326 CHAPTER 13. NORMS

Similarly, for y ∈ Y with basis {w1, · · · ,wm}, and yi its components with respect to thisbasis,

|y| ≡

(m∑i=1

|yi|2)1/2

For A ∈ L (X,Y ) , the space of linear mappings from X to Y,

||A|| ≡ sup{|Ax| : |x| ≤ 1}. (13.3)

The first thing to show is that the two norms, ||·|| and |·| , are equivalent. This meansthe conclusion of the following theorem holds.

Theorem 13.0.4 Let (X, ||·||) be a finite dimensional normed linear space and let |·| bedescribed above relative to a given basis, {v1, · · · ,vn} . Then |·| is a norm and there existconstants δ,∆ > 0 independent of x such that

δ ||x|| ≤ |x| ≤∆ ||x|| . (13.4)

Proof: All of the above properties of a norm are obvious except the second, the triangleinequality. To establish this inequality, use the Cauchy Schwarz inequality to write

|x+ y|2 ≡n∑

i=1

|xi + yi|2 ≤n∑

i=1

|xi|2 +n∑

i=1

|yi|2 + 2Re

n∑i=1

xiyi

≤ |x|2 + |y|2 + 2

(n∑

i=1

|xi|2)1/2( n∑

i=1

|yi|2)1/2

= |x|2 + |y|2 + 2 |x| |y| = (|x|+ |y|)2

and this proves the second property above.It remains to show the equivalence of the two norms. By the Cauchy Schwarz inequality

again,

||x|| ≡

∣∣∣∣∣∣∣∣∣∣

n∑i=1

xivi

∣∣∣∣∣∣∣∣∣∣ ≤

n∑i=1

|xi| ||vi|| ≤ |x|

(n∑

i=1

||vi||2)1/2

≡ δ−1 |x| .

This proves the first half of the inequality.Suppose the second half of the inequality is not valid. Then there exists a sequence

xk ∈ X such that ∣∣xk∣∣ > k

∣∣∣∣xk∣∣∣∣ , k = 1, 2, · · · .

Then define

yk ≡ xk

|xk|.

It follows ∣∣yk∣∣ = 1,

∣∣yk∣∣ > k

∣∣∣∣yk∣∣∣∣ . (13.5)

Letting yki be the components of yk with respect to the given basis, it follows the vector(yk1 , · · · , ykn

)