Worked Representations on Noncompact Spaces

Choosing Test Functions on the Real Line

On a noncompact space such as \(\mathbb{R}\), the choice of test-function space determines which measures a representation theorem can describe. The space \(C_0(\mathbb{R})\) consists of continuous real-valued functions that vanish at infinity; with the supremum norm, it is suited to bounded functionals and finite measures. The space \(C_c(\mathbb{R})\) consists of continuous real-valued functions with compact support. It can test measures that are finite on compact sets even when their total mass is infinite.

We use the Riesz Representation Theorems for \(C_0(\mathbb{R})\) and \(C_c(\mathbb{R})\), established earlier in the course. In the first case, a positive bounded linear functional on \(C_0(\mathbb{R})\) is integration against a unique finite regular Borel measure. In the second case, a positive linear functional on \(C_c(\mathbb{R})\) is integration against a unique locally finite regular Borel measure. Here “locally finite” means finite on every compact set. The measures in these statements are Borel measures; they are not required to be defined on the Lebesgue sigma-algebra.

Definition: A real linear functional \(L\) on either test-function space is positive if \(L(f)\ge 0\) whenever \(f\ge 0\). The supremum norm is \(\|f\|_\infty=\sup_{x\in\mathbb{R}}|f(x)|\). A functional on \(C_0(\mathbb{R})\) is bounded if some finite constant \(C\) satisfies \(|L(f)|\le C\|f\|_\infty\) for every \(f\in C_0(\mathbb{R})\).

For a finite measure, integration is bounded because \(\left|\int f\,d\mu\right|\le\|f\|_\infty\mu(\mathbb{R})\). For a locally finite measure and \(f\in C_c(\mathbb{R})\), the integral is finite: the function vanishes outside a compact set, and the measure of that set is finite. This does not imply boundedness in the supremum norm on all of \(C_c(\mathbb{R})\). The examples show how these distinctions guide representation problems.

Finite Measures: Atoms and Densities

Worked Example: Recovering a Finite Atomic Measure

Suppose \(L:C_0(\mathbb{R})\to\mathbb{R}\) is given by \(L(f)=2f(-1)+3f(2)\). Find its representing measure and verify the representation.

For \(a\in\mathbb{R}\), the Dirac measure \(\delta_a\) assigns mass \(1\) to a Borel set containing \(a\), and mass \(0\) otherwise; its integral satisfies \(\int f\,d\delta_a=f(a)\). Thus the candidate measure is \(\mu=2\delta_{-1}+3\delta_2\). It is finite, with total mass \(2+3=5\), and for every \(f\in C_0(\mathbb{R})\),

\int f\,d\mu=2\int f\,d\delta_{-1}+3\int f\,d\delta_2 =2f(-1)+3f(2)=L(f).

To see how test functions isolate the two weights, define \(\phi_a(x)=\max(1-|x-a|,0)\). This is continuous with compact support \([a-1,a+1]\), so it belongs to \(C_0(\mathbb{R})\). We have \(\phi_{-1}(-1)=1\) and \(\phi_{-1}(2)=0\), giving \(L(\phi_{-1})=2(1)+3(0)=2\). Also \(\phi_2(-1)=0\) and \(\phi_2(2)=1\), giving \(L(\phi_2)=2(0)+3(1)=3\). These evaluations recover the two masses. The representation theorem gives uniqueness, and the bound \(|L(f)|\le 2|f(-1)|+3|f(2)|\le5\|f\|_\infty\) verifies boundedness directly.

Check: applying the proposed integral to an arbitrary \(f\) gives exactly \(2f(-1)+3f(2)\), the original functional. The computed total mass \(5\) also agrees with the bound obtained by evaluating the two point masses.

Worked Example: A Finite Measure Given by a Density

On \(C_0(\mathbb{R})\), consider \(L(f)=\int_{\mathbb{R}} f(x)e^{-|x|}\,dx\), where \(dx\) denotes Lebesgue integration. Identify the measure and compute its total mass.

The measure is \(d\mu(x)=e^{-|x|}\,dm(x)\), meaning \(\mu(E)=\int_E e^{-|x|}\,dm(x)\) for Borel \(E\). The density is nonnegative and integrable. Splitting the integral at zero gives

\mu(\mathbb{R}) =\int_{-\infty}^{0}e^x\,dx+\int_{0}^{\infty}e^{-x}\,dx =1+1=2.

