Xi}i∈N be a countable family of Banach spaces and fix 1 ≤ p ≤ ∞. 

Let X = 

p Xi. 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, we 

often denote this space by X1 × · · · × Xn.


Exercise 4.28. Let X and Y be normed vector spaces. Define T : B(X, Y ) × X → Y by 

T (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 that 

M 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 linear 

operator. Now, F is one-dimensional, and a linear operator is entirely determined by what 

it does to the basis elements. In particular, {1} is a basis for F, so L ∗ (c) = L ∗ (c1) = c L ∗ (1). 

Hence range(L ∗ ) = span{L ∗ (1)}. If L ∗ (1) = 0 then L ∗ is the zero operator, and hence L is 

the zero operator as well (why?). Otherwise, range(L ∗ ) must be one-dimensional, and hence 

is 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 H 

onto H/ ker(L), then since k ∈ ker(L) we have 

Therefore 

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

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

= {cL ∗ (1) + ker(L) : c ∈ F} 

= span{L ∗ (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 the 

following tool. In abstract algebra, the group version of the next result is called the First 

Isomorphism Theorem or the Homomorphism Theorem. In the group setting, an isomorphism 

is a bijective homomorphism. In the vector space setting, an isomorphism is a linear 

bijection.


Exercise 5.3 (Isomorphism Theorem). Let X and Y be vector spaces, and let ϕ: X → Y 

be 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 ∈ ℓ p : x(2k) = 0 for every k}. 

Prove that M is a closed subspace of ℓ p , and that ℓ p /M is isometrically isomorphic to ℓ 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 statements 


(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. Then 

by the Isomorphism Theorem, there is a linear bijection ψ : X/ ker(µ) → F. Since F is 

one-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 there 

exists a linear bijection ψ : X/M → F. Set ϕ = ψ ◦ π, then ψ : X → F. If f ∈ M, then 

π(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 implies 

f + 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 linear 

functionals, then 

Proof. ⇐. Trivial. 

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

⇒. Suppose that ker(µ) = ker(ν). Since µ = 0, there exists some f ∈ X such that 

µ(f) = 0, and by rescaling, we can assume that µ(f) = 1. Since f /∈ ker(µ) = ker(ν), we 

have ν(f) = 0. Given any g ∈ X, we have 

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

µ g − µ(g)f = µ(g) − µ(g) · µ(f) = 0, 

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 closed 

or is dense in X.


Proof. We are given that X/M is one-dimensional. Let π be the canonical projection of X 

onto 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 = M 

and 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: 

Proof. ⇒. Exercise. 

µ is continuous ⇐⇒ ker(µ) is closed. 

⇐. 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, there 

exists a linear bijection ψ : X/ ker(µ) → F. Since X/ ker(µ) is a one-dimensional normed 

linear space, and linear map from X/ ker(µ) into another normed space is continuous by 

Exercise 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 |Lf|. 

f=1 

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 Hölder’s Inequality. 

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

1 

p 

+ 1 

p ′ = 1, then 

∀ f ∈ L p (X), ∀ g ∈ L p′ 

(X), fg1 ≤ fp gp ′. 

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 together 

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 function 

d(·, ·) → 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 

(b) A subset U ⊆ X is open if 

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

∀ 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 using 

the preceding definition.

FUNCTIONAL ANALYSIS LECTURE NOTES CHAPTER 3. BANACH ...

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