The Pairing and Its Operator Norm
Let \((X,\mathcal{M},\mu)\) be a sigma-finite measure space; completeness is not required. Sigma-finiteness means that \(X\) is a countable union of measurable sets of finite measure. We work with complex-valued measurable functions, identifying functions that agree \(\mu\)-almost everywhere. The space \(L^1(\mu)\) consists of these equivalence classes \(f\) for which \(\int |f|\,d\mu<\infty\), with norm \(\|f\|_1=\int |f|\,d\mu\).
The integral defining \(T_g(f)\) exists because \(|fg|\le\|g\|_\infty|f|\) almost everywhere. If \(f\) is changed on a null set, \(fg\) is changed only on a null set, so the value of the integral is unchanged. Thus \(T_g\) is defined on equivalence classes, not just on particular representatives. Its operator norm is \(\|T_g\|=\sup\{|T_g(f)|:\|f\|_1\le1\}\).
Essential Bounds and the Exact Norm
Proof. By the definition of the infimum, for each positive integer \(n\) there is a bound \(C_n<M+1/n\) such that \(|g|\le C_n\) almost everywhere. Therefore \(|g|\le M+1/n\) outside a null set \(N_n\). The union \(N=\bigcup_nN_n\) is null, and outside \(N\) all these bounds hold. Letting \(n\) tend to infinity gives \(|g|\le M\) there. Now suppose \(0\le a<M\) and \(\{|g|>a\}\) were null. Then \(|g|\le a\) almost everywhere, which would imply \(M\le a\) by the definition of \(M\), a contradiction. Hence \(\{|g|>a\}\) has positive measure. \(\square\)
Proof. Linearity follows from the linearity of the complex Lebesgue integral. Write \(M=\|g\|_\infty\). The lemma gives \(|g|\