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Contents 1. Normed vector spaces 1.1. Examples of normed vector spaces 1.2. Continuous linear operators 1.3. Equivalent norms 1.4. Matrix norms 1.5. Banach spaces

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1. Normed vector spaces Recall that a vector space over a field K is a set X together with two operations, X × X → X, (x, y) 7→ x + y

addition:

scalar multiplication: K × X → X, (λ, x) 7→ λ x which are required to satisfy certain axioms. The elements of X are called points or vectors and the elements of K are called scalars. Definition 1.1. A normed vector space is a pair (X, k.k) consisting of a vector space X (over R or C) and a mapping k.k : X → R, x 7→ kxk such that (i) kxk = 0 if and only if x = 0 (ii) kλ xk = |λ| kxk for all scalars λ and all x ∈ X (iii) kx + yk ≤ kxk + kyk for all x, y ∈ X

(Triangle inequality)

The mapping k.k is called a norm on X and kxk is called the norm of x. The norm of a vector is non-negative and we can think of kxk as the length of a vector x. The mapping d : X × X → R given by d(x, y) = kx − yk

for all x, y ∈ X

defines a metric on X. In this way every normed vector space can be regarded as a metric space.

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1.1. Examples of normed vector spaces. Example 1.2. The Euclidean norm on Rn . kxk2 =

√

x·x=

q

x21 + · · · + x2n

where x = (x1 , . . . , xn ) ∈ Rn . Example 1.3. Other norms on Rn . (a) The 1-norm. kxk1 = |x1 | + · · · + |xn | (b) The ∞-norm. kxk∞ = max {|x1 |, . . . , |xn |} where x = (x1 , . . . , xn ) ∈ Rn . Example 1.4. The complex numbers. (C, |.|) is a normed vector space where |.| is the modulus, |z| =

p x2 + y 2 ,

for all z = x + i y ∈ C

Example 1.5. Let C[0, 1] be the set of all continuous real-valued functions f : [0, 1] → R. Addition and scalar multiplication of functions can be defined pointwise: If f and g are real-valued functions on [0, 1] then (f + g)(x) = f (x) + g(x)

for all x ∈ [0, 1]

If λ is a scalar then (λ f )(x) = λ (f (x))

for all x ∈ [0, 1]

With these operations C[0, 1] is a vector space over R. The following define two different norms on C[0, 1],

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(a) kf k1 =

Z

1

|f (x)| dx

0

(b) kf k∞ = sup |f (x)| x∈[0,1]

for all f ∈ C[0, 1].

Example 1.6. Let c0 be the set of all sequences (xj )∞ j=1 of real numbers which converge to 0. Addition and scalar multiplication of sequences can be defined componentwise: ∞ If x = (xj )∞ j=1 and y = (yj )j=1 then

x + y = (xj + yj )∞ j=1 If λ is a scalar then λx = (λ xj )∞ j=1 With these operations c0 is a vector space over R. We can define a norm on c0 by kxk = sup |xj | j

for all points x = (xj )∞ j=1 in c0 .

Example 1.7. Let `2 be the set of all sequences (xj )∞ j=1 of real numbers which are square-summable. i.e. ∞ X

x2j < ∞

j=1

Addition and scalar multiplication of sequences can be defined componentwise as in the previous example. With these operations `2 is a vector space

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over R. We can define a norm on `2 by

kxk2 =

∞ X j=1

2 for all points x = (xj )∞ j=1 in ` .

x2j

! 12

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1.2. Continuous linear operators. Let X and Y be vector spaces over the same field K. A mapping T : X → Y is called a linear operator if T (x + y) = T (x) + T (y) T (λ x) = λ T (x) for all x, y ∈ X and for all scalars λ. Now suppose X and Y are normed vector spaces. Since we can regard X and Y as metric spaces, it makes sense to consider continuous linear operators. The set of all continuous linear operators T : X → Y is denoted L(X, Y ). We can define operations of addition and scalar multiplication on L(X, Y ) as follows: if S, T ∈ L(X, Y ) then define S + T ∈ L(X, Y ) by (S + T )(x) = S(x) + T (x)

for all x ∈ X

if T ∈ L(X, Y ) and λ is a scalar then define λ T ∈ L(X, Y ) by (λ T )(x) = λ(T (x))

for all x ∈ X

With these operations L(X, Y ) is a vector space (over the same field as X and Y ).

Theorem 1.8. Let T : X → Y be a linear operator between normed vector spaces (X, k.k) and (Y, k.k). Then T is continuous if and only if there exists a real number M such that kT (x)k ≤ M kxk

for all x ∈ X.

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We can define a norm on L(X, Y ) by kT kop = inf {M : kT (x)k ≤ M kxk for all x ∈ X} for all T ∈ L(X, Y ). As a consequence of Theorem 1.8, a linear operator T is continuous if and only if kT kop < ∞. It will be convenient to have the following descriptions of the operator norm.

Theorem 1.9. If T ∈ L(X, Y ) then kT kop =

sup kT (x)k kxk≤1

= sup x6=0

=

kT (x)k kxk

sup kT (x)k kxk=1

Note that

(i) kT kop = 0 if and only if T = 0 (ii) kλ T kop = |λ| kT kop (iii) kS + T kop ≤ kSkop + kT kop Thus (L(X, Y ), k.kop ) is a normed vector space. If X is a normed vector space then we denote by L(X) the set of all continuous linear operators T : X → X. (i.e. L(X) = L(X, X)). We can define multiplication on L(X) as follows: if S, T ∈ L(X) then define ST ∈ L(X) by (ST )(x) = S(T (x))

for all x ∈ X

With these operations L(X) is an example of an associative algebra.

