327

is a unit vector in Fn. By the Heine Borel theorem, there exists a subsequence, still denotedby k such that (

yk1 , · · · , ykn)→ (y1, · · · , yn) .

It follows from 13.5 and this that for

y =

n∑i=1

yivi,

0 = limk→∞

∣∣∣∣yk∣∣∣∣ = lim

k→∞

∣∣∣∣∣∣∣∣∣∣

n∑i=1

yki vi

∣∣∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣∣∣

n∑i=1

yivi

∣∣∣∣∣∣∣∣∣∣

but not all the yi equal zero. The last equation follows easily from 13.1 and∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

n∑i=1

yki vi

∣∣∣∣∣∣∣∣∣∣−∣∣∣∣∣∣∣∣∣∣

n∑i=1

yivi

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣ ≤

∥∥∥∥∥n∑

i=1

(yki − yi

)vi

∥∥∥∥∥ ≤n∑

i=1

∣∣yki − yi∣∣ ∥vi∥

This contradicts the assumption that {v1, · · · ,vn} is a basis and proves the second half ofthe inequality. ■

Definition 13.0.5 Let (X, ||·||) be a normed linear space and let {xn}∞n=1 be a sequence ofvectors. Then this is called a Cauchy sequence if for all ε > 0 there exists N such that ifm,n ≥ N, then

||xn − xm|| < ε.

This is written more briefly as

limm,n→∞

||xn − xm|| = 0.

Definition 13.0.6 A normed linear space, (X, ||·||) is called a Banach space if it is com-plete. This means that, whenever, {xn} is a Cauchy sequence there exists a unique x ∈ Xsuch that limn→∞ ||x− xn|| = 0.

Corollary 13.0.7 If (X, ||·||) is a finite dimensional normed linear space with the field ofscalars F = C or R, then (X, ||·||) is a Banach space.

Proof: Let {xk} be a Cauchy sequence. Then letting the components of xk with respectto the given basis be

xk1 , · · · , xkn,

it follows from Theorem 13.0.4, that (xk1 , · · · , xkn

)is a Cauchy sequence in Fn and so(

xk1 , · · · , xkn)→ (x1, · · · , xn) ∈ Fn.

Thus, letting x =∑n

i=1 xivi, it follows from the equivalence of the two norms shown abovethat

limk→∞

∣∣xk − x∣∣ = lim

k→∞

∥∥xk − x∥∥ = 0. ■