A Hilbert space is a special case of a Banach space. In a Hilbert space, the norm comes from an inner product. This is described next. It is really a generalization of the familiar dot product from calculus.
Definition 13.2.1 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.
 (13.7) 
 (13.8) 
For a,b ∈ ℂ and x,y,z ∈ X,
 (13.9) 
Note that 13.8 and 13.9 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.
Proof: Let ω ∈ ℂ,ω = 1, and ω(x,y) = (x,y) = Re(x,yω). Let

If y = 0 there is nothing to prove because

and so

This yields

Since this inequality holds for all t ∈ ℝ, it follows from the quadratic formula that

This yields the conclusion and proves the theorem.
Proof: All the axioms are obvious except the triangle inequality. To verify this,
The following lemma is called the parallelogram identity.
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 13.10 holds in the case that x = 0.
Definition 13.2.6 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.
In Hilbert space, one can define a projection map onto closed convex nonempty sets.
Theorem 13.2.8 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.
Proof: Consider uniqueness. Suppose that z_{1} and z_{2} are two elements of K such that for i = 1,2,
 (13.11) 
for all y ∈ K. Also, note that since K is convex,

Therefore, by the parallelogram identity,
Now let λ = inf{x−y : y ∈ K} and let y_{n} be a minimizing sequence. This means

Then by the parallelogram identity, and convexity of K,

Let Px = y.
Corollary 13.2.9 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
 (13.12) 
for all y ∈ K.
Before proving this, consider what it says in the case where the Hilbert space is ℝ^{n}.
The inequality 13.12 is an example of a variational inequality and this corollary characterizes the projection of x onto K as the solution of this variational inequality.
Proof of Corollary: Let z ∈ K and let y ∈ K also. Since K is convex, it follows that if t ∈

Furthermore, every point of K can be written in this way. (Let t = 1 and y ∈ K.) Therefore, z = Px if and only if for all y ∈ K and t ∈ [0,1],

for all t ∈ [0,1] and y ∈ K if and only if for all t ∈

If and only if for all t ∈
 (13.13) 
Now this is equivalent to 13.13 holding for all t ∈

for all t ∈
Corollary 13.2.10 Let K be a nonempty convex closed subset of a Hilbert space, H. Then the projection map, P is continuous. In fact,

Proof: Let x,x^{′}∈ H. Then by Corollary 13.2.9,

Hence

This proves the corollary.
The next corollary is a more general form for the Brouwer fixed point theorem. This was discussed in exercises and elsewhere earlier. However, here is a complete proof.
Corollary 13.2.11 Let f : K → K where K is a convex compact subset of ℝ^{n}. Then f has a fixed point.
Proof: Let K ⊆B

Now the equation also requires x ∈ K and so P
Definition 13.2.12 Let H be a vector space and let U and V be subspaces. U ⊕ V = H if every element of H can be written as a sum of an element of U and an element of V in a unique way.
The case where the closed convex set is a closed subspace is of special importance and in this case the above corollary implies the following.
Corollary 13.2.13 Let K be a closed subspace of a Hilbert space, H, and let x ∈ H. Then for z ∈ K,z = Px if and only if
 (13.14) 
for all y ∈ K. Furthermore, H = K ⊕ K^{⊥} where

and
 (13.15) 
Proof: Since K is a subspace, the condition 13.12 implies Re(x−z,y) ≤ 0 for all y ∈ K. Replacing y with −y, it follows Re(x−z,−y) ≤ 0 which implies Re(x−z,y) ≥ 0 for all y. Therefore, Re(x−z,y) = 0 for all y ∈ K. Now let

This shows that z = Px, if and only if 13.14.
For x ∈ H, x = x − Px + Px and from what was just shown, x − Px ∈ K^{⊥} and Px ∈ K. This shows that K^{⊥} + K = H. Is there only one way to write a given element of H as a sum of a vector in K with a vector in K^{⊥}? Suppose y + z = y_{1} + z_{1} where z,z_{1} ∈ K^{⊥} and y,y_{1} ∈ K. Then
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 13.2.14 Let H be a Hilbert space and let f ∈ H^{′}. Then there exists a unique z ∈ H such that
 (13.16) 
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
