518 CHAPTER 19. HILBERT SPACES

Proof: All the axioms are obvious except the triangle inequality. To verify this,

||x+ y||2 ≡ (x+ y,x+ y)≡ ||x||2 + ||y||2 +2Re(x,y)

≤ ||x||2 + ||y||2 +2 |(x,y)|≤ ||x||2 + ||y||2 +2 ||x|| ||y||= (||x||+ ||y||)2.

The following lemma is called the parallelogram identity.

Lemma 19.1.4 In an inner product space,

||x+ y||2 + ||x− y||2 = 2||x||2 +2||y||2.

The proof, a straightforward application of the inner product axioms, is left to thereader.

Lemma 19.1.5 For x ∈ H, an inner product space,

||x||= sup||y||≤1

|(x,y)| (19.1.4)

Proof: By the Cauchy Schwarz inequality, if x ̸= 0,

||x|| ≥ sup||y||≤1

|(x,y)| ≥(

x,x||x||

)= ||x|| .

It is obvious that 19.1.4 holds in the case that x = 0.

Definition 19.1.6 A Hilbert space is an inner product space which is complete. Thus aHilbert space is a Banach space in which the norm comes from an inner product as de-scribed above.

In Hilbert space, one can define a projection map onto closed convex nonempty sets.

Definition 19.1.7 A set, K, is convex if whenever λ ∈ [0,1] and x,y∈K, λx+(1−λ )y∈K.

Theorem 19.1.8 Let K be a closed convex nonempty subset of a Hilbert space, H, and letx ∈H. Then there exists a unique point Px ∈ K such that ||Px−x|| ≤ ||y−x|| for all y ∈ K.

Proof: Consider uniqueness. Suppose that z1 and z2 are two elements of K such thatfor i = 1,2,

||zi− x|| ≤ ||y− x|| (19.1.5)

for all y ∈ K. Also, note that since K is convex,

z1 + z2

2∈ K.