Consequently \(|L(f)|\le 2\|f\|_\infty\), so \(L\) is bounded on \(C_0(\mathbb{R})\). For a direct test of the formula, \(f(x)=e^{-|x|}\) is continuous and vanishes at infinity. Substituting it gives

L(f)=\int_{\mathbb{R}} e^{-2|x|}\,dx =\int_{-\infty}^{0}e^{2x}\,dx+\int_{0}^{\infty}e^{-2x}\,dx =\frac12+\frac12=1.

This check uses the density formula itself and confirms that the test function is integrated against \(e^{-|x|}\), not against Lebesgue measure alone. The total mass remains \(2\), as computed by integrating the density over the whole line.

When Is Integration Bounded?

The distinction between finite and merely locally finite measures can be expressed as an exact criterion. For the statement below, \(\mu\) is a regular Borel measure on \(\mathbb{R}\), and integration of a real-valued test function is understood in the usual signed sense; it is finite for compactly supported functions because \(\mu\) is finite on compact sets.

Proposition: Let \(\mu\) be a locally finite regular Borel measure on \(\mathbb{R}\). The functional \(L(f)=\int f\,d\mu\) on \(C_c(\mathbb{R})\) is bounded in the supremum norm if and only if \(\mu(\mathbb{R})<\infty\). In that case its norm is \(\mu(\mathbb{R})\). If \(\mu\) is finite, the same integration formula defines a bounded functional on \(C_0(\mathbb{R})\) with norm \(\mu(\mathbb{R})\).

Proof. First suppose \(\mu(\mathbb{R})<\infty\). For \(f\in C_c(\mathbb{R})\), and also for \(f\in C_0(\mathbb{R})\),

\left|\int f\,d\mu\right| \le \int |f|\,d\mu \le \|f\|_\infty\mu(\mathbb{R}).

Thus the norm is at most \(\mu(\mathbb{R})\). If \(\mu(\mathbb{R})=0\), this bound says the functional is zero, so its norm equals the total mass. If \(\mu(\mathbb{R})>0\), fix \(\varepsilon>0\). By inner regularity, there is a compact set \(K\subset\mathbb{R}\) with \(\mu(K)>\mu(\mathbb{R})-\varepsilon\). Choose a continuous compactly supported function \(\psi\) such that \(0\le\psi\le1\) and \(\psi=1\) on \(K\); such a function can be made by taking a piecewise linear cutoff equal to \(1\) on a bounded interval containing \(K\). Then \(\|\psi\|_\infty\le1\) and \(\int\psi\,d\mu\ge\mu(K)>\mu(\mathbb{R})-\varepsilon\). Since this holds for every \(\varepsilon>0\), the norm is at least \(\mu(\mathbb{R})\). It therefore equals \(\mu(\mathbb{R})\).

Conversely, suppose \(\mu(\mathbb{R})=\infty\). The sets \([-n,n]\) increase to \(\mathbb{R}\), so continuity from below for measures gives \(\mu([-n,n])\uparrow\infty\). For every \(n\), choose a continuous compactly supported cutoff \(\psi_n\) with \(0\le\psi_n\le1\) and \(\psi_n=1\) on \([-n,n]\). Then \(\|\psi_n\|_\infty\le1\), while \(L(\psi_n)\ge\mu([-n,n])\). These values are unbounded as \(n\) increases, so no finite constant can bound \(|L(f)|\) by a constant times \(\|f\|_\infty\) on \(C_c(\mathbb{R})\). This proves both directions. \(\square\)

Locally Finite Measures of Infinite Mass

Worked Example: Lebesgue Measure Acts on \(C_c(\mathbb{R})\)

Let \(\mu=m\), Lebesgue measure on the Borel subsets of \(\mathbb{R}\), and define \(L(f)=\int f\,dm\) for \(f\in C_c(\mathbb{R})\). This is a positive linear functional: if \(f\ge0\), then \(\int f\,dm\ge0\); and each integral is finite because \(f\) is bounded and supported on a compact set of finite Lebesgue measure. The \(C_c\) representation is Lebesgue measure itself, which is regular and locally finite, but has \(\mu(\mathbb{R})=\infty\).

