From Measure Domination to a Functional
A measure can be used to turn a function into a number by integration. To make this construction a functional on \(L^2\), however, two issues must be addressed: the integral must be finite for every \(L^2\) function, and it must depend only on the function’s almost-everywhere equivalence class. A simple domination condition on measures resolves both issues and gives a quantitative bound.
Throughout, let \((X,\mathcal{M})\) be a measurable space, let \(\lambda\) and \(\nu\) be finite positive measures on \((X,\mathcal{M})\), and assume \(\nu(E)\le\lambda(E)\) for every \(E\in\mathcal{M}\). No completeness assumption is needed. Functions in this tutorial are complex-valued and measurable; real-valued functions are included as a special case. The space \(L^2(\lambda)\) consists of equivalence classes of functions with \(\int_X |f|^2\,d\lambda<\infty\), where two functions represent the same class when they agree \(\lambda\)-a.e.
Domination Controls Integrals
First, the setwise inequality \(\nu\le\lambda\) also controls integrals. This is the step that allows an estimate under \(\nu\) to be related to the \(L^2(\lambda)\) norm.
Proof. For a nonnegative simple function \(\varphi=\sum_{j=1}^k a_j\chi_{E_j}\), with disjoint measurable \(E_j\) and \(a_j\ge0\), the definition of the simple-function integral gives \(\int\varphi\,d\nu=\sum_j a_j\nu(E_j)\le\sum_j a_j\lambda(E_j)=\int\varphi\,d\lambda\). For general nonnegative measurable \(u\), its Lebesgue integral is the supremum of \(\int\varphi\) over nonnegative simple \(\varphi\le u\). Each such \(\varphi\) satisfies \(\int\varphi\,d\nu\le\int\varphi\,d\lambda\le\int u\,d\lambda\). Taking the supremum on the left proves the inequality, including when the right side is infinite. \(\square\)
In particular, applying the proposition to \(|f|^2\) shows that an \(L^2(\lambda)\) function also belongs to \(L^2(\nu)\). Since \(\nu(X)\) is finite, Cauchy–Schwarz then makes \(f\) integrable with respect to \(\nu\). The next theorem records the full conclusion, including independence from the choice of representative.
The Bounded Measure Functional
Proof. Let \(f\in L^2(\lambda)\) be a measurable representative. By the proposition, \(\int_X|f|^2\,d\nu\le\int_X|f|^2\,d\lambda<\infty\). Cauchy–Schwarz on the finite measure space \((X,\mathcal{M},\nu)\), applied to \(|f|\) and the constant function \(1\), gives \[ \int_X|f|\,d\nu\le\left(\int_X|f|^2\,d\nu\right)^{1/2}\nu(X)^{1/2}<\infty. \] Thus \(f\) is integrable with respect to \(\nu\), so its complex integral is defined. Combining these inequalities yields \[ \left|\int_X f\,d\nu\right| \le\int_X|f|\,d\nu \le\sqrt{\nu(X)}\left(\int_X|f|^2\,d\nu\right)^{1/2} \le\sqrt{\nu(X)}\|f\|_2. \] The assumption on sets, applied to \(X\), gives \(\nu(X)\le\lambda(X)\), so the stated second bound follows.
If \(f=g\) \(\lambda\)-a.e., then \(N=\{x:f(x)\ne g(x)\}\) is measurable and \(\lambda(N)=0\). Domination gives \(\nu(N)=0\). Since \(f\) and \(g\) are both \(\nu\)-integrable and agree outside this \(\nu\)-null set, their integrals are equal. Thus \(T\) is well-defined on equivalence classes. Linearity follows from linearity of the integral for integrable complex-valued functions. The displayed estimate gives boundedness and the operator norm bound. If \(\nu(X)=0\), then \(\nu(E)=0\) for all measurable \(E\), so every \(\nu\)-integral is zero. \(\square\)
The estimate \(\sqrt{\nu(X)}\) is often sharper than \(\sqrt{\lambda(X)}\). It does not claim that the operator norm always equals either quantity: it is a guaranteed upper bound obtained directly from domination and Cauchy–Schwarz.
Worked Example: A Functional on a Three-Point Space
Let \(X=\{a,b,c\}\) with \(\mathcal{M}\) the set of all subsets. Give the points \(\lambda\)-masses \(2,1,3\), respectively, and \(\nu\)-masses \(1,\tfrac12,0\), respectively. These are finite measures, and \(\nu(E)\le\lambda(E)\) for every \(E\), since the inequality holds at each point. For \(f(a)=2\), \(f(b)=-1\), and \(f(c)=3\), compute \(T(f)\) and check the theorem’s estimate.
On a finite space the integral is the sum of the point values multiplied by their masses. Therefore \[ T(f)=2\cdot1+(-1)\cdot\tfrac12+3\cdot0=\tfrac32. \] The \(L^2(\lambda)\) norm is computed from \[ \|f\|_2^2=|2|^2\cdot2+|-1|^2\cdot1+|3|^2\cdot3=8+1+27=36, \] so \(\|f\|_2=6\). Also, \(\nu(X)=1+\tfrac12+0=\tfrac32\), and \[ \int_X|f|^2\,d\nu=4\cdot1+1\cdot\tfrac12+9\cdot0=\tfrac92. \] The sharper intermediate estimate gives \(|T(f)|\le\sqrt{\tfrac32}\sqrt{\tfrac92}=\tfrac{3\sqrt3}{2}\). Indeed, \(\tfrac32\le\tfrac{3\sqrt3}{2}\), since \(1\le\sqrt3\). The theorem’s coarser norm estimate also gives \(|T(f)|\le\sqrt{\tfrac32}\cdot6=3\sqrt6\). The direct point-mass sum verifies the value \(\tfrac32\) independently.
