LECTURE NOTE 1: NORMED VECTOR SPACE 1. Vector Space

LECTURE NOTE 1: NORMED VECTOR SPACE

WEI XIONG

1.1. vector space.

Definition 1.1. Vector space Let X be a non-empty set and K be R or C. If there exists two mappings: (x, y) ∈ X × X → (x + y) ∈ X and (α, x) ∈ K × X → αx ∈ X called respectively addition and scalar multiplication, that together satisfy the following properties:

• x + y = y + x for all x, y, z ∈ X • x + (y + z) = (x + y) + z for all x, y, z ∈ X • There exists a zero vector in θ s.t. x + θ = x, ∀x ∈ X • Given any x ∈ X, there exists an element of X denoted by (−x) s.t. x + (−x) = θ • α(x + y) = αx + αy and (α + β)x = αx + βx • α(βx) = (αβ)x and 1x = x

Observation 1.2. We can define metric on any non-empty set. Thus a metric space may not be avector space.

1.2. Quasi-norm and Norm on vector space. Suppose that (X, d) is a vector space with metric d s.t.

• d(x + z, y + z) = d(x, y), ∀x, y, z ∈ X

• d(xn, x) → 0 ⇒ d(αxn, αx) → 0, ∀α ∈ K

• αn → α ⇒ d(αnx, αx) → 0, ∀x ∈ X

Definition 1.3. Quasi-norm We define ∥∥ : X → R : ∥x∥ = d(x, θ), ∀x ∈ X. Then it satisfies:

• ∥x∥ ≥ 0, equality holds if and only if x = θ • ∥x + y∥ ≤ ∥x∥ + ∥y∥ • ∥x∥ = ∥ − x∥ • ∥ ∥ ∥ ∥ limαn→0 αnx = 0, lim∥xn∥→0 αxn = 0

Definition 1.4. F ∗ space and F space

Given a vector space X with a quasi-norm, we define convergence w.r.t. ∥xn − x∥ → 0. Then, the space is called a F ∗ space. Further, if the space is complete, it is called a Frechet space or F space. 1 Definition 1.5. Norm Let X be a vector space. A norm on X is any mapping ∥.∥ : x → R that satisfies

• ∥x∥ ≥ 0 equality if and only if x = 0 • ∥αx∥ = ∥α∥∥x∥ for all x ∈ X, α ∈ K • ∥x + y∥ ≤ ∥x∥ + ∥y∥ for all x, y ∈ X

Definition 1.6. B∗ space and Banach space Normed vector space is called a B∗ space and a Banach space if it is also complete.

Definition 1.7. Equivalence of norms

We say that ∥.∥2 is stronger than ∥.∥1 if ∥xn∥2 → 0 ⇒ ∥xn∥1 → 0.

Two norms are said to be equivalent if ∥.∥1 is stronger than ∥.∥2 and vice versa.

Theorem 1.8. ∥.∥2 is stronger than ∥.∥1 if and only if there exists some constant c > 0 s.t. ∥.∥1 ≤

c∥.∥2, ∀x ∈ X.

∥.∥1, ∥.∥2 are equivalent if and only if there exists constants c1, c2 > 0 s.t. c1∥x∥1 ≤ ∥x∥2 ≤ c2∥x∥1, ∀x ∈ X.

Proof. The second part is simply the corollary of the first part. So we focus on the proof of the first partof the theorem.

⇒: ∥x1∥ ≤ c∥x∥2 → 0.

⇐: We prove by contradiction. Suppose that we cannot find c > 0 s.t. ∥∥1 ≤ c∥∥2. Then, ∀n ∈ N, we can

xn 1 find xn ∈ X s.t. ∥xn∥1 ≥ n∥xn∥2. Let yn =: where ∥yn∥1 = 1 for all n and 0 ≤ ∥yn∥2 ≤ . Thus, ∥xn∥1 n

∥yn∥2 → 0 but ∥yn∥1 does not converges to 0 which is a contradiction. □

Let X be a normed vector space of dimension n < ∞. Then we can find a base {e1, ..., en} s.t. ∀x ∈ X, there n exists an unique expression, x = ξ1e1 +...+ξnen. So, we can define a one-one mapping T x : x ∈ X → ξ ∈ K . ∑ ∥ n ∥ n → We can also define a function p(ξ) = j=1 ξjej : K R.

Theorem 1.9. A vector space X is of dimension n < ∞. Then X and Kn are algebra isometric and topological isometric.

Proof. We know that X and Y are algebra isometric if and only if dimX = dimY . The second part of theorem is a corollary of next lemmas. □

∑ | | n | |2 1/2 Let ξ =: [ j=1 ξj ] Lemma 1.10. 1 p(ξ) is uniformly continuous w.r.t. ξ. ∑ | | | |∥ n ξj ∥ | | ξ ̸ 2 p(ξ) = ξ j=1 |ξ| ej = ξ p( |ξ| ), ξ = θ 2 Proof. Let ξ, η ∈ Kn.

