Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In Wikipedia, there are three versions of Riesz theorems:

1 The Hilbert space representation theorem for the (continuous) dual space of a Hilbert space;

2 The representation theorem for positive linear functionals on $C_c(X)$, where $X$ is a locally compact Hausdorff space;

3 The representation theorem for the dual of $C_0(X)$, where $X$ is a locally compact Hausdorff space.

I was wondering

  1. if none of the three versions is more general than the others, in the sense that no one can be derived from another?
  2. when two or three of them can coincide?

Thanks and regards!

share|cite|improve this question
In general, a theorem which describes (some class of) linear functionals on a certain space is sometimes called a Riesz representation theorem. For instance, the fact that the dual of $L^p$ is $L^q$ (under the suitable assumptions) is mentioned as a/the Riesz representation theorem in some real analysis texts. – Mark Jul 7 '11 at 21:25
@Mark: I don't think this general version is correct as you state it. The reason why they are all called Riesz representation theorem is because Riesz identified these dual spaces in his foundational works on functional analysis. The $L^2$ and $L^p$-versions are often called Fréchet-Riesz because of two 1907 Comptes Rendus notes in issue 144 (one by F. and one by R.), and the measure version is often called Riesz-Markov (sometimes Kakutani is added) for similar reasons. – t.b. Jul 8 '11 at 0:58
I dunno, googling "a Riesz representation theorem" and "Riesz type theorem" yields all sorts of results of this form (e.g. and – Mark Jul 8 '11 at 3:23
up vote 15 down vote accepted

These are three different theorems and there's no relation between 1) and the others except in the case when $X$ is finite, where all three theorems coincide (since $\ell^2(X)$, $C_0(X)$ and $C_c(X)$ then are the same topological vector space). Note however that $C_0(X)$ and $C_c(X)$ are never Hilbert spaces (unless the locally compact space $X$ is empty or a point), so 2) and 3) can't have a direct relation to 1).

Since positivity implies continuity 2) can be interpreted as characterizing continuous linear functionals on $C_c{(X)}$ as well, and I'm addressing this version below.

The results 2) and 3) are closely related and often 3) is proved as a corollary of 2).

Note that the space $C_c(X)$ is dense in $C_{0}(X)$ with respect to the sup-norm. So a continuous linear functional on $C_c(X)$ (= a signed Radon measure) extends (uniquely) to a continuous linear functional $C_{0}(X)$ (= a signed bounded Radon measure) if and only if it is of bounded variation. Moreover, 2) and 3) coincide if $X$ is compact (and there are also proofs of 3) reducing it to that case).

share|cite|improve this answer
@Theo: Thanks! In your first sentence, by "when $X$ is finite", do you mean 1 and the others coincide for $L_2(X)$? – Tim May 3 '11 at 16:20
@Tim: If $X$ is a finite set with $n$ elements, then these are all the same (rather trivial) theorem on $\mathbb{R}^n$ or $\mathbb{C}^n$. – t.b. May 3 '11 at 16:22
@Theo: Thanks! On $L_2$ space as a Hilbert space with the usual inner product, will the three coincide? – Tim May 3 '11 at 16:24
@Tim: No, I'm not saying that. What I'm saying is: If $X$ has $n$ elements, then $\ell^2(X)$, $C_0(X)$ and $C_c(X)$ are all isomorphic to the same topological vector space, namely $\mathbb{R}^n$ or $\mathbb{C}^n$ (since there is only one Hausdorff topological vector space of dimension $n$ over $\mathbb{R}$ or $\mathbb{C}$. In general, $L^2(X)$ depends on the choice of a measure on $X$ and the Riesz theorem for $L^2$ is ultimately an abstract theorem on Hilbert spaces. While the two theorems on $C_0$ and $C_c$ are theorems on these two latter spaces (which are never Hilbert spaces). – t.b. May 3 '11 at 16:31
@Tim: Ah, maybe I misunderstood you. Yes, I was speaking of $L^2(X)$ when $X$ is finite (and equipped with counting measure, say). – t.b. May 3 '11 at 16:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.