Worked Representations on Compact Intervals

Reading a Functional as a Measure

On a compact interval, a positive linear functional on continuous functions can be represented by integration against a finite measure. In explicit problems, the functional is often already written as a sum of point evaluations and a density integral. The practical task is to identify which part of the measure each term describes, keep atoms separate from the density, and evaluate the resulting representation without losing endpoint or set-membership details.

Let $a<b$ be finite real numbers. Write $C([a,b])$ for the real-valued continuous functions on $[a,b]$, with norm $\|f\|_\infty=\sup_{x\in[a,b]}|f(x)|$. A linear functional $L:C([a,b])\to\mathbb R$ is positive if $L(f)\ge0$ whenever $f(x)\ge0$ for every $x$. Let $x_1,\ldots,x_N$ be distinct points of $[a,b]$, let $c_j\ge0$, and let $w:[a,b]\to[0,\infty]$ be Lebesgue measurable with $\int_a^b w\,dm<\infty$. Here $m$ is Lebesgue measure. The expression

L(f)=\sum_{j=1}^{N}c_j f(x_j)+\int_a^b f(x)w(x)\,dm(x)

is defined for every $f\in C([a,b])$. The integral is finite because $|f|\le\|f\|_\infty$ and $w$ is integrable. Each point evaluation and the density integral are positive linear functionals, so $L$ is positive and bounded.

Definition: For $x\in[a,b]$, the Dirac measure $\delta_x$ on the Borel sets of $[a,b]$ is given by $\delta_x(E)=1$ if $x\in E$, and $\delta_x(E)=0$ if $x\notin E$. The measure associated with the displayed functional is $$ \nu=\sum_{j=1}^{N}c_j\delta_{x_j}+w\,m, $$ where $(w\,m)(E)=\int_Ew\,dm$ for Borel $E\subseteq[a,b]$. It is a finite Borel measure, and $$ \int_{[a,b]} f\,d\nu=\sum_{j=1}^{N}c_jf(x_j)+\int_a^b fw\,dm=L(f). $$

The measure space in this representation is $([a,b],\mathcal B_{[a,b]},\nu)$, with $\nu$ finite. The density is Lebesgue measurable, and its integral over a Borel set is well-defined. The density measure $w\,m$ is regular on the compact interval, as is each Dirac measure; hence their finite sum is a finite regular Borel measure. The equality above follows by linearity of integration. The Riesz representation theorem for compact spaces says that every positive linear functional on $C([a,b])$ has a unique finite regular Borel representing measure; we use that standard theorem and its uniqueness, established in the earlier material on the Riesz representation theorem.

Worked Evaluations

Worked Example: A Density and Two Atoms on [0,1]

Define $L:C([0,1])\to\mathbb R$ by $L(f)=3f(0)+2f(1/2)+\int_0^1(1+x)f(x)\,dm(x)$. Find its representing measure and compute $L(1)$, $L(x)$, and $L(x^2)$.

The coefficients are nonnegative, and $1+x\ge0$ is integrable on $[0,1]$. Thus the functional is positive, and its representing measure is $\nu=3\delta_0+2\delta_{1/2}+(1+x)m$, where the density term is restricted to $[0,1]$. For the constant test function, the point terms sum to $3+2=5$, while the density contributes $\int_0^1(1+x)\,dm=1+1/2=3/2$. Therefore $L(1)=5+3/2=13/2$.

For $f(x)=x$, the point terms are $3\cdot0+2\cdot(1/2)=1$. The density contribution is $\int_0^1(x+x^2)\,dm=1/2+1/3=5/6$. Hence $L(x)=1+5/6=11/6$.

For $f(x)=x^2$, the point terms are $3\cdot0^2+2\cdot(1/2)^2=1/2$. The density contribution is $\int_0^1(x^2+x^3)\,dm=1/3+1/4=7/12$. Thus $L(x^2)=1/2+7/12=13/12$. Check: integrating each test function against $3\delta_0+2\delta_{1/2}+(1+x)m$ gives, respectively, $13/2$, $11/6$, and $13/12$, the same values obtained directly from $L$.

Worked Example: Set Values and Endpoint Atoms on [-1,1]

Let $L(f)=2f(-1)+f(1)+\int_{-1}^1(1+x)f(x)\,dm(x)$. Find the representing measure, its values on $[-1,0]$ and $(-1,0)$, and $L(x)$.

The density $1+x$ is nonnegative on $[-1,1]$ and has integral $2$, so the representing measure is the finite positive Borel measure $\nu=2\delta_{-1}+\delta_1+(1+x)m$. On $[-1,0]$, the atom at $-1$ is included and the atom at $1$ is excluded. The density contribution is $\int_{-1}^0(1+x)\,dm=[x+x^2/2]_{-1}^0=0-(-1+1/2)=1/2$. Consequently $\nu([-1,0])=2+1/2=5/2$.

On $(-1,0)$, neither atom is included. Removing or adding the endpoints does not change the density integral, since a finite set has Lebesgue measure zero. Therefore $\nu((-1,0))=1/2$. In particular, $\nu(\{-1,0\})=2$: the density contributes zero on these two points, and only the atom at $-1$ contributes.

For $f(x)=x$, the point terms are $2(-1)+1(1)=-1$. The density term is $\int_{-1}^1(x+x^2)\,dm=0+2/3=2/3$, because $x$ is odd on the symmetric interval and $\int_{-1}^1x^2\,dm=2/3$. Hence $L(x)=-1+2/3=-1/3$. Check: the same point and density contributions in $\int x\,d\nu$ are $-1$ and $2/3$, so the represented integral is also $-1/3$.

