Functional Norm and Total Mass

From a Functional Norm to Total Mass

A positive linear functional on continuous functions records the mass of a measure through integration. Its operator norm measures the largest value the functional can produce on functions bounded by one. For a positive functional on $C_0(X)$, these two quantities agree: the norm is the total mass of its representing measure. On a noncompact space, the proof cannot simply test the functional on the constant function $1$, because that function may not vanish at infinity. Instead, compactly supported functions approximate the total mass from below.

Throughout, $X$ is a locally compact Hausdorff space, and $\mathcal{B}_X$ is its Borel sigma-algebra. The space $C_0(X)$ consists of real-valued continuous functions that vanish at infinity: for each $\varepsilon>0$, the set $\{x:|f(x)|\ge\varepsilon\}$ is compact. Its norm is $\|f\|_\infty=\sup_{x\in X}|f(x)|$. A linear functional $\Lambda:C_0(X)\to\mathbb{R}$ is positive if $\Lambda(f)\ge0$ whenever $f(x)\ge0$ for every $x$. Its operator norm is $\|\Lambda\|=\sup_{\|f\|_\infty\le1}|\Lambda(f)|$.

We use the Riesz representation theorem for $C_0(X)$: a positive bounded linear functional $\Lambda$ has a unique finite Radon measure $\mu$ on $(X,\mathcal{B}_X)$ such that $\Lambda(f)=\int_X f\,d\mu$ for every $f\in C_0(X)$. Here Radon means, in particular, that $\mu$ is inner regular on Borel sets; for $X$, this gives $\mu(X)=\sup\{\mu(K):K\subseteq X\text{ compact}\}$. The theorem gives the representing measure; the result below identifies its total mass directly from the operator norm.

The Norm–Mass Identity

Theorem: Let $X$ be locally compact Hausdorff, let $\Lambda:C_0(X)\to\mathbb{R}$ be a positive bounded linear functional, and let $\mu$ be its finite Radon representing measure on $(X,\mathcal{B}_X)$. Then $$ \|\Lambda\|=\mu(X). $$

For the reverse inequality, let $K\subseteq X$ be compact. The standard Urysohn cutoff fact for locally compact Hausdorff spaces says there is a $\varphi\in C_c(X)$ with $0\le\varphi\le1$ everywhere and $\varphi=1$ on $K$; here $C_c(X)$, the continuous compactly supported functions, is a subset of $C_0(X)$. Thus $\|\varphi\|_\infty\le1$, and $$ \|\Lambda\|\ge|\Lambda(\varphi)|=\int_X\varphi\,d\mu\ge\int_K\varphi\,d\mu=\mu(K). $$ Positivity of $\varphi$ makes $\Lambda(\varphi)\ge0$, and $\varphi=1$ on $K$ gives the final equality. Taking the supremum over compact $K$ and using inner regularity yields $\|\Lambda\|\ge\mu(X)$. This also covers $\mu(X)=0$, since the first inequality gives $\|\Lambda\|\le0$ and a norm is nonnegative. If $X$ is empty, both $C_0(X)$ and its only measure have value zero, so the identity holds there as well. Combining the two bounds proves the claim. $\square$

The proof uses two distinct ingredients: the integral bound controls every test function from above, while inner regularity and cutoffs supply test functions whose integrals come arbitrarily close to the total mass. No maximum need be attained in the noncompact case.

Worked Computations

Worked Example: Two Atoms on the Real Line

On $X=\mathbb{R}$ with its Borel sigma-algebra, let $\mu=2\delta_{-1}+3\delta_2$, where $\delta_a$ assigns mass one to the point $a$. Define $\Lambda(f)=\int f\,d\mu=2f(-1)+3f(2)$ for $f\in C_0(\mathbb{R})$. This measure is finite and Radon, and $\Lambda$ is positive because $f\ge0$ implies both terms are nonnegative.

The mass is $\mu(\mathbb{R})=2+3=5$, so the theorem gives $\|\Lambda\|=5$. To check the lower bound with a single function, set $$ f(x)=\max(1-|x+1|,0)+\max(1-|x-2|,0). $$ The first summand is supported on $[-2,0]$, and the second on $[1,3]$; these intervals are disjoint. Each summand lies between zero and one, so $f\in C_c(\mathbb{R})$ and $\|f\|_\infty=1$. Substitution gives $f(-1)=1$ and $f(2)=1$, hence $\Lambda(f)=2(1)+3(1)=5$. The upper bound also checks directly: for any $\|g\|_\infty\le1$, $|2g(-1)+3g(2)|\le2|g(-1)|+3|g(2)|\le2+3=5$. Thus the computed norm is exactly five.

Worked Example: An Exponential Density on the Real Line

Let $\mu$ be the measure on $(\mathbb{R},\mathcal{B}_{\mathbb{R}})$ with density $e^{-|x|}$ relative to Lebesgue measure $m$, and let $\Lambda(f)=\int_{\mathbb{R}}f(x)e^{-|x|}\,dx$. The density is nonnegative, so $\Lambda$ is positive. Its total mass is $$ \mu(\mathbb{R})=\int_{-\infty}^{\infty}e^{-|x|}\,dx =2\int_0^\infty e^{-x}\,dx =2[-e^{-x}]_0^\infty=2. $$ The norm–mass theorem therefore gives $\|\Lambda\|=2$.

