Hilbert Representation for Measure Functionals

From a Bounded Functional to a Vector

A Hilbert-space representation theorem turns an abstract bounded linear functional into an inner product with a fixed vector. For measure theory, the useful case is a functional defined by integration against a finite measure \(\nu\) that is dominated by another finite measure \(\lambda\). The Hilbert-space result supplies a function \(g\in L^2(\lambda)\); evaluating the functional on indicator functions then identifies \(\nu(E)\) with \(\int_E g\,d\lambda\). The domination assumption will also force \(g\) to lie between zero and one almost everywhere.

We first work with a real Hilbert space \(H\), whose inner product is written \(\langle\cdot,\cdot\rangle\), and a bounded real-linear functional \(T:H\to\mathbb R\). Boundedness means that some \(C<\infty\) satisfies \(|T(x)|\le C\|x\|\) for every \(x\in H\). The representation is an equality for every \(x\in H\), not merely almost everywhere; the almost-everywhere qualification enters when \(H=L^2(\lambda)\), whose elements are equivalence classes of functions.

The proof uses projection onto the closed subspace \(\ker T\). We give the projection argument as well, so that the representation does not depend on an unproved existence claim.

Proof. Put \(d=\inf_{y\in K}\|x-y\|\), and choose \(y_n\in K\) with \(\|x-y_n\|\to d\). Since \((y_n+y_m)/2\in K\), the definition of \(d\) and the parallelogram identity give

The right side tends to zero as \(n,m\to\infty\). Thus \((y_n)\) is Cauchy; completeness of \(H\) and closedness of \(K\) give a limit \(p\in K\). Continuity of the norm gives \(\|x-p\|=d\). For \(z\in K\) and real \(t\), \(p+tz\in K\), so minimality implies \(\|x-p-tz\|^2\ge\|x-p\|^2\). Expanding yields \(-2t\langle x-p,z\rangle+t^2\|z\|^2\ge0\) for every real \(t\), which is possible only if \(\langle x-p,z\rangle=0\). If \(p,q\in K\) both have this orthogonality property, then \(p-q\in K\) and \(x-p\) is orthogonal to \(p-q\); likewise \(x-q\) is orthogonal to \(p-q\). Subtracting gives \(\|p-q\|^2=0\), so \(p=q\). \(\square\)

Proof. If \(T=0\), take \(g=0\). Suppose \(T\ne0\), let \(K=\ker T\), and choose \(x\in H\) with \(T(x)\ne0\). The subspace \(K\) is closed because \(T\) is continuous: if \(x_n\in K\) and \(x_n\to x\), then \(T(x)=\lim_nT(x_n)=0\). By the projection lemma, write \(x=p+u\), with \(p\in K\) and \(u\) orthogonal to \(K\). Since \(T(u)=T(x)-T(p)=T(x)\ne0\), \(u\ne0\). For any \(y\in H\), the vector \(y-\frac{T(y)}{T(u)}u\) lies in \(K\), because applying \(T\) gives zero. Orthogonality of \(u\) to \(K\) therefore gives \(\langle y,u\rangle=\frac{T(y)}{T(u)}\|u\|^2\). Set \(g=\frac{T(u)}{\|u\|^2}u\); then \(T(y)=\langle y,g\rangle\) for every \(y\). If \(g_1,g_2\) both represent \(T\), then \(\langle y,g_1-g_2\rangle=0\) for every \(y\); choosing \(y=g_1-g_2\) proves \(g_1=g_2\). Finally, Cauchy–Schwarz gives \(|T(y)|\le\|y\|\|g\|\), so \(\|T\|\le\|g\|\). Taking \(y=g/\|g\|\), which is defined since \(g\ne0\), gives \(|T(y)|=\|g\|\), and hence \(\|T\|=\|g\|\). \(\square\)

Applying the Theorem to Dominated Measures

Let \((X,\mathcal M)\) be a measurable space, and let \(\lambda\) and \(\nu\) be finite positive measures on it. Here finite means \(\lambda(X)<\infty\) and \(\nu(X)<\infty\). Assume \(\nu(E)\le\lambda(E)\) for every \(E\in\mathcal M\). This relation is called domination of \(\nu\) by \(\lambda\). We use the real Hilbert space \(L^2(\lambda)\), complete as established in the earlier material on \(L^2\) completeness, with inner product \(\langle f,h\rangle=\int fh\,d\lambda\). Its elements are equivalence classes of real-valued measurable functions with finite integral of the square, where functions equal \(\lambda\)-a.e. represent the same element.