= c0|ξ − η|

∑ | | | |∥ n xij ∥ ̸ □ p(ξ) = ξ j=1 |ξ| ej = RHS if ξ = θ

n We then define an (intermediate) norm onX: ∥∥T : x ∈ X → T x = ξ ∈ K → |T x| ∈ R.

Lemma 1.11. Let X be a normed vector space of dimension n. Then its norm ∥∥ is equivalent to ∥∥T . ∪ n Proof. Note B(0, 1) X/B(0, 1) is open.(The ball induced by metric). So S1 =: {ξ ∈ K : |ξ| = 1} is compact. Thus p(ξ) attains its maximum c2 and minimum c1, i.e., c1 ≤ p(ξ) ≤ c2, ∀ξ ∈ S1. According to the second part of last lemma, we know that

n c1|ξ| ≤ p(ξ) ≤ c2|ξ|, ∀ξ ∈ K

We proceed to show that c1 > 0 by contradiction. ∗ ∗ ∗ ∈ ∗ ∗ ∗ Suppose that c1 = 0. Then there exists ξ = (ξ1 , .., ξn) S1 s.t. p(ξ ) = 0. ξ1 e1 + ... + ξn = θ if and only if ξ∗ = 0 which is a contradiction. □

Corollary: Let X be a vector space of finite dimension n. Then, all norms defined above are equivalent.

Theorem 1.12. Finite normed vector space must be complete. Any sub-space of finite dimensional normed vector space is closed.

(n) Proof. Let (xn) be a Cauchy sequence in finite normed vector space X and (ξ ) be the corresponding n (n) (m) (n) sequence in K . Then |ξ − ξ | = |xn − xm|T ≤ c2∥xn − xm∥ → 0 as m,n tend to infinity. So, (ξ ) is a Cauchy sequence in Kn and thus converges to some ξ ∈ Kn. Using the same argument, we show that −1 (n) ∥xn − T (ξ)∥ ≤ c1|ξ − ξ| → 0. □

2. Approximation problem

Given a set of function {ϕ1, ..., ϕn}, we want to approximate some function f by the linear combination of them. The performance is evaluated by some loss function, e.g., the Fourier series. 3 ∗ n Given a B space X and a finite subset {e1, ..., en} of X, we want to find (λ1, ..., λn) ∈ K s.t. for a fixed x ∈ X, ∑n ∑n ∥x − λjej∥ = min ∥x − ajej∥ a∈Kn j=1 j=1 ∑ n The element j=1 λjej is called a best approximation. We proceed to show that there exists at least one best approximation and it is unique under certain assumption. WLOG, we assume that ei are linearly independent. Theorem 2.1. Existence of the best approximation Let X be a normed vector space. Then, ∀x ∈ X, we can find the optimal coefficients s.t.

∑n ∑n ∥x − λjej∥ = min ∥x − ajej∥ a∈Kn j=1 j=1

Proof. We define the functions ∑n n F (a) = ∥x − ajej∥a ∈ K j=1 ∑n P (a) = ∥ ajej∥ j=1 F (a) is continuous and F (a) ≥ P (a) − ∥x∥, ∀a ∈ Kn. We can verify that P (a) is a norm on Kn and thus n there exists c1 > 0 s.t. P (a) ≥ c1|a|, ∀a ∈ K according to the lemma 1.11. So, F (a) → ∞ as |a| → ∞. (Note ∥x∥ is fixed). Thus we may select M > 0 s.t. F (a) ≥ M, ∀|a| > N and consider F (a) on the compact set B(0,N). So, F attains its minimum. □ Theorem 2.2. Existence of the best approximation Let M be a normed vector sub-space of X and M is of finite dimension. Then, for all x ∈ X, the approximation of x exists.

Proof. The problem is equivalent to that find the best approximation in span{e1, ..., en}. □

If ei are linearly dependent, then the best approximation is not unique. So we suppose that ei are linearly independent. Definition 2.3. Strictly convex A normed vector space X is said to be strictly convex if ∀x, y ∈ X, x ≠ y, we have

∥x∥ = ∥y∥ = 1 ⇒ ∥αx + βy∥ < 1

holds ∀α, β > 0, α + β = 1

Theorem 2.4. Let X be a strictly convex normed vector space. Given a set of linearly independent vectors

{e1, ..., en} of X, then, ∀x ∈ X, its best approximation is unique. 4 Proof. Let M = span(e1, ..., en). Suppose that there exists y, z ∈ M, y ≠ z and ∥x − z∥ = ∥x − y∥ = d0 =

d(x, M). Then, if d0 > 0 1 1 ∥x − (αy + βz)∥ = ∥α(x − y) + β(x − z)∥ d0 d0 x − y x − z = ∥α( ) + β( )∥ < 1 d0 d0

i.e., ∥x − (αy + βz)∥ < d0 which contradicts the fact that d0 is minimum.

If d0 = 0, then the best approximation is exactly x itself. □

3. Equivalent condition of B∗ space of finite dimension

3.1. Riesz Lemma. Lemma 3.1. F.Riesz

Let X be a normed vector space. Let X0 ⊂ X and X0 ≠ X be a closed sub-space. Then, ∀0 < ϵ < 1, ∃y ∈ X

s.t. ∥y∥ = 1 and ∥y − x∥ ≥ 1 − ϵ, ∀x ∈ X0.

