As explained earlier, a normed linear space is a vector space X with a norm. This is a map
Then, as discussed earlier, this is a metric space if d
Definition 23.1.1 Let X be a vector space. An inner product is a mapping from X × X to ℂ if X is complex and from X × X to ℝ if X is real, denoted by (x,y) which satisfies the following.
 (23.1) 
 (23.2) 
For a,b ∈ ℂ and x,y,z ∈ X,
 (23.3) 
Note that 23.2 and 23.3 imply (x,ay + bz) = a(x,y) + b(x,z). Such a vector space is called an inner product space.
The Cauchy Schwarz inequality is fundamental for the study of inner product spaces. The proof is identical to that given earlier in Theorem 11.4.4 on Page 730.
Also, as earlier, the norm given by the inner product, really is a norm.
The following lemma is called the parallelogram identity. It was also discussed earlier.
The proof, a straightforward application of the inner product axioms, is left to the reader.
Proof: By the Cauchy Schwarz inequality, if x≠0,

It is obvious that 23.4 holds in the case that x = 0.
Definition 23.1.6 A Hilbert space is an inner product space which is complete. Thus a Hilbert space is a Banach space in which the norm comes from an inner product as described above. Often people use the symbol
In Hilbert space, one can define a projection map onto closed convex nonempty sets.
This was done in the problems beginning with Problem 8 on Page 777.
Theorem 23.1.8 Let K be a closed convex nonempty subset of a Hilbert space, H, and let x ∈ H. Then there exists a unique point Px ∈ K such that Px − x≤y − x for all y ∈ K.
Corollary 23.1.9 Let K be a closed, convex, nonempty subset of a Hilbert space, H, and let x ∈ H. Then for z ∈ K,z = Px if and only if
 (23.5) 
for all y ∈ K.
Here we present an easier version which will be sufficient for what is needed. First is a simple lemma which is interesting for its own sake.
Proof: Consider the following:
Theorem 23.1.11 Let H be a Hilbert space and let M be a closed subspace. Then if w ∈ H, there exists a unique Pw ∈ H such that

for all y ∈ M. Thus Pw is the point of M closest to w and it is unique. Then z ∈ M equals Pw if and only if

Also the map w → Pw is a linear map which satisfies
Proof: Let λ ≡ inf

and so Pw ≡ z is a closest point. There can be only one closest point because if z_{i} works, then by the parallelogram identity again,

Now for the characterization: For z ∈ M, letting y ∈ M consider z + t
 (*) 
Then for z to be the closest point to w, one needs the above to be minimized when t = 0. Taking a derivative, this requires that

for any y ∈ M. But this is the same as saying that Re
Conversely, if
As to P being linear, for y ∈ M arbitrary,

Also,

By uniqueness, P
Finally,

which yields the desired estimate. ■
Note that the operator norm of P equals 1.

Now pick w ∈ M with
Definition 23.1.12 If A : X → Y is linear where X,Y are two normed linear spaces, then A is said to be in ℒ

In case that Y = F the field of scalars, equal to either ℝ or ℂ, ℒ
Thus P the projection map in the above is in ℒ
There is a general easy result about ℒ
Theorem 23.1.13 Let X and Y be two normed linear spaces and let L : X → Y be linear (L(ax + by) = aL(x) + bL(y) for a,b scalars and x,y ∈ X). The following are equivalent
a.) L is continuous at 0
b.) L is continuous
c.) There exists K > 0 such that
Proof: a.)⇒b.) Let x_{n} → x. It is necessary to show that Lx_{n} → Lx. But (x_{n} − x) → 0 and so from continuity at 0, it follows

so Lx_{n} → Lx. This shows a.) implies b.).
b.)⇒c.) Since L is continuous, L is continuous at 0. Hence Lx_{Y } < 1 whenever x_{X} ≤ δ for some δ. Therefore, suppressing the subscript on the ,

Hence

c.)⇒a.) follows from the inequality given in c.). ■
The following theorem is called the Riesz representation theorem for the dual of a Hilbert space. If z ∈ H then define an element f ∈ H^{′} by the rule
Theorem 23.1.14 Let H be a Hilbert space and let f ∈ H^{′}. Then there exists a unique z ∈ H such that
 (23.8) 
for all x ∈ H.
Proof: Letting y,w ∈ H the assumption that f is linear implies

which shows that yf(w) − f(y)w ∈ f^{−1}

Thus, solving for f

Let z = f(w)w∕w^{2}. This proves the existence of z. If f
If R : H → H^{′} is defined by Rx