Worked Example: A Trigonometric Test Function

On $[0,\pi]$, define $L(f)=4f(\pi/3)+\int_0^\pi(1+\sin x)f(x)\,dm(x)$. Find the representing measure and evaluate $L(1)$, $L(\sin x)$, and $L(\cos x)$.

Since $1+\sin x\ge0$ on $[0,\pi]$, the functional is positive. Its measure is $\nu=4\delta_{\pi/3}+(1+\sin x)m$. For $f=1$, the atom contributes $4$, and the density contributes $\int_0^\pi(1+\sin x)\,dm=\pi+2$. Thus $L(1)=\pi+6$.

For $f(x)=\sin x$, the point term is $4\sin(\pi/3)=4(\sqrt3/2)=2\sqrt3$. The density term is $\int_0^\pi(\sin x+\sin^2x)\,dm=2+\pi/2$. Therefore $L(\sin x)=2\sqrt3+2+\pi/2$. For $f(x)=\cos x$, the point term is $4\cos(\pi/3)=4(1/2)=2$. The density term is $\int_0^\pi(\cos x+\sin x\cos x)\,dm=0+0=0$, since the antiderivatives are $\sin x$ and $\sin^2x/2$, respectively, and both have equal endpoint values at $0$ and $\pi$. Thus $L(\cos x)=2$. Check: the atom contributes $2$ to $\int\cos x\,d\nu$, and the density contributes zero, giving the same result.

What the Representation Guarantees

Proposition: Under the assumptions above, the measure $\nu=\sum_{j=1}^N c_j\delta_{x_j}+w\,m$ is finite and represents $L$ on $C([a,b])$. Its total mass and the operator norm of $L$ satisfy $$ \nu([a,b])=L(1)=\sum_{j=1}^N c_j+\int_a^b w\,dm=\|L\|. $$

Proof. Each $c_j\delta_{x_j}$ is a finite Borel measure of mass $c_j$, and $w\,m$ has mass $\int_a^b w\,dm<\infty$. Their finite sum is therefore finite, and its total mass is the sum of those masses. The representation identity follows from linearity of integration, so applying it to the constant function $1$ gives $\nu([a,b])=L(1)$. For any $f\in C([a,b])$ with $\|f\|_\infty\le1$, positivity of $\nu$ gives $|L(f)|=|\int f\,d\nu|\le\int|f|\,d\nu\le\int1\,d\nu=\nu([a,b])$. Taking the supremum over such $f$ shows $\|L\|\le\nu([a,b])$. The function $f=1$ has $\|1\|_\infty=1$ and $L(1)=\nu([a,b])$, so the supremum is at least $\nu([a,b])$. The two inequalities prove equality. $\square$

Proposition: If $x_1,\ldots,x_N$ are distinct and $w$ is integrable, then the mass of the representing measure at each $x_j$ is exactly $c_j$. Consequently the point-mass coefficients are determined by the measure. If two such representations define the same measure, their densities agree $m$-almost everywhere after their point masses have been matched.

Proof. Fix $j$. By distinctness, $\delta_{x_i}(\{x_j\})=0$ for $i\ne j$, while $\delta_{x_j}(\{x_j\})=1$. Also $m(\{x_j\})=0$, so $\int_{\{x_j\}}w\,dm=0$. Hence $\nu(\{x_j\})=c_j$. If two representations give the same measure, evaluating both on each singleton gives equality of the coefficients at every listed point (a point not listed in one representation has mass zero there). After these atomic measures are subtracted, the density measures agree. Uniqueness of the Radon–Nikodym derivative, a standard result, then implies equality of the densities $m$-almost everywhere. $\square$

The interval assumption matters: on $[a,b]$, every continuous real-valued function is bounded, so integration against a finite measure is defined for every $f\in C([a,b])$. For the degenerate interval $[a,a]=\{a\}$, Lebesgue measure restricted to the interval is zero; any density integral vanishes, and all point evaluations at $a$ combine into a single mass at $a$.

A Common Mistake: Losing the Atoms

A density measure $w\,m$ assigns zero mass to every singleton, but a Dirac mass does not. For the measure in the first example, the incorrect calculation $\nu(\{1/2\})=\int_{\{1/2\}}(1+x)\,dm=0$ omits the point evaluation term. The correct calculation is $\nu(\{1/2\})=2+\int_{\{1/2\}}(1+x)\,dm=2+0=2$. The density integral alone describes only the absolutely continuous part of the measure; it does not encode point evaluations. Conversely, when computing $\nu(E)$, an atom contributes only if its location belongs to $E$. Thus $-1\in[-1,0]$, but $-1\notin(-1,0)$, exactly as the second example illustrates.

These representations are useful because they separate two different sources of mass. The density controls mass spread across sets according to Lebesgue measure, while the Dirac terms concentrate mass at specified points. For continuous test functions, the representation reduces every calculation to a finite sum of function values and an ordinary Lebesgue integral. For Borel sets, it reduces the calculation to checking which atom locations are included and integrating the density over the set.

Check Your Understanding

For each question, identify which part of the representing measure contributes before carrying out the calculation.

On $[0,1]$, let $L(f)=f(1)+\int_0^1 2x f(x)\,dm(x)$. Write the representing measure and compute $L(1)$ and $L(x)$.
For $\nu=3\delta_0+(1+x)m$ on $[0,1]$, compute $\nu((0,1])$ and $\nu(\{0\})$. State which endpoints are included in each set.
Prove directly from positivity of a finite representing measure that $|L(f)|\le L(1)\|f\|_\infty$ for every real-valued $f\in C([a,b])$.
Why does changing an integrable density at a single point leave its density measure unchanged, while changing a Dirac coefficient changes the represented functional?
If two finite positive regular Borel measures on $[a,b]$ have the same integrals against every $f\in C([a,b])$, what standard uniqueness result identifies them?

Worked Representations on Compact Intervals

What You'll Learn