To check that this functional is not bounded in the supremum norm, for \(R>0\) define

f_R(x)= \begin{cases} 1, & |x|\le R,\\ R+1-|x|, & R<|x|<R+1,\\ 0, & |x|\ge R+1. \end{cases}

This function is continuous, compactly supported, nonnegative, and has \(\|f_R\|_\infty=1\). Its integral is the area of a rectangle of width \(2R\) and height \(1\), plus two triangular tails, each of base \(1\) and height \(1\):

L(f_R)=2R+\frac12+\frac12=2R+1.

For \(R=3\), the support is \([-4,4]\), the plateau contributes \(6\), and the two tails contribute \(1/2\) each, so \(L(f_3)=6+1/2+1/2=7\), despite \(\|f_3\|_\infty=1\). As \(R\) increases, \(2R+1\) is unbounded. This agrees with the proposition: Lebesgue measure is locally finite but has infinite total mass, so its integration functional on \(C_c(\mathbb{R})\) is not bounded in the supremum norm.

Worked Example: A Locally Finite Atomic Measure of Infinite Mass

Consider the Borel measure \(\mu=\sum_{k\in\mathbb{Z}}\delta_k\), which places mass \(1\) at every integer. It has infinite total mass because there are infinitely many distinct unit point masses. It is locally finite: a compact subset of \(\mathbb{R}\) is bounded and therefore contains only finitely many integers, so its measure is finite.

For \(f\in C_c(\mathbb{R})\), the functional is \(L(f)=\sum_{k\in\mathbb{Z}}f(k)\). Only finitely many terms can be nonzero, since the support of \(f\) is compact. For the triangular test function \(\phi_0(x)=\max(1-|x|,0)\), the only integers in its support are \(-1,0,1\); its values there are \(0,1,0\), respectively. Hence \(L(\phi_0)=0+1+0=1\), matching \(\int\phi_0\,d\mu\).

This example is atomic rather than density-based, yet it has the same essential domain distinction as Lebesgue measure: the integral is finite for every compactly supported continuous function, but the measure is not finite on the whole line. Thus its positive \(C_c\) functional is represented by a locally finite measure, not by a finite measure arising from a bounded \(C_0\) functional.

A Common Misstep

A frequent error is to infer that a functional is bounded because its value is finite on every test function. For Lebesgue measure, each individual \(f\in C_c(\mathbb{R})\) has a finite integral. But the functions \(f_R\) above all have supremum norm \(1\), while their integrals equal \(2R+1\). Finiteness separately for each input is not a uniform bound over the unit ball.

There is a related domain error: applying the finite-measure \(C_0\) representation to Lebesgue integration on the whole real line. The function \(g(x)=1/\sqrt{1+x^2}\) is continuous and vanishes at infinity, but its Lebesgue integral is infinite. Indeed, for \(x\ge1\), \(1+x^2\le2x^2\), so \(g(x)\ge1/(\sqrt{2}x)\), whose integral on \([1,\infty)\) diverges. Thus integration against Lebesgue measure does not give a finite-valued functional on all of \(C_0(\mathbb{R})\). The \(C_c\) theorem is the appropriate representation result.

Takeaway: For positive representations on \(\mathbb{R}\), first identify the domain and boundedness. A bounded functional on \(C_0(\mathbb{R})\) corresponds to a finite regular Borel measure; a positive functional on \(C_c(\mathbb{R})\) can correspond to a locally finite regular Borel measure with infinite total mass.

Check Your Understanding

For each problem, identify the relevant test-function space and justify the distinction between finite and locally finite measure.

What measure represents \(L(f)=4f(0)+f(3)\) on \(C_0(\mathbb{R})\), and what is its total mass?
For \(d\mu(x)=e^{-2|x|}\,dm(x)\), compute \(\mu(\mathbb{R})\) by splitting the integral at zero. Is integration bounded on \(C_0(\mathbb{R})\)?
Explain why Lebesgue integration defines a finite value for each \(f\in C_c(\mathbb{R})\) but is not bounded in the supremum norm on that space.
Show that \(\sum_{k\in\mathbb{Z}}\delta_k\) is finite on every compact set but has infinite total mass.
State the condition on a locally finite regular Borel measure \(\mu\) that makes \(f\mapsto\int f\,d\mu\) bounded in the supremum norm on \(C_c(\mathbb{R})\).