Calculus of Real and Complex Variables

2.5 Equivalence Of Norms

As mentioned above, it makes absolutely no difference which norm you decide to use on ℝ^p. This holds in general finite dimensional normed spaces and is shown here. Of course the main interest here is where the normed linear space is

but it is no harder to present the more general result where you have a finite dimensional vector space V which has a norm. If you have not seen such things, just let V be ℝ^p in what follows or consider the problems at end of the chapter.

The following fundamental lemma comes from the extreme value theorem for continuous functions defined on a compact set. Let

∥ ∥ ∥∥∑ ∥∥ |||| −1 |||| f (α ) ≡ ∥∥ αivi∥∥ ≡ θ α i

Then it is clear that f is a continuous function. This is because α →∑ _iα_iv_i is a continuous map into V and from the triangle inequality x →

is continuous as a map from V to ℝ.

Proof: These numbers exist thanks to the extreme value theorem, Theorem 2.2.25. It cannot be that δ = 0 because if it were, you would have

= 1 but ∑ _j=1ⁿα_kv_j = 0 which is impossible since is linearly independent. The first of the above inequalities follows from

∥∥ −1 α ∥∥ ( α ) δ ≤ ∥∥θ |α|∥∥ = f |α-| ≤ Δ

the second follows from observing that θ⁻¹α is a generic vector v in V . ■

Now we can draw several conclusions about

for V finite dimensional.

Proof: First note that the inequalities 2.15 and 2.16 show that both θ⁻¹ and θ are continuous. Thus these take convergent sequences to convergent sequences.

Let

_k=1^∞ be a Cauchy sequence. Then from 2.16, _k=1^∞ is a Cauchy sequence. Thanks to Theorem 2.2.25, it converges to some β ∈ F^p. It follows that lim_k→∞θ⁻¹θw_k = lim_k→∞w_k = θ⁻¹β ∈ V . This shows completeness.

Next let K be a closed and bounded set. Let

⊆ K. Then ⊆ θK which is also a closed and bounded set thanks to the inequalities 2.15 and 2.16. Thus there is a subsequence still denoted with k such that θw_k → β ∈ F^p. Then as just done, w_k → θ⁻¹β. Since K is closed, it follows that θ⁻¹β ∈ K.

Finally, why are the only compact sets those which are closed and bounded? Let K be compact. If it is not bounded, then there is a sequence of points of K,

_m=1^∞ such that ≥ m. It follows that it cannot have a convergent subsequence because the points are further apart from each other than 1/2. Hence K is not sequentially compact and consequently it is not compact. It follows that K is bounded. If K is not closed, then there exists a limit point k which is not in K. (Recall that closed means it has all its limit points.) By Theorem 2.1.9, there is a sequence of distinct points having no repeats and none equal to k denoted as _m=1^∞ such that k^m → k. Then this sequence fails to have a subsequence which converges to a point of K. Hence K is not sequentially compact. Thus, if K is compact then it is closed and bounded.

The last part is identical to the proof in Theorem 2.2.25. You just take a convergent subsequence of a minimizing (maximizing) sequence and exploit continuity. ■

Next is the theorem which states that any two norms on a finite dimensional vector space are equivalent. In particular, any two norms on ℝ^p are equivalent.

Proof: In Lemma 2.5.2, let δ,Δ go with

and , go with . Then using the inequalities of this lemma,

ˆ ˆ ||v|| ≤ Δ |θv| ≤ Δ-|||v||| ≤ ΔΔ-|θv| ≤ Δ-Δ-||v|| ˆδ ˆδ δ ˆδ

and so

ˆδ ˆΔ --||v|| ≤ |||v ||| ≤-||v|| Δ δ

Thus the norms are equivalent. ■