8.4. EQUIVALENCE OF NORMS 191

Lemma 8.4.4 Let Ki be a nonempty compact set in F. Then P≡∏ni=1 Ki is compact in Fn.

Proof: Let {xk} be a sequence in P. Taking a succession of subsequences as in theproof of Corollary 8.4.2, there exists a subsequence, still denoted as {xk} such that if xi

k isthe ith component of xk, then limk→∞ xi

k = xi ∈ Ki. Thus if x is the vector of P whose ith

component is xi,

limk→∞

|xk−x| ≡ limk→∞

(n

∑i=1

∣∣xik− xi∣∣2)1/2

= 0

It follows that P is sequentially compact, hence compact.A set K in Fn is said to be bounded if it is contained in some ball B(0,r).

Theorem 8.4.5 A set K ⊆ Fn is compact if it is closed and bounded. If f : K→ R, then fachieves its maximum and its minimum on K.

Proof: Say K is closed and bounded, being contained in B(0,r). Then if x ∈ K, |xi|< rwhere xi is the ith component. Hence K ⊆ ∏

ni=1 D(0,r) , a compact set by Lemma 8.4.4.

By Proposition 7.6.8, since K is a closed subset of a compact set, it is compact. The lastclaim is just the extreme value theorem, Theorem 7.7.1.

Definition 8.4.6 Let {v1, · · · ,vn} be a basis for V where (V, ||·||) is a finite dimensionalnormed vector space with field of scalars equal to either R or C. Define θ : V → Fn asfollows.

θ

(n

∑j=1

α jv j

)≡ α ≡(α1, · · · ,αn)

T

Thus θ maps a vector to its coordinates taken with respect to a given basis.

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

f (α)≡

∥∥∥∥∥∑iα ivi

∥∥∥∥∥≡ ∥∥θ−1

α∥∥

Then it is clear that f is a continuous function defined on Fn. This is because α → ∑i α iviis a continuous map into V and from the triangle inequality x→∥x∥ is continuous as a mapfrom V to R.

Lemma 8.4.7 There exists δ > 0 and ∆≥ δ such that

δ = min{ f (α) : |α|= 1} , ∆ = max{ f (α) : |α|= 1}

Also,

δ |α| ≤∥∥θ−1

α∥∥≤ ∆ |α| (8.4.21)

δ |θv| ≤ ∥v∥ ≤ ∆ |θv| (8.4.22)