Mathematical Tools for Physics

6—Vector Spaces 157

How much bigger than zero the left side is will depend on the parameterλ. To find the smallest value that the
left side can have you simply differentiate. Letλ=x+iyand differentiate with respect toxandy, setting the
results to zero. This gives (see problem 5 )

λ=

〈

~v,~u

〉/〈

~v,~v

〉

. (17)

Substitute this value into the above inequality ( 16 )

〈

~u,~u

〉

+

∣

∣

〈

~u,~v

〉∣

∣^2

〈

~v,~v

〉 −

∣

∣

〈

~u,~v

〉∣

∣^2

〈

~v,~v

〉 −

∣

∣

〈

~u,~v

〉∣

∣^2

〈

~v,~v

〉 ≥ 0. (18)

This becomes
∣
∣

〈

~u,~v

〉∣

∣^2 ≤

〈

~u,~u

〉〈

~v,~v

〉

(19)

This isn’t quite the result I wanted, because Eq. ( 14 ) is written differently. It refers to a norm and I haven’t
established that the square root of

〈

~v,~v

〉

isa norm. When I do, then the square root of this is the inequality I
want.

For a couple of examples of this inequality, take specific scalar products. First the common directed line
segments:
〈
~u,~v

〉

=~u.~v=uvcosθ, so |uvcosθ|^2 ≤|u|^2 |v|^2

∣

∣

∣

∣

∣

∫b

a

dxf(x)*g(x)

∣

∣

∣

∣

∣

2

≤

[∫

b

a

dx|f(x)|^2

][∫

b

a

dx|g(x)|^2

]

The first of these is familiar, but the second is not, though when you look at it from the general vector space
viewpoint they are essentially the same.

Norm from a Scalar Product

The equation ( 4 ),‖~v‖=

√〈

~v,~v

〉

, defines a norm. Properties one and two for a norm are simple to check. (Do