Here is an explicit check that compactly supported tests approach this value. For $R>0$, define $f_R(x)=1$ when $|x|\le R$, $f_R(x)=R+1-|x|$ when $R<|x|<R+1$, and $f_R(x)=0$ when $|x|\ge R+1$. This continuous compactly supported function satisfies $0\le f_R\le1$, with $\|f_R\|_\infty=1$. By symmetry and the substitution $t=x-R$ in the tail, $$ \Lambda(f_R)=2\left(\int_0^R e^{-x}\,dx+\int_R^{R+1}(R+1-x)e^{-x}\,dx\right) =2\left(1-e^{-R}+e^{-R}\int_0^1(1-t)e^{-t}\,dt\right). $$ Compute the remaining integral explicitly: $$ \int_0^1(1-t)e^{-t}\,dt=(1-e^{-1})-(1-2e^{-1})=e^{-1}. $$ Consequently $\Lambda(f_R)=2(1-e^{-R}+e^{-(R+1)})$, which tends to $2$ as $R\to\infty$. For example, at $R=1$ the value is $2(1-e^{-1}+e^{-2})$. The upper bound $|\Lambda(g)|\le\int|g|\,d\mu\le2$ holds for every $\|g\|_\infty\le1$; the displayed sequence of tests approaches that bound. This independently verifies the norm.

Worked Example: A Density on a Compact Interval

Take $X=[0,1]$ with its usual topology and Borel sigma-algebra, and define $\mu(E)=\int_E(1+x)\,dm(x)$ for Borel $E\subseteq[0,1]$. The measure is finite and Radon. For $f\in C([0,1])=C_0(X)$, set $\Lambda(f)=\int_0^1f(x)(1+x)\,dx$. The constant function $1$ belongs to $C_0(X)$ because $X$ is compact.

Its total mass is $$ \mu([0,1])=\int_0^1(1+x)\,dx=[x+x^2/2]_0^1=1+\frac12=\frac32. $$ The theorem gives $\|\Lambda\|=3/2$. Indeed, $\|1\|_\infty=1$ and $\Lambda(1)=3/2$, so the norm is at least $3/2$; for every $f$ with $\|f\|_\infty\le1$, $|\Lambda(f)|\le\int_0^1(1+x)\,dx=3/2$, giving the matching upper bound. As a separate substitution check, for $f(x)=x$ one obtains $\Lambda(x)=\int_0^1(x+x^2)\,dx=1/2+1/3=5/6$, which is below the norm $3/2$.

A Noncompact-Space Pitfall

On a noncompact space, it is incorrect to treat $1+f$ and $1-f$ as elements of $C_0(X)$ merely because they are nonnegative. For a concrete check, take $X=\mathbb{R}$ and $f=0$. Then $1+f=1$, and the set $\{x:|1|\ge1/2\}=\mathbb{R}$ is not compact. Thus $1\notin C_0(\mathbb{R})$, and a functional whose domain is $C_0(\mathbb{R})$ cannot be applied to $1$.

The valid positivity argument uses functions that remain in the domain. If $f\in C_0(X)$ is real-valued, then $|f|\in C_0(X)$, and both $|f|+f$ and $|f|-f$ are nonnegative members of $C_0(X)$. Positivity therefore gives $\Lambda(|f|)+\Lambda(f)\ge0$ and $\Lambda(|f|)-\Lambda(f)\ge0$, hence $|\Lambda(f)|\le\Lambda(|f|)$. This argument is legitimate but, by itself, does not identify $\Lambda(|f|)$ with a bound involving total mass. The representing-measure proof supplies that bound by integration. On a compact space the constant function does belong to $C_0(X)$, as the interval example illustrates; the noncompact case requires the cutoff and inner-regularity argument instead.

The identity $\|\Lambda\|=\mu(X)$ turns a functional norm into a concrete measure quantity. It also provides a quick consistency check for a proposed representing measure: the norm must be its total mass, not merely a bound on that mass. In noncompact settings, compactly supported functions recover the mass through approximation even when no constant test function is available.

Check Your Understanding

For each question, distinguish the norm bound from the approximation that gives the matching lower bound.

For a positive functional represented by a finite Radon measure $\mu$, which inequality gives $\|\Lambda\|\le\mu(X)$? State the hypotheses needed for that estimate.
Why does inner regularity on $X$, together with compactly supported cutoffs, give the reverse inequality?
On $\mathbb{R}$, let $\mu=4\delta_0$. Find $\mu(\mathbb{R})$ and the norm of $f\mapsto\int f\,d\mu$, and describe a function in $C_c(\mathbb{R})$ that attains the norm.
For the density $e^{-|x|}$ on $\mathbb{R}$, compute the total mass and explain why testing only compactly supported functions can still recover it.
Give a concrete reason that the constant function $1$ cannot generally be used as a test function for a functional on $C_0(\mathbb{R})$.

What You'll Learn