Kenneth Kuttler

4.4. EQUIVALENCE OF NORMS 111

Definition 4.4.6 Let {v1, · · · ,vn} be a basis for V where (V,∥·∥) is a finite dimen-sional normed vector space with field of scalars equal to either R or C. Define θ : V → Fn

as follows.

∑j=1

α jv j

)≡α≡ (α1, · · · ,αn)

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−1α

∥∥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 amap from V to R.

Lemma 4.4.7 There exists δ > 0 and ∆≥ δ such that

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

Also,

δ |α| ≤∥∥θ−1α

∥∥≤ ∆ |α| (4.21)δ |θv| ≤ ∥v∥ ≤ ∆ |θv| (4.22)

Proof: These numbers exist thanks to Theorem 4.4.5. It cannot be that δ = 0 because ifit were, you would have |α|= 1 but ∑

nj=1 αkv j = 0 which is impossible since {v1, · · · ,vn}

is linearly independent. The first of the above inequalities follows from δ ≤∥∥∥θ−1 α|α|

∥∥∥ =f(

α|α|

)≤ ∆. The second follows from observing that θ

−1α is a generic vector v in V . ■Note that these inequalities yield the fact that convergence of the coordinates with re-

spect to a given basis is equivalent to convergence of the vectors. More precisely, to saythat limk→∞v

k = v is the same as saying that limk→∞ θvk = θv. Indeed,

δ |θvn−θv| ≤ ∥vn−v∥ ≤ ∆ |θvn−θv|

Now we can draw several conclusions about (V,∥·∥) for V finite dimensional.

Theorem 4.4.8 Let (V,∥·∥) be a finite dimensional normed linear space. Then thecompact sets are exactly those which are closed and bounded. Also (V,∥·∥) is complete. IfK is a closed and bounded set in (V,∥·∥) and f : K→R, then f achieves its maximum andminimum on K.

Proof: First note that the inequalities 4.21 and 4.22 show that both θ−1 and θ are

continuous. Thus these take convergent sequences to convergent sequences.Let {wk}∞

k=1 be a Cauchy sequence. Then from 4.22, {θwk}∞

k=1 is a Cauchy sequence.Thanks to Theorem 4.4.5, it converges to some β ∈ Fn. It follows that limk→∞ θ

−1θwk =

limk→∞wk = θ−1β ∈V . This shows completeness.

4.4. EQUIVALENCE OF NORMS 111Definition 4.4.6 Let {v,,---,v,} be a basis for V where (V,||-||) is a finite dimen-sional normed vector space with field of scalars equal to either R or C. Define 0: V — F"as follows.n_, = T0 y QA jVj =a= (Q1,-°- , An)j=lThus 0 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. LetYi aj;iThen it is clear that f is a continuous function defined on F”. This is because a@ > )°; 00;is a continuous map into V and from the triangle inequality « — ||a|| is continuous as amap from V to R.f(a)= = \|o- ‘aLemma 4.4.7 There exists & > 0 and A> 6 such that6=min{f(a@):|a|=1}, A=max{f (a): |a|=1}Also,dljal < ||@-'al| <Ala| (4.21)6|Ov| <_ |lv|| <AlOv| (4.22)Proof: These numbers exist thanks to Theorem 4.4.5. It cannot be that 6 = 0 because ifit were, you would have |a| = 1 but );_ @vj = 0 which is impossible since {v1,--- , Un}is linearly independent. The first of the above inequalities follows from 6 < |e Tal | =f ( 2) < A. The second follows from observing that @~'a is a generic vector v in V. llNote that these inequalities yield the fact that convergence of the coordinates with re-spect to a given basis is equivalent to convergence of the vectors. More precisely, to saythat lim;_,.. v§ = v is the same as saying that lim,_,.. 9v‘ = Ov. Indeed,6 |Ov, — Bv| < |lvn —v|| < A|Ov, — Ov|Now we can draw several conclusions about (V, ||-||) for V finite dimensional.Theorem 4.4.8 Let (V,||-||) be a finite dimensional normed linear space. Then thecompact sets are exactly those which are closed and bounded. Also (V,||-||) is complete. IfK is aclosed and bounded set in (V,||-||) and f : K > R, then f achieves its maximum andminimum on K.Proof: First note that the inequalities 4.21 and 4.22 show that both @~! and @ arecontinuous. Thus these take convergent sequences to convergent sequences.Let {w;};_, be a Cauchy sequence. Then from 4.22, {@w,;};_, is a Cauchy sequence.Thanks to Theorem 4.4.5, it converges to some ( € F”. It follows that lim,_,.. g7! Ow, =limy_5.. we = 9 |B EV. This shows completeness.