5. Finite Dimensional Normed Vector Spaces

Michael Richard

April 21, 2006

5.1 Some Definitions

1. A vector space (over R) consists of a set V and operations:

(a) Vector Addition:

V × V → V

(x, y) 7→ xy

(Which is commutatative, associative, zeroes and inverses)(b) Scalar Multiplication with distribution laws:

R× V → V

(λ, x) 7→ λx

Where 1 · v = v and 0 · v = 0

2. A norm on a vector space V is a function

|| · || : V 7→ [0,∞)

with the properties

(a) Positive Definite

||x|| = 0 ⇔ x = 0

(b) Homogeneous

||λ · x|| = |λ| · ||x|| ∀ λ ∈ R, x ∈ V

(c) Triangle inequality

||x + y|| ≤ ||x||+ ||y|| ∀ x, y ∈ V

The pair (V, || · ||) is called a normed vector space

3. Let (V, ||·||) be a normed vector space. A sequence (xn)n∈N in V convergesto x ∈ V if

limn→∞||xn − x|| = 0

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5.2 Remark

If A ⊂ V , we still have the notions of openness, closedness, compactness andcompleteness. Recall that

1. A is said to be open if, for each point x ∈ A, there exists some r > 0 suchthat Br(x) ⊆ A

2. A is said to be closed if each convergent sequence xn ∈ A converges to apoint x ∈ A

3. A is said to be compact if every sequence in xn ∈ A has a convergentsubsequence xnk

convering to a point x ∈ A

4. A is said to be complete if every Cauchy sequence in A converges to apoint in A

5.3 Remark

Let (V, || · ||) be a normed vector space where V is finite-dimensional. Thismeans there exists a basis (v1, v2, ..., vn) ∈ V such that

1. (v1, v2, ..., vn) are linearly independent that is

0 = λ1v1 + λ2v2 + ... + λnvn ⇒ λi = 0 ∀ i

2. Each v ∈ V can be written as

v = λ1v1 + λ2v2 + ... + λnvn (uniquely)

Fixing such a basis (v1, v2, ..., vn) we can define a mapping T : V → Rn

λ1v1 + λ2v2 + ... + λnvn 7→ (λ1, λ2, ..., λn)

Where T is linear and bijective

Putting||(λ1, λ2, ..., λn)|| := ||λ1v1 + λ2v2 + ... + λnvn||

defines a norm on Rn

Therefore, understanding all n-dimensional normed vector spaces is equiva-lent to understanding all norms on Rn.

Example: There exist many different norms on Rn.

||(λ1, λ2, ..., λn)||1 := |λ1|+ |λ2|+ ... + |λn|

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||(λ1, λ2, ..., λn)||2 :=√

λ21 + λ2

2 + ... + λ2n

||(λ1, λ2, ..., λn)||∞ := max(|λ1|, |λ2|, ...|λn|)

More generally,

||(λ1, λ2, ..., λn)||p := (|λ1|p + ... + |λn|p)1/p

5.4 Definition

Let V be a vector space and let || · ||1 and || · ||2 be two norms.

We say that ||·||1 and ||·||2 are equivalent if there exists c1, c2 > 0 such that:

||x||1 ≤ c1||x||2 and

||x||2 ≤ c2||x||1 ∀ x ∈ V

5.5 Proposition

Let || · ||1 and || · ||2 be two equivalent norms on V .

1. For a sequence (xn)n in V , we have

(xn)n converges w.r.t. || · ||1 ⇔ (xn)n converges w.r.t. || · ||2

2. For a subset A ⊂ V we have

A open w.r.t. || · ||1 ⇔ A is open w.r.t. || · ||2

A closed w.r.t. || · ||1 ⇔ A is closed w.r.t. || · ||2

A compact w.r.t. || · ||1 ⇔ A is compact w.r.t. || · ||2

A complete w.r.t. || · ||1 ⇔ A is complete w.r.t. || · ||2

Proof

1. ||xn − x||1 ≤ c1||xn − x||2

Note that ||xn − x||2 → 0 ⇒ ||xn − x||1 → 0The same argument can be used to show ⇐.

