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)
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
(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
M^{perp} 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 M^{perp}
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
comes from an inner product.

**Proof** of the Riesz lemma:

Consider the null space
N = N(), which is a closed subspace.
If N = H, then
is just the zero function, and g = 0.
This is the trivial case. Otherwise, There must be a nonzero vector in
N^{perp}. In fact, N^{perp} is one-dimensional, since if there were two linearly independent vectors
z_{1,2} in N^{perp}, then we can choose numbers
a and b, different from 0, such that

(a z_{1}-b z_{2}) =
a(z_{1})
- b(z_{2}) = 0.
But then a z_{1}-b z_{2} belongs to N and
N^{perp}. This can only happen if
z_{1}-b z_{2}=0, which is a contradiction. Conclusion: The entire space N^{perp} consists of the
multiples of a particular non-zero vector z_{1}.
Now scale z_{1} so that (z_{1}) is a real number and then let

g := (z_{1}) z_{1}/||z_{1}||.
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)
N. Notice also that
<x, g> = a <x, g> + 0,
because g belongs to N^{perp} and (x-a g) to N.
We calculate:
(x) = a (g) + 0
= < a g + (x-a g), g> = < x, g>

QED

Link to
Chapter XV
Chapter XVI
Table of Contents
Evans Harrell's home page
Jim Herod's home page