The Riesz representation theorem

Linear Methods of Applied Mathematics

Evans M. Harrell II and James V. Herod*

*(c) Copyright 2000 by Evans M. Harrell II and James V. Herod. All rights reserved.

version of 15 March 2000

The Riesz representation theorem

Theorem XV.2. (The Riesz representation theorem for Hilbert space.) If {E, < , >} is a Hilbert space, then these are equivalent:

(a) Phi is a continuous, linear function from E to R (or the complex numbers C), and

(b) there is a unique member v of E such that Phi(x) = < x, v > ; for each x in E.

First, let us formalize the

Projection lemma.

Let H be a Hilbert space and M a closed subspace. Let Mperp consist of the vectors which are orthogonal to all the vectors of M. Given any vector x in H, there is a unique vector y in M and a unique z in Mperp such that x = y + z. The function associating y to x is a linear projection operator, as is the function associating z to x.

The Riesz lemma, stated in words, claims that every continuous linear functional Phi comes from an inner product.

Proof of the Riesz lemma:

Consider the null space N = N(Phi), which is a closed subspace. If N = H, then Phi is just the zero function, and g = 0. This is the trivial case. Otherwise, There must be a nonzero vector in Nperp. In fact, Nperp is one-dimensional, since if there were two linearly independent vectors z1,2 in Nperp, then we can choose numbers a and b, different from 0, such that

Phi(a z1-b z2) = aPhi(z1) - bPhi(z2) = 0. But then a z1-b z2 belongs to N and Nperp. This can only happen if z1-b z2=0, which is a contradiction. Conclusion: The entire space Nperp consists of the multiples of a particular non-zero vector z1.

Now scale z1 so that Phi(z1) is a real number and then let

g := Phi(z1) z1/||z1||. Of course, it remains to verify that this g fills the bill. Because of the projection lemma, we may write any x in H as x = a g + (x-a g), where (x-a g) belongs to N. Notice also that <x, g> = a <x, g> + 0, because g belongs to Nperp and (x-a g) to N.

We calculate: Phi(x) = a Phi(g) + 0 = < a g + (x-a g), g> = < x, g>

                                                                                                                        QED

Link to
  • Chapter XV
  • Chapter XVI
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