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2. This is clear. Note that for ”Open”, each ball in || · ||1 norm containsa ball in || · ||2, and vice versa. Similar arguments can be used forclosed, compact and complete.

Thus, all topological statements are the same w.r.t. equivalent norms.

5.6 Theorem

Let V be a finite dimensional vector space. Then all norms on V areequivalent.

Proof By Remark 5.3, it suffices to consider V = Rn.

We show that any norm || · || on Rn is equivalent to special norm || · ||∞given by

||(λ1, λ2, ..., λn)||∞ := max(|λ1|, |λ2|, ...|λn|)

Let (e1, e2, ..., en) be a standard basis of Rn

where e1 = (1, 0, 0, ..., 0), e2 = (0, 1, 0, ..., 0) etc.

||(λ1, ..., λn)|| = ||n∑

i=1

λiei||

≤n∑

i=1

||λiei|| where ||λiei|| = |λi|||ei||

≤ n ·maxi|λi| ·maxi||ei||= c1 · ||(λ1, ..., λn)||∞ with c1 := n ·maxi=1,...,n||ei||

Now define a functionf : Rn → Rx 7→ f(x) := ||x||⇒ continuous w.r.t. || · ||∞ - norm, since

|f(x)− f(y)| = |||x|| − ||y|||≤ ||x− y||≤ c1||x− y||∞ by above

Now let’s restrict f to

S := {x ∈ Rn|||x||∞ = 1}

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Heine Borel ⇒ S is compact, for || · ||∞ - norm

The Extreme Value Theorem (EVT) ⇒ there exists a minimum of f on S.

In other words, EV T ⇒ ∃ a ∈ S : f(a) ≤ f(x) ∀ x ∈ S.

Put c2 := f(a) = ||a||

(c2 = 0 ⇒ ||a|| = 0 = a ⇒ ||a||∞ = 0 ⇒ a /∈ S), so c2 > 0

This means||x|| ≥ c2 ∀ x with ||x||∞ = 1

y 6= 0 ∈ R → x :=y

||y||∞⇒ ||x||∞ = 1

⇒ ||x|| ≥ c2 ⇒ ||y|| ≥ c2||y||∞

⇒ ||y||∞ ≤ 1c2||y||

5.7 Corollary

Let V be a finite dimensional normed vector space

1. V is complete

2. A ⊂ V is compact ⇔ A is closed and bounded

5.8 Definition

A complete normed vector space is called a Banach Space.

5.9 Example

Every finite dimensional normed vector space is a Banach space, in par-ticular, Rn with norms

||(λ1, ..., λn||p := p√|λ1|p + ... + |λn|p

is, for any 1 ≤ p < ∞, a Banach Space.

We can also include p = ∞

||(λ1, ..., λn||∞ := max(|λ1|, ..., |λn|)

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5.10 Example

Let (V, || · ||) be a normed vector space and prove that

|||x|| − ||y||| ≤ ||x− y|| ∀ x, y ∈ V

and use this to show that the mapping

|| · || : V → R

x 7→ ||x||

is continuous.

In other words, we want to show that

||x|| − ||y|| ≤ ||x− y|| if ||x|| ≥ ||y|| and

||y|| − ||x|| ≤ ||x− y|| if ||y|| ≥ ||x||

First, x = x− y + y

Then, ||x|| = ||(x− y) + y||≤ ||x− y||+ ||y|| By the Triangle Inequality, now rearrange

||x|| − ||y|| ≤ ||x− y||

We do a similar computation for when ||y|| ≥ ||x||, using y = y − x + xand the fact that ||x− y|| = ||y − x||

Now, show that ∀ ε > 0, ∃ δ > 0, x, x0 ∈ V such that ||x − x0|| < δ ⇒|||x|| − ||x0||| < ε

We know that |||x|| − ||x0||| ≤ ||x− x0||

Let ε > 0 be given, and let δ = ε

Then, for any ||x− x0|| < δ,

|||x|| − ||x0||| ≤ ||x− x0|| < δ = ε and so |||x|| − ||x0||| < ε

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