Number Theory: An Introduction to Mathematics

72 I The Expanding Universe of Numbers

Thus‖v‖≥0, with equality if and only ifv=O. Evidently

‖αv‖=|α|‖v‖ for allα∈Fandv∈V.

Inner products and norms are connected bySchwarz’s inequality:

|〈u,v〉| ≤ ‖u‖‖v‖ for allu,v∈V,

with equality if and only ifuandvare linearly dependent. For the proof we may sup-
pose thatuandvare linearly independent, since it is easily seen that equality holds if
u=λvorv=λufor someλ∈F. Then, for allα,β∈F, not both 0,

0 <〈αu+βv,αu+βv〉=|α|^2 〈u,u〉+αβ ̄〈u,v〉+ ̄αβ〈u,v〉+|β|^2 〈v,v〉.

If we chooseα=〈v,v〉andβ=−〈u,v〉, this takes the form

0 <‖u‖^2 ‖v‖^4 − 2 ‖v‖^2 |〈u,v〉|^2 +|〈u,v〉|^2 ‖v‖^2 ={‖u‖^2 ‖v‖^2 −|〈u,v〉|^2 }‖v‖^2.

Hence

|〈u,v〉|^2 <‖u‖^2 ‖v‖^2 ,

as we wished to show. We follow common practice by naming the inequality after
Schwarz (1885), but (cf.§4) it had already been proved forRnby Cauchy (1821) and
forC(I)by Bunyakovskii (1859).
It follows from Schwarz’s inequality that

‖u+v‖^2 =‖u‖^2 + 2 R〈u,v〉+‖v‖^2

≤‖u‖^2 + 2 |〈u,v〉| +‖v‖^2 ≤{‖u‖+‖v‖}^2.

Thus

‖u+v‖≤‖u‖+‖v‖ for allu,v∈V,

with strict inequality ifuandvare linearly independent.
It now follows thatV acquires the structure of a metric space if we define the
distance betweenuandvby

d(u,v)=‖u−v‖.

In the caseV=Rnthis is theEuclidean distance

d(x,y)=

(∑n

j= 1

|ξj−ηj|^2

) 1 / 2

,

and in the caseV=C(I)it is theL^2 -norm

d(f,g)=

(∫b

a

|f(t)−g(t)|^2 dt

) 1 / 2

.