Proof that Finite-Dimensional Normed Vector Spaces are Closed | Assignments Theories of Communication

ECE-542 Digital Communication Theory

Spring 2004

Homework Assignment 4 solutions

Problem 1. Reconstruct the proof done in class which shows that any ﬁnite-dimensional normed

vector space is closed. Be precise, mathematically descriptive of the concepts, and justify your

steps thoroughly along the way.

We prove this by induction on the dimension. First note that by a norm, || · ||,wemeana

mapping from the vector space to the reals such that (1) ||u||≥0 for any vector uand ||u|| =0

only if u=0;(2)||αu|| =|α|||u||, for any scalar αand vector u; and (3) ||u+v|| ≤ ||u|| +||v|| for

any vectors uand v.

Induction initialization:

Assume that the dimension is 1. We need to show that the one-dimensional space, say M1,is

closed. That is, if limn→∞ ||x−xn|| =0andxn∈M1, then x∈M1. Note that we can write

xn=anv1, where v1spans M1and anis a scalar (real or complex). We will show that the sequence

anis bounded, which means that it has a convergent subsequence ani. (This is a consequence of the

completeness of the real/comples ﬁeld, which is called the Bolzano Wierstrass Theorem). Assume

therefore that ani→a0, for some scalar a0. It follows that limi→∞ ||xni−a0v1|| = 0. But since

xnconverges to x, any subsequence of it also converges to x. Thus, limi→∞ ||x−xni|| = 0. Hence

x=a0v1∈M1. It remains to prove that anis bounded. Note that ||xn|| ≤ ||x−xn|| +||x|| <∞,

and hence |an|=||xn||/||v1|| <∞.

Induction hypothesis:

Any k-dimensional space Mk,k≥1, is closed.

Proof of the induction step:

We show that any (k+ 1)-dimensional space Mk+1 , is closed. That is, if xn∈Mk+1 is conver-

gent to x, then x∈Mk+1. We can write xn=a1

1u1+...+ak+1

nuk+1, where u1,...,u

k+1 span

Mk+1. We now show that the sequence ak+1

nis bounded. We prove this by contradiction: Suppose

that ak+1

nis unbounded, and without loss of generality, assume that ak+1

n→∞. Observe that

xn/ak+1

n=a1

nu1/ak+1

n+...+ak

nu1/ak+1

n+uk+1.Now||xn/ak+1

n|| → 0 since ak+1

n→∞.Thus,

nu1/ak+1

n+...+ak

nuk/ak+1

n→−uk+1. But since a1

nu1/ak+1

n+...+ak

nuk/ak+1

nis in a k-dimensional

space spanned by u1,...,u

k, which is closed (by the induction hypothesis), its limit must be in

the space spanned by u1,...,u

kand not uk+1, which is a contradiction. Hence, ak+1

nmust be

bounded. Consequently, it has a convergent subsequence ak+1

ni, with a limit say ak+1

0. Observe that

niu1+...+ak

niuk=xni−ak+1

niuk+1, which converges to x−ak+1

0uk+1.Buta1

niu1+...+ak

niuk

is in a k-dimensional space spanned by u1,...,u

k, its limit, which is x−ak+1

0uk+1,mustalsobe

in the same space, namely, x−ak+1

0uk+1 =b1u1+...+bkuk, for some scalars b1,...,b

k+1.This

implies that x=b1u1+...+bkuk+ak+1

0uk+1.Thus,x∈Mk+1 as desired.

From the text: 3.29, 3.32, 3.33, 3.38 (pick up solution from my oﬃce).

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ECE-542 Digital Communication Theory Spring 2004 Homework Assignment 4 solutions

Problem 1. Reconstruct the proof done in class which shows that any finite-dimensional normed vector space is closed. Be precise, mathematically descriptive of the concepts, and justify your steps thoroughly along the way.

We prove this by induction on the dimension. First note that by a norm, || · ||, we mean a mapping from the vector space to the reals such that (1) ||u|| ≥ 0 for any vector u and ||u|| = 0 only if u = 0; (2) ||αu|| = |α|||u||, for any scalar α and vector u; and (3) ||u + v|| ≤ ||u|| + ||v|| for any vectors u and v.

Induction initialization: Assume that the dimension is 1. We need to show that the one-dimensional space, say M 1 , is closed. That is, if limn→∞ ||x − xn|| = 0 and xn ∈ M 1 , then x ∈ M 1. Note that we can write xn = anv 1 , where v 1 spans M 1 and an is a scalar (real or complex). We will show that the sequence an is bounded, which means that it has a convergent subsequence ani. (This is a consequence of the completeness of the real/comples field, which is called the Bolzano Wierstrass Theorem). Assume therefore that ani → a 0 , for some scalar a 0. It follows that limi→∞ ||xni − a 0 v 1 || = 0. But since xn converges to x, any subsequence of it also converges to x. Thus, limi→∞ ||x − xni || = 0. Hence x = a 0 v 1 ∈ M 1. It remains to prove that an is bounded. Note that ||xn|| ≤ ||x − xn|| + ||x|| < ∞, and hence |an| = ||xn||/||v 1 || < ∞.

Induction hypothesis: Any k-dimensional space Mk, k ≥ 1, is closed.

Proof of the induction step: We show that any (k + 1)-dimensional space Mk+1, is closed. That is, if xn ∈ Mk+1 is conver- gent to x, then x ∈ Mk+1. We can write xn = a^11 u 1 +... + ak n+1 uk+1, where u 1 ,... , uk+1 span Mk+1. We now show that the sequence ak n+1 is bounded. We prove this by contradiction: Suppose that ak n+1 is unbounded, and without loss of generality, assume that ak n+1 → ∞. Observe that xn/ak n+1 = a^1 nu 1 /ak n+1 +... + aknu 1 /ak n+1 + uk+1. Now ||xn/ak n+1 || → 0 since ak n+1 → ∞. Thus, a^1 nu 1 /ak n+1 +.. .+aknuk/ak n+1 → −uk+1. But since a^1 nu 1 /ak n+1 +.. .+aknuk/ak n+1 is in a k-dimensional space spanned by u 1 ,... , uk, which is closed (by the induction hypothesis), its limit must be in the space spanned by u 1 ,... , uk and not uk+1, which is a contradiction. Hence, ak n+1 must be bounded. Consequently, it has a convergent subsequence ak n+1i , with a limit say ak 0 +1. Observe that a^1 ni u 1 +... + akni uk = xni − ak n+1i uk+1, which converges to x − ak 0 +1 uk+1. But a^1 ni u 1 +... + akni uk is in a k-dimensional space spanned by u 1 ,... , uk, its limit, which is x − ak 0 +1 uk+1, must also be in the same space, namely, x − ak 0 +1 uk+1 = b 1 u 1 +... + bkuk, for some scalars b 1 ,... , bk+1. This implies that x = b 1 u 1 +... + bkuk + ak 0 +1 uk+1. Thus, x ∈ Mk+1 as desired.

From the text: 3.29, 3.32, 3.33, 3.38 (pick up solution from my office).

Proof that Finite-Dimensional Normed Vector Spaces are Closed, Assignments of Theories of Communication

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