Benchmark Functionals on Compact Intervals

Three Basic Ways to Build a Positive Functional

Continuous functions on a compact interval can be tested in several concrete ways: at one point, at finitely many points with nonnegative weights, or across the interval by integration against a nonnegative density. Each method gives a linear rule that assigns a real number to every continuous real-valued function. These examples are useful benchmarks: they make positivity and linearity explicit, and they show that a positive functional need not be an average.

Throughout, let \(a,b\in\mathbb{R}\) with \(a<b\), and let \(C([a,b])\) denote the real vector space of continuous real-valued functions on \([a,b]\). A function \(f\in C([a,b])\) is nonnegative if \(f(x)\ge0\) for every \(x\in[a,b]\). A linear functional is a linear map \(L:C([a,b])\to\mathbb{R}\).

The constant function with value \(1\) is denoted by \(1\). Point evaluation is the simplest example. For a fixed \(c\in[a,b]\), set \(L_c(f)=f(c)\). Linearity follows from evaluating sums and scalar multiples at the same point, and \(L_c(f)\ge0\) whenever \(f\ge0\). In measure language, this is integration against the Dirac measure \(\delta_c\), defined on Borel sets \(E\subseteq[a,b]\) by \(\delta_c(E)=1\) if \(c\in E\) and \(0\) otherwise.

Finite Weighted Sums of Evaluations

Choose finitely many points \(c_1,\ldots,c_N\in[a,b]\) and real weights \(w_1,\ldots,w_N\ge0\). The rule

is linear, because each evaluation is linear and finite sums preserve linearity. It is positive, because if \(f\ge0\), then every term \(w_j f(c_j)\ge0\), so their sum is nonnegative. This rule corresponds to the finite measure \(\sum_{j=1}^{N}w_j\delta_{c_j}\). Repeated points are allowed; their weights simply contribute separate terms at the same location. Zero weights are also allowed. If every weight is zero, \(L\) is the zero functional.

Proof. For \(f,g\in C([a,b])\) and scalars \(\alpha,\beta\in\mathbb{R}\), evaluation gives \((\alpha f+\beta g)(c_j)=\alpha f(c_j)+\beta g(c_j)\) for each \(j\). Summing these equalities after multiplication by \(w_j\) proves \(L(\alpha f+\beta g)=\alpha L(f)+\beta L(g)\), so \(L\) is linear. If \(f\ge0\) at every point, then \(f(c_j)\ge0\); since \(w_j\ge0\), each product \(w_jf(c_j)\ge0\), and hence \(L(f)\ge0\). Finally, \(1(c_j)=1\) for every \(j\), giving \(L(1)=\sum_{j=1}^{N}w_j\). If \(N=0\), the sum is empty, \(L\) is the zero functional, and the same conclusions hold with value \(0\). \(\square\)

Integration Against a Nonnegative Density

A third construction averages function values over the interval with a density that may vary from point to point. Let \(\mathcal{L}\) be the Lebesgue sigma-algebra on \([a,b]\), and let \(m\) be Lebesgue measure there. Suppose \(q:[a,b]\to\mathbb{R}\) is \(\mathcal{L}\)-measurable, \(q\ge0\) \(m\)-a.e., and \(\int_{[a,b]}q\,dm<\infty\). For every \(f\in C([a,b])\), define

This integral is finite: continuity on the compact interval implies that \(f\) is bounded, so \(|f(x)q(x)|\le \|f\|_\infty q(x)\) \(m\)-a.e.; the right-hand side is integrable. The integral does not depend on the values of \(q\) on a null set. Linearity follows from linearity of the Lebesgue integral for integrable real-valued functions. If \(f\ge0\) everywhere, then \(f q\ge0\) \(m\)-a.e., so \(L_q(f)\ge0\).

Proof. Fix \(f\in C([a,b])\). Since \([a,b]\) is compact, \(M=\sup_{x\in[a,b]}|f(x)|\) is finite. Thus \(|f q|\le Mq\) \(m\)-a.e., and \(\int Mq\,dm=M\int q\,dm<\infty\). Therefore \(fq\) is integrable and \(L_q(f)\) is a finite real number. For \(f,g\in C([a,b])\) and \(\alpha,\beta\in\mathbb{R}\), pointwise linearity gives \((\alpha f+\beta g)q=\alpha fq+\beta gq\); linearity of the integral then gives \(L_q(\alpha f+\beta g)=\alpha L_q(f)+\beta L_q(g)\). If \(f\ge0\) everywhere, then \(fq\ge0\) \(m\)-a.e., so its integral is nonnegative. Finally, \(1q=q\) \(m\)-a.e., and hence \(L_q(1)=\int q\,dm\). \(\square\)

Combining the Constructions and Avoiding a Common Error

Positive functionals can be added and multiplied by nonnegative scalars without losing positivity. More explicitly, if \(L_1,L_2:C([a,b])\to\mathbb{R}\) are positive linear functionals and \(\alpha,\beta\ge0\), then \(L=\alpha L_1+\beta L_2\) is linear and positive: for \(f\ge0\), \(L(f)=\alpha L_1(f)+\beta L_2(f)\ge0\). The same reasoning allows a finite mixture of point evaluations and density integrals. This observation constructs more examples; it does not claim that every positive functional must be presented in one of these particular finite forms.

Common mistake: treating positivity as normalization. In the three-point example, the weights sum to \(4\), not \(1\). A wrong calculation divides the weighted sum \(12\) by \(4\), reports \(L(f)=3\), and describes this as the functional value. That division defines a different rule, namely the normalized weighted average. For the functional actually specified, the formula is \(L(f)=\sum_jw_jf(c_j)\), with no division. Substitution gives \(L(f)=12\), and substituting the constant function gives \(L(1)=1/2+2+3/2=4\). Positivity requires nonnegative outputs on nonnegative inputs; it does not impose \(L(1)=1\).

These constructions provide practical tests for formulas on \(C([a,b])\). First check that the proposed rule is defined and finite for every continuous input. Then verify linearity from the formula, and verify positivity using the nonnegative weights or density. For a density, the essential conditions are integrability and nonnegativity almost everywhere; pointwise nonnegativity at every point is stronger than necessary. For a finite weighted sum, the weights need not add to one, and nodes may lie at either endpoint.