̸ ∈ ∥ − ∥ Proof. Since X0 = X, we can find x X/X0. Let d0 = d(x, X0) =: infy∈X0 x y . Since X0 is closed,

∈ ≤ ∥ − ∥ d0 d0 > 0. Thus there exists y0 X0 s.t. d0 x y0 < 1−ϵ by the definition of infimum. − Let y = x y0 → ∥y∥ = 1. ∥x−y0∥

∀z ∈ X0, we have x − y ∥x − (y + z∥x − y ∥)∥ d ∥y − z∥ = ∥ 0 − z∥ = 0 0 ≥ 0 = 1 − ϵ ∥x − y0∥ ∥x − y0∥ d0/(1 − ϵ) where the last inequality follows d ∥x − y ∥ ≤ 0 0 1 − ϵ

∥x − (y0 + z∥x − y0∥)∥ ≥ d0

since y0 ∈ X0, z ∈ X0 → (y0 + z∥x − z0∥) ∈ X0. □ Theorem 3.2. Riesz

Let X be a normed vector space. It is of finite dimension if and only if S1 =: {x ∈ X : ∥x∥ = 1} is sequentially compact.

∑ ⇐ → n n → Proof. Suppose that X is a normed vector space of finite dimension. Let T : X K : x = i=1 ξiei ξ.

Then the norm ∥∥T is equivalent to ∥∥,i.e.,

c1|T x| ≤ ∥x∥ ≤ c2|T x|, ∀x ∈ X

n The first inequality implies that T (S1) is bounded in K and thus is sequentially compact. ∀(xn) ⊂ S1, we → ∈ → −1 ∈ can find T xnk η T (S1). Then, we cliam that xnk T (η) X since

∥ − −1 ∥ ≤ | − | → xnk T (η) T xnk η 0 5 where the linearity follows

x − y = (ξ1 − η1)e1 + ... + (ξn − ηn)en

⇒ Suppose that S1 is sequentially compact. We prove by contradiction. So, we will assume that dim X = ∞. { }∞ ⊂ We thus can find en n=1 X are linearly independent. Let Xn = span(e1, e2, ..., en). We have:

Xn is a closed space of X. (dim Xn < ∞)

Xn ⊂ Xn+1 and Xn ≠ Xn+1 1 ∀ ∈ ∈ ∩ ≥ 1 Let ϵ = 2 . By Riesz Lemma, n N, we can find xn S1 Xn+1 s.t. d(xn,Xn) 2 so 1 ∥x − x ∥ ≥ , ∀n ≠ m n m 2

Thus the sequence (xn) has no convergent sub-sequence which is a contradiction. □ Theorem 3.3. B∗ space is of finite dimension if and only if any bounded set A of X if sequentially compact.

Proof. It suffices to prove that S1 is sequentially compact if and only if any bounded set A of X if sequentially

compact.. ⇒ Trivially since S1 is bounded.

⇐ Suppose S1 is sequentially compact. According to Riesz theorem, we know that dim X < ∞ → ∥∥ ↔ ∥∥T . ∞ ⊂ Let A be an arbitrary bounded subset of X and let (xn)n=1 A. A is bounded implies that TA is bounded → → □ and thus we can find T xnk T x. So, xnk x which implies that A is sequentially compact.

3.2. What is the meaning of equivalence of norm?

Theorem 3.4. Let X be a normed vector space and let ∥∥a ↔ ∥∥b be two norms on X.

• A sequence converges in da if and only if it converges in db where di(x, y) =: ∥x − y∥i.

• (X, da) is complete if and only if (X, db) is complete.

• F is closed in (X, da) if and only if F is closed in (X, db)

Proof. The first result follows directly from the definition of equivalence. Based on the first part oftheorem, the proof of remaining is trivially. □

Theorem 3.5. Let X be a finite dimensional vector space with base {v1, ..., vn}. (n) A sequence (xn) converges if and only if the coordinate sequence (xi ) in R or C converges.

∥∥T is equivalent to all norms on X.

Proof. The second part has been proved above. So, based on this result, we shall see that a sequence

converges in ∥∥ if and only if it converges in ∥∥T . According to the triangle inequality of common norm in Rn, we know that a sequence converges in Rn if and only if the coordinate sequences converge. □

4. Quotient space

∗ Let X be a B space and X0 be a closed subspace of X. Definition 4.1. Quotient space

∀x, y ∈ X, if x − y ∈ X0, we say that x and y are equivalent. Regarding the equivalent classes [x] as new 6 vectors, we get a new vector space, denoted by

X/X0

Proposition:

1 Let [x] ∈ X/X0. Then, x ∈ [x] if and only if [x] = x + X0.

2 [x] + [y] =: x + y + X0

3 α[x] =: αx + X0 4 ∥[x]∥ =: inf{∥z∥ : z ∈ [x]}

5 Let x be an arbitrary element of [x]. ∥[x]∥ = inf{∥x − x0∥ : x0 ∈ X0}

6 If X is a B space, then, X/X0 is also complete.