11.2. THE GRAM SCHMIDT PROCESS 277

which verifies 11.2.Next suppose 11.2. Is it true that 11.3 follows? Let z ∈ M be arbitrary and let |θ| =

1, θ̄ (x− w,w − z) = |(x− w,w − z)|. Then let

p (t) ≡ |x− w + tθ (w − z)|2 = |x− w|2 + 2Re (x− w, tθ (w − z)) + t2 |w − z|2

= |x− w|2 + 2Re tθ̄ (x− w, (w − z)) + t2 |w − z|2

= |x− w|2 + 2t |(x− w, (w − z))|+ t2 |w − z|2

Then p has a minimum when t = 0 and so p′ (0) = 2 |(x− w, (w − z))| = 0 which shows11.3. This proves the first part of the theorem since z is arbitrary.

It only remains to verify that w given in 11.4 satisfies 11.3 and is the only point of Mwhich does so.

First, could there be two minimizers? Say w1, w2 both work. Then by the above char-acterization of minimizers,

(x− w1, w1 − w2) = 0

(x− w2, w1 − w2) = 0

Subtracting gives (w1 − w2, w1 − w2) = 0. Hence the minimizer is unique.Finally, it remains to show that the given formula works. Letting {e1, · · · , em} be an

orthonormal basis for M, such a thing existing by the Gramm Schmidt process,(x−

m∑i=1

(x, ei) ei, ek

)= (x, ek)−

m∑i=1

(x, ei) (ei, ek)

= (x, ek)−m∑i=1

(x, ei) δik

= (x, ek)− (x, ek) = 0

Since this inner product equals 0 for arbitrary ek, it follows that(x−

m∑i=1

(x, ei) ei, z

)= 0

for every z ∈M because each such z is a linear combination of the ei. Hence∑m

i=1 (x, ei) eiis the unique minimizer. ■

Example 11.2.5 Consider X equal to the continuous functions defined on [−π, π] and letthe inner product be given by ∫ π

−π

f (x) g (x)dx

It is left to the reader to verify that this is an inner product. Letting ek be the functionx→ 1√

2πeikx, define

M ≡ span({ek}nk=−n

).

Then you can verify that

(ek, em) =

∫ π

−π

(1√2πe−ikx

)(1√2πemix

)dx =

1

2π

∫ π

−π

ei(m−k)x = δkm