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Theorem 1.10. If S, T ∈ L(X) then kST kop ≤ kSkop kT kop Example 1.11. The unilateral shift operator on `2 is defined by T : `2 → `2 ,

(x1 , x2 , x3 , . . .) 7→ (0, x1 , x2 , x3 , . . .)

This is a continuous linear operator with kT kop = 1. Example 1.12. Let C 1 [0, 1] denote the set of continuous real-valued functions f : [0, 1] → R which have continuous derivatives

df dx

: [0, 1] → R. Then

C 1 [0, 1] is a vector space with pointwise operations of addition and scalar multiplication (the same operations as C[0, 1]). The supremum norm kf k∞ = sup |f (x)| x∈[0,1]

defines a norm on C 1 [0, 1] and C[0, 1]. The differentiation operator D : C 1 [0, 1] → C[0, 1], is a linear operator which is not continuous.

f 7→

df dx

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1.3. Equivalent norms. Definition 1.13. Two norms k.k1 and k.k2 on a vector space X are said to be equivalent if there exist positive real numbers a, b such that akxk1 ≤ kxk2 ≤ bkxk1 for all x ∈ X. Theorem 1.14. Let X be a vector space. The following statements are equivalent: (i) k.k1 and k.k2 are equivalent norms on X, (ii) k.k1 and k.k2 generate the same topology on X (i.e. the same open sets), (iii) A sequence (xj ) in X converges to a point x with respect to k.k1 if and only if (xj ) converges to x with respect to k.k2 . The above result tells us that if we want to check continuity of a mapping or convergence of a sequence then we are allowed to swap our norm for an equivalent one. This can make calculations easier. Theorem 1.15. Let X be a finite dimensional vector space. Then all norms on X are equivalent. Example 1.16. Let C[0, 1] denote the set of continuous real-valued functions f : [0, 1] → R. Then the norms kf k1 =

Z

1

|f (x)| dx

0

and kf k∞ = sup |f (x)| x∈[0,1]

are not equivalent on C[0, 1]. Using Theorem 1.15 we can prove the following:

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Theorem 1.17. Let (X, k.k) and (Y, k.k) be normed vector spaces and let T : X → Y be a linear operator. If X is finite dimensional then T is continuous.

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1.4. Matrix norms. The collection Mm,n (R) of all (m × n)-matrices with entries in R is a vector space over R. There are many ways to put a norm on Mm,n (R), here we consider some of the most frequently used matrix norms. Example 1.18. The Frobenius norm. If we regard the entries of a matrix A = (aij ) as coordinates for a point in Euclidean space Rmn then we arrive at n m X X

kAkF =

a2ij

! 12

=

i=1 j=1

p

trace(At A)

We can regard an (m × n)-matrix as a continuous linear operator from Rn to Rm . In the following three cases we use the operator norm on L(Rn , Rm ) while varying the norms on Rn and Rm . Example 1.19. The 1-norm. kAk1 =

sup kA(x)k1 kxk1 =1

=

m X

max

1≤j≤n

|aij |

i=1

i.e. the maximum of the absolute column sums. Example 1.20. The ∞-norm. kAk∞ = =

sup kA(x)k∞

kxk∞ =1

max

1≤i≤m

n X

|aij |

j=1

i.e. the maximum of the absolute row sums. Example 1.21. The spectral norm (or 2-norm). kAk2 =

sup kA(x)k2 kxk2 =1

=

p largest eigenvalue of At A

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1.5. Banach spaces. Definition 1.22. A Banach space is a normed vector space (X, k.k) which is complete (with respect to the metric d(x, y) = kx − yk). Theorem 1.23. Every finite dimensional normed vector space is a Banach space. In light of Theorem 1.23 we now consider some infinite dimensional Banach spaces. Definition 1.24. Let (X, d) and (Y, d0 ) be metric spaces. Let (fn )∞ n=1 be a sequence of mappings fn : X → Y . Then (fn )∞ n=1 is said to converge pointwise to a mapping f : X → Y if for each x0 ∈ X, given any  > 0 there exists N ∈ N such that d0 (fn (x0 ), f (x0 )) <  for all n ≥ N (Note that N depends on the point x0 ). The sequence (fn )∞ n=1 is said to converge uniformly to f : X → Y if given any  > 0 there exists N ∈ N such that d0 (fn (x), f (x)) <  for all n ≥ N, and for all x ∈ X (In this case N does not depend on any point). Example 1.25. For each n ∈ N consider the continuous function fn : (0, ∞) → R,

x 7→

1 nx

Then the sequence (fn )∞ n=1 converges pointwise but not uniformly to the zero function f : (0, ∞) → R,

x 7→ 0

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Lemma 1.26. (Uniform Limit Theorem) Let (X, d) and (Y, d0 ) be metric spaces. Let (fn )∞ n=1 be a sequence of continuous mappings fn : X → Y . If (fn )∞ n=1 converges uniformly to f : X → Y then f is continuous. Example 1.27. For each n ∈ N consider the continuous function fn : [0, 1] → R,

x 7→ xn

Then the sequence (fn )∞ n=1 converges pointwise but not uniformly to the function f : [0, 1] → R, Notice that f is not continuous.

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