Proof. For an \(L^2(\lambda)\) class represented by \(f\), define \(T(f)=\int f\,d\nu\). First this is well-defined. If \(f=h\) \(\lambda\)-a.e., then the set where they differ is contained in a measurable \(\lambda\)-null set, and domination makes every such set \(\nu\)-null as well. Thus the integrals agree. The integrals are finite: domination implies \(\int |f|\,d\nu\le\int |f|\,d\lambda\), and finiteness of \(\lambda\) and Cauchy–Schwarz give \(\int |f|\,d\lambda\le\lambda(X)^{1/2}\|f\|_2\). The integral comparison follows first for nonnegative simple functions from \(\nu(E)\le\lambda(E)\), and then for all nonnegative measurable functions by their increasing simple approximations and Monotone Convergence. Therefore \(T\) is real-linear and \(|T(f)|\le\lambda(X)^{1/2}\|f\|_2\), so it is bounded. The Hilbert-space representation gives a unique \(g\in L^2(\lambda)\) with \(T(f)=\int fg\,d\lambda\) for all \(f\in L^2(\lambda)\). For each \(E\in\mathcal M\), \(\chi_E\in L^2(\lambda)\), since \(\int\chi_E^2\,d\lambda=\lambda(E)\le\lambda(X)<\infty\). Applying the representation to \(\chi_E\) yields \(\nu(E)=\int_Eg\,d\lambda\).

It remains to establish the bounds on \(g\). Choose a measurable representative. For each positive integer \(n\), let \(A_n=\{g\le-1/n\}\). If \(\lambda(A_n)>0\), then \(\int_{A_n}g\,d\lambda\le-\lambda(A_n)/n<0\), contradicting \(\int_{A_n}g\,d\lambda=\nu(A_n)\ge0\). Thus every \(A_n\) is null, and \(\{g<0\}=\bigcup_{n\ge1}A_n\) is null. Next let \(B_n=\{g\ge1+1/n\}\). If \(\lambda(B_n)>0\), then \(\nu(B_n)=\int_{B_n}g\,d\lambda\ge(1+1/n)\lambda(B_n)>\lambda(B_n)\), contradicting domination. Hence each \(B_n\) is null, and \(\{g>1\}=\bigcup_{n\ge1}B_n\) is null. Consequently \(0\le g\le1\) \(\lambda\)-a.e. Uniqueness up to \(\lambda\)-a.e. equality follows from uniqueness of the representing element in \(L^2(\lambda)\). \(\square\)

Why the Domination Hypothesis Matters

A common mistake is to define \(T(f)=\int f\,d\nu\) on \(L^2(\lambda)\) without checking that the value depends only on the \(\lambda\)-a.e. equivalence class. For example, take \(X=[0,1]\), \(\lambda=m\), and \(\nu=\delta_0\), the unit point mass at zero. The functions \(f=0\) and \(h=\chi_{\{0\}}\) represent the same element of \(L^2(m)\), because \(\{0\}\) has Lebesgue measure zero. But \(\int f\,d\delta_0=0\), whereas \(\int h\,d\delta_0=1\). Thus the proposed functional has two different values on the same \(L^2(m)\) element; it is not well-defined. The correction is to check domination first: here \(\delta_0(\{0\})=1\) while \(m(\{0\})=0\), so \(\delta_0\) is not dominated by \(m\).

The proposition is deliberately limited to positive finite measures with \(\nu\le\lambda\). Finiteness makes the displayed \(L^2\)-to-\(L^1\) bound available, and domination both makes the functional well-defined and yields the pointwise bounds on its density. The construction is the Hilbert-space step in the finite Radon–Nikodym argument; the next step develops the full theorem. In particular, the result here does not assert a density for an arbitrary measure, nor does it replace the hypotheses needed for signed or complex measures.