Positive Linear Functionals on C_c(X)

Positive Functionals and Pointwise Order

Many familiar ways of assigning a number to a function preserve inequalities. If $f\le g$, then evaluation at a point satisfies $f(x_0)\le g(x_0)$; integration against a nonnegative density does as well. These examples motivate positivity for linear functionals. The key consequence is that a pointwise bound on functions gives a bound on their functional values. On a locally compact space, a common compactly supported function can control all functions whose supports lie in a specified compact set.

Let $X$ be a locally compact Hausdorff space. Write $C_c(X)$ for the real vector space of continuous real-valued functions $f:X\to\mathbb{R}$ whose support is compact, where $\operatorname{supp} f$ is the closure of $\{x\in X:f(x)\ne0\}$. Inequalities between functions in $C_c(X)$ mean pointwise inequalities at every $x\in X$. No measure is assumed on $X$ in the definitions below.

Definition: A linear functional on $C_c(X)$ is a map $L:C_c(X)\to\mathbb{R}$ such that $L(af+bg)=aL(f)+bL(g)$ for all $f,g\in C_c(X)$ and $a,b\in\mathbb{R}$. It is positive if $L(f)\ge0$ whenever $f(x)\ge0$ for every $x\in X$. Positivity permits $L(f)=0$ even when $f$ is nonzero and nonnegative.

For a compact set $K\subseteq X$, let $C_K(X)=\{f\in C_c(X):\operatorname{supp}f\subseteq K\}$. A compactly supported cutoff lemma for locally compact Hausdorff spaces states that for every compact $K$ there is an $\eta\in C_c(X)$ with $0\le\eta\le1$ everywhere and $\eta=1$ on $K$. This is the standard Urysohn cutoff result; only its existence is needed here. In particular, $\eta\ge1$ on $K$, as needed for the bound below.

Order Preservation and the Compact-Support Bound

Proposition: Let $X$ be a locally compact Hausdorff space and $L:C_c(X)\to\mathbb{R}$ a positive linear functional. If $f,g\in C_c(X)$ satisfy $f\le g$ everywhere, then $L(f)\le L(g)$. Also, for every $f\in C_c(X)$, $|L(f)|\le L(|f|)$, where $|f|(x)=|f(x)|$.

Proof. Since $g-f$ is continuous with compact support and $g-f\ge0$ everywhere, positivity gives $L(g-f)\ge0$. Linearity yields $L(g-f)=L(g)-L(f)$, so $L(f)\le L(g)$. Pointwise, $-|f|\le f\le |f|$. Applying the order inequality twice gives $-L(|f|)\le L(f)\le L(|f|)$. Positivity also gives $L(|f|)\ge0$, so these inequalities imply $|L(f)|\le L(|f|)$. $\square$

Theorem: Let $X$ be a locally compact Hausdorff space, $L:C_c(X)\to\mathbb{R}$ positive and linear, and $K\subseteq X$ compact. Choose $\eta\in C_c(X)$ with $\eta\ge1$ on $K$ and $\eta\ge0$ everywhere. For $f\in C_K(X)$, define $\|f\|_\infty=\sup_{x\in X}|f(x)|$. Then $$ |L(f)|\le \|f\|_\infty L(\eta). $$

Proof. If $f$ is identically zero, then linearity gives $L(f)=L(0)=0$, and the claimed inequality holds. Otherwise, $f$ is bounded because it is continuous and has compact support. Set $c=\|f\|_\infty$. On $K$, $|f(x)|\le c\le c\eta(x)$. Outside $K$, $f(x)=0$, so $|f(x)|\le c\eta(x)$ there as well, since $\eta\ge0$. Thus $|f|\le c\eta$ everywhere. The proposition and order preservation give $|L(f)|\le L(|f|)\le cL(\eta)$. As $c=\|f\|_\infty$, this is the desired bound. $\square$

The constant $L(\eta)$ depends on the chosen controlling function, but once $\eta$ is fixed, the estimate holds for every $f$ supported in $K$. In particular, if $f,g\in C_K(X)$, then $f-g\in C_K(X)$, and the same argument gives $|L(f)-L(g)|\le L(\eta)\|f-g\|_\infty$. Thus the restriction of $L$ to functions supported in $K$ is continuous in the supremum norm.

Worked Examples

Worked Example: Evaluation at a Point

Take $X=\mathbb{R}$ with its usual topology and define $L(f)=f(0)$ for $f\in C_c(\mathbb{R})$. Linearity follows from $(af+bg)(0)=af(0)+bg(0)$. If $f\ge0$ everywhere, then $L(f)=f(0)\ge0$, so $L$ is positive.

For $K=[-1,1]$, use the continuous compactly supported function $\eta(x)=1$ when $|x|\le1$, $\eta(x)=2-|x|$ when $1<|x|<2$, and $\eta(x)=0$ when $|x|\ge2$. It satisfies $\eta\ge1$ on $K$, and $L(\eta)=\eta(0)=1$. Therefore $|f(0)|\le\|f\|_\infty$ for every $f$ supported in $K$.

For a direct check, let $f(x)=\max(1-|x|,0)$. Its support is $[-1,1]$, $\|f\|_\infty=1$, and $L(f)=f(0)=1$. The estimate reads $1\le1$, with equality. Also, $L(-f)=-1$; positivity does not say that $L$ is nonnegative on functions that take negative values.

