Riesz-Markov theorem: any positive linear functional is continuous

borel-measuresfunctional-analysisgeneral-topologymeasure-theoryriesz-representation-theorem

Below is the Riesz–Markov theorem that I take from here.

Riesz–Markov theorem: Let $X$ be a locally compact Hausdorff space and $C_0(X)$ the space of continuous compactly supported functionals on $X$. For any positive linear functional $\varphi$ on $C_0(X)$, there is a unique non-negative Radon measure $\mu$ on $X$ such that
$$
\varphi (f)=\int f \mathrm d \mu \quad \forall f \in C_0(X).
$$

Let

  • $E:= C_0 (X)$ be endowed with the supremum norm $\| \cdot \|_\infty$.
  • $E'$ be the dual space of $E$.
  • $\mathcal M$ be the space of all non-negative Radon measures on $X$.

We define a map
$$
T: \mathcal M \to \mathbb R^{E}, \mu \mapsto \left (f \mapsto \int f \mathrm d \mu \right).
$$

Clearly, $T(\mu)$ is linear and positive. Moreover,
$$
\sup_{f \in E} \frac{\int f \mathrm d \mu}{\|f\|_\infty} = \mu(X) <\infty \quad \forall \mu \in \mathcal M.
$$

So $T(\mu)$ is continuous and thus belongs to $E'$ for all $\mu$. Let $\varphi: E \to \mathbb R$ be a positive linear functional. By Riesz-Markov theorem, there is unique $\mu \in \mathcal M$ such that $\varphi = T(\mu) \in E'$. Hence $\varphi$ is continuous.

This in turn implies that any positive linear functional on $E$ is continuous.

Could you please confirm if my understanding is correct?

Best Answer

There are some deep relations between:

  1. order structures (e.g. pointwise order of functions),

  2. analytic structures (e.g. the Banach space structure of $C_0(X)$), and

  3. algebraic structures (e.g. pointwise multiplication of functions).

The question under discussion is an example of the relationship between (1) and (2), while (2) and (3) are linked e.g. via the Theorem asserting that any linear functional $\varphi $ on $C_0(X)$ that preserves multiplication, that is, $\varphi (fg)=\varphi (f)\varphi (g)$, is necessarily also continuous!

Here is a direct proof of the fact that every positive linear functional on $C_0(X)$ is continuous:

Arguing by contradiction, suppose that for every $n\in {\mathbb N}$, there is $f_n\in C_0(X)$, with $\|f_n\|\leq 1$ and $\varphi (f_n)\geq 4^n$.

Letting $f_n^+$ be the positive part of $f_n$ (that is, $f_n^+(x) = \max\{f_n(x),0\}$), we have that $$ f_n^+\geq f_n \Rightarrow \varphi (f_n^+)\geq \varphi (f_n)\geq 4^n, $$ and clearly $\|f_n^+\|\leq 1$. Setting $f = \sum_{n=0}^\infty 2^{-n}f_n^+$, we then have for all $N\in {\mathbb N}$ that $$ g_N := \sum_{n=0}^N2^{-n}f_n^+ \leq f, $$ so $\varphi (g_N) \leq \varphi (f)$. However $$ \varphi (g_N)= \sum_{n=0}^N2^{-n}\varphi (f_n^+) \geq \sum_{n=0}^N2^{-n}4^n = \sum_{n=0}^N2^n\to \infty , $$ a contradiction!

EDIT: Bob's comment and answer below are excellent points and indeed the present answer depends on interpreting $C_0(X)$ as the space of continuous functions vanishing at $\infty$ as opposed to compactly supported. The latter in not complete in the uniform norm so there is no reason for the series representing $f$ to converge. As pointed out, everything is OK with the former, I believe also standard, interpretation of $C_0(X)$.

In case $\varphi$ is instead defined on the space of continuous compactly supported functions, more commonly denoted $C_c(X)$, a similar result may be proved, namely that $\varphi$ is continuous wrt the so called inductive limit topology. This essentialy means that, for each compact subset $K$ of $X$, the restriction of $\varphi$ is continuous on the subspace $C_K(X)$, formed by the continuous functions on $X$, vanishing on the complement of $K$, and equipped with the sup norm.

In the link to the Riesz–Markov theorem provided by the OP there are actually two main results, one for $C_c(X)$ and another for $C_0(X)$. Only the second one involves a necessarily finite measure so the statement by the OP that "$\mu(X)<\infty$" is not compatible with the definition of $C_0(X)$ as the compactly supported functions. I therefore suggest the OP edit the question to resolve this issue.

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