12.3. RIESZ REPRESENTATION THEOREM, ADJOINT MAP 313

From 12.3 and 12.4,

|u|2 = d2 +∑i, j

Gi jziz j = d2 +zT G(v1, ...,vn)z = d2 +yTz

Then from 12.3 and 12.4,(G(v1, ...,vn) 0

yT 1

)(z

d2

)=

(y

|u|2

)

By Cramer’s rule,

d2 =

det

(G(v1, ...,vn) y

yT |u|2

)det(G(v1, ...,vn))

≡ det(G(v1, ...,vn,u))det(G(v1, ...,vn))

This proves the interesting approximation theorem.

Theorem 12.2.3 Suppose {v1, ...,vn} is a linearly independent set of vectors in H an innerproduct space. Then if u ∈ H, and d is the distance to V ≡ span(v1, ...,vn) , then d2 =det(G(v1,...,vn,u))det(G(v1,...,vn))

.

12.3 Riesz Representation Theorem, Adjoint MapThe next theorem is one of the most important results in the theory of inner product spaces.It is called the Riesz representation theorem.

Theorem 12.3.1 Let f ∈L (H,F) where H is a Hilbert space and f is continuous. Recallthat in finite dimensions, this is automatic. Then there exists a unique z ∈ H such that forall x ∈ H, f (x) = (x,z) .

Proof: First I will verify uniqueness. Suppose z j works for j = 1,2. Then for all x ∈H,

0 = f (x)− f (x) = (x,z1− z2)

and so z1 = z2.If f (H) = 0, let z = 0 and this works. Otherwise, let u /∈ f−1 (0) which is a closed

subspace of H. Let w = u−Pu ̸= 0. Then

f ( f (w)x− f (x)w) = f (w) f (x)− f (x) f (w) = 0

and so from Theorem 12.1.7,

0 = ( f (w)x− f (x)w,w) = f (w)(x,w)− f (x)(w,w)

It follows that for all x, f (x) =(

x, f (w)w|w|2

). ■

This leads to the following important definition.