Worked Example: Integration Against Lebesgue Measure

On the measure space $(\mathbb{R},\mathcal{L},m)$, where $m$ is complete Lebesgue measure and is sigma-finite but not finite, define $L(f)=\int_{\mathbb{R}} f\,dm$ for $f\in C_c(\mathbb{R})$. This integral is finite: a continuous function with compact support is bounded and vanishes outside a compact set, which has finite Lebesgue measure. Linearity of the Lebesgue integral gives linearity of $L$, and $f\ge0$ everywhere implies $L(f)\ge0$, so $L$ is positive.

For $K=[-1,1]$, take the $\eta$ from the preceding example. Its integral is the area of a rectangle of width $2$ and height $1$, plus two triangular areas, each with base $1$ and height $1$: $$ L(\eta)=\int_{\mathbb{R}}\eta\,dm=2+\frac12+\frac12=3. $$ Hence $\left|\int f\,dm\right|\le3\|f\|_\infty$ whenever $\operatorname{supp}f\subseteq[-1,1]$.

For $f(x)=\max(1-|x|,0)$, direct integration gives $$ L(f)=2\int_0^1(1-x)\,dx =2\left[x-\frac{x^2}{2}\right]_0^1 =2\left(1-\frac12\right)=1. $$ Here $\|f\|_\infty=1$, so the bound is $1\le3$. As a check, the graph of $f$ is a triangle with base $2$ and height $1$, whose area is $\tfrac12\cdot2\cdot1=1$, agreeing with the integral.

Worked Example: Weighted Evaluations on a Discrete Space

Let $X=\mathbb{Z}$ with the discrete topology. This is locally compact and Hausdorff, and its compact subsets are finite. Every function on $X$ is continuous, and $C_c(X)$ consists exactly of the functions with finite support. Define $L(f)=2f(0)+3f(2)$. This is linear. If $f\ge0$ at every integer, then $2f(0)+3f(2)\ge0$, so $L$ is positive.

Take $K=\{0,2\}$ and let $\eta$ be $1$ at $0$ and $2$, and $0$ at every other integer. This belongs to $C_c(X)$, is nonnegative, and equals $1$ on $K$. Thus $L(\eta)=2+3=5$. Now choose $f$ with $f(0)=1$, $f(2)=-2$, and $f(k)=0$ for all other $k$. Then $\operatorname{supp}f=K$, $\|f\|_\infty=2$, and $$ L(f)=2(1)+3(-2)=2-6=-4,\qquad |L(f)|=4\le 2\cdot5=10. $$ The pointwise absolute value is $|f(0)|=1$, $|f(2)|=2$, and zero elsewhere, so $L(|f|)=2(1)+3(2)=8$. The proposition's intermediate inequality checks as $4\le8$.

Why the Fixed Compact Set Matters

The compact-support estimate is uniform over all functions supported in one fixed $K$, but it does not assert a single supremum-norm bound for $L$ on all of $C_c(X)$. For Lebesgue integration on $\mathbb{R}$, consider, for each integer $n\ge2$, the continuous function $$ f_n(x)= \begin{cases} x, & 0\le x\le1,\\ 1, & 1\le x\le n-1,\\ n-x, & n-1\le x\le n,\\ 0, & x<0\text{ or }x>n. \end{cases} $$ At the shared endpoints the formulas agree, including when $n=2$; hence $f_n$ is continuous. Its support is $[0,n]$ and $\|f_n\|_\infty=1$. The integral is the area of two triangles of area $1/2$ and a rectangle of area $n-2$: $$ \int_{\mathbb{R}} f_n\,dm=\frac12+(n-2)+\frac12=n-1. $$ For example, at $n=2$ this is $1$, and at $n=5$ it is $4$. Since $n-1$ grows without bound while every $\|f_n\|_\infty=1$, no finite constant $C$ can satisfy $|L(f)|\le C\|f\|_\infty$ for all $f\in C_c(\mathbb{R})$. The supports $[0,n]$ do not lie in one fixed compact set. There is therefore no conflict with the compact-support theorem.

A common mistake is to infer from positivity that a functional is bounded by the supremum norm with one global constant. For the integration functional, that would require $n-1\le C$ for every $n\ge2$, by the functions just computed, which is impossible for finite $C$. The correct conclusion is local in the support: fix $K$, choose a nonnegative compactly supported $\eta$ that is at least $1$ on $K$, and use $L(\eta)$ as the bound. Positivity matters because it lets us apply $L$ to the pointwise inequalities; linearity then turns those inequalities into numerical ones.

Point evaluation and integration illustrate two different ways a positive functional can encode information about test functions. This order-based estimate is useful in the study of such functionals, but by itself it does not identify every positive functional with a measure. The bound established here is the control needed on each fixed compact region; further representation results require additional arguments.

Check Your Understanding

For each question, distinguish the pointwise order assumption from the compact-support assumption.

Show that the zero functional on $C_c(X)$ is positive and linear, even though it is not strictly positive on every nonzero nonnegative function.
If $L$ is positive and $f,g\in C_c(X)$ satisfy $f\le g$, derive $L(f)\le L(g)$ directly from positivity and linearity.
Suppose $K$ and $\eta$ satisfy the hypotheses of the compact-support theorem. Explain why $|f|\le\|f\|_\infty\eta$ holds both on $K$ and outside $K$ for $f\in C_K(X)$.
For the discrete-space example, compute $L(|f|)$ for the specified $f$, and compare it with $|L(f)|$.
For the Lebesgue integration example, why do the functions $f_n$ not contradict a bound proved for functions supported in one fixed compact set?

Positive Linear Functionals on C_c(X)

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