Worked Example: The Bound Can Be Attained
Again take \(X=\{a,b\}\) with all subsets measurable, but now let \(\nu=\lambda\), with masses \(\lambda(\{a\})=2\) and \(\lambda(\{b\})=1\). The space has total measure \(3\). For the function \(f(a)=f(b)=1/\sqrt3\), compute its norm and the functional value.
The norm calculation is \[ \|f\|_2^2=\tfrac13\cdot2+\tfrac13\cdot1=1, \] so \(\|f\|_2=1\). Since \(\nu=\lambda\), \[ T(f)=\tfrac1{\sqrt3}\cdot2+\tfrac1{\sqrt3}\cdot1=\tfrac3{\sqrt3}=\sqrt3. \] The theorem gives \(\|T\|\le\sqrt{\nu(X)}=\sqrt3\), while this unit-norm function has \(|T(f)|=\sqrt3\). Hence \(\|T\|=\sqrt3\) in this example. Substitution into the original finite-sum formula gives the same value, \(2/\sqrt3+1/\sqrt3=\sqrt3\), checking the calculation.
Why Absolute Continuity Alone Is Not Enough
Domination is stronger than absolute continuity. The condition \(\nu\ll\lambda\) means that \(\lambda(E)=0\) implies \(\nu(E)=0\); it does not require \(\nu(E)\le\lambda(E)\). Absolute continuity does guarantee that changing a function on a \(\lambda\)-null set does not change its \(\nu\)-integral, but it need not provide the \(L^2\) bound proved above.
Worked Example: An Absolutely Continuous Measure with an Unbounded Functional
Let \(X=(0,1)\) with its Lebesgue sigma-algebra and let \(\lambda\) be Lebesgue measure. Define \(\nu(E)=\int_E \tfrac14x^{-3/4}\,d\lambda(x)\). This is a finite measure absolutely continuous with respect to \(\lambda\). Its total mass is \[ \nu(X)=\tfrac14\int_0^1x^{-3/4}\,dx=\tfrac14\cdot4=1. \] But it is not dominated by \(\lambda\): on \(E=(0,1/16)\), \[ \nu(E)=\tfrac14\cdot4\cdot(1/16)^{1/4}=\tfrac12, \qquad \lambda(E)=\tfrac1{16}, \] and \(\tfrac12>\tfrac1{16}\).
For each positive integer \(n\), let \(f_n=\sqrt n\,\chi_{(0,1/n)}\). These functions are measurable, and \[ \|f_n\|_2^2=n\lambda((0,1/n))=n\cdot\tfrac1n=1. \] Their integrals against \(\nu\) are \[ \int_X f_n\,d\nu =\tfrac{\sqrt n}{4}\int_0^{1/n}x^{-3/4}\,dx =\tfrac{\sqrt n}{4}\cdot4(1/n)^{1/4} =n^{1/4}. \] Thus unit-norm functions have functional values \(n^{1/4}\), which are unbounded as \(n\) increases. For example, at \(n=16\) the value is \(16^{1/4}=2\), while the norm remains \(1\). Direct substitution into the integral formula confirms this value. This example shows concretely that \(\nu\ll\lambda\), even for finite measures, is not by itself sufficient for boundedness on \(L^2(\lambda)\).
A Common Mistake and the Role of the Estimate
A tempting but incorrect argument is: “\(\nu\ll\lambda\), so \(\nu\le\lambda\).” These statements have different meanings. Absolute continuity only transfers null sets; setwise domination compares the measures of every measurable set. In the last example, \(\nu\ll\lambda\) holds because \(\nu\) is defined by integrating a density, yet on \(E=(0,1/16)\) the calculation gives \(\nu(E)=1/2>1/16=\lambda(E)\). Applying the theorem’s norm estimate there would therefore use a hypothesis that fails. The sequence \(f_n\) confirms that no boundedness conclusion follows from absolute continuity alone.
Under the domination hypothesis, the result is useful because it constructs a bounded linear functional using only integration and a norm estimate. The estimate also states exactly which control was obtained: \(|T(f)|\le\sqrt{\nu(X)}\|f\|_2\). It does not identify a representing vector in \(L^2(\lambda)\); that is a separate Hilbert-space question.
Check Your Understanding
For each question, keep separate the assumptions \(\nu\ll\lambda\) and \(\nu(E)\le\lambda(E)\).
- Why does \(\nu(E)\le\lambda(E)\) for every measurable \(E\) imply \(\int u\,d\nu\le\int u\,d\lambda\) for every nonnegative measurable \(u\)?
- Which step in the theorem’s proof ensures that \(\int f\,d\nu\) is finite for \(f\in L^2(\lambda)\)?
- Explain why two representatives that agree \(\lambda\)-a.e. have the same \(\nu\)-integral under the domination hypothesis.
- In the two-point example with masses \(2\) and \(1\), why does the constant function with value \(1/\sqrt3\) show that the norm bound is attained?
- For the density \(\tfrac14x^{-3/4}\), verify that the sequence \(\sqrt n\,\chi_{(0,1/n)}\) has \(L^2(\lambda)\) norm \(1\) and that its \(\nu\)-integral is \(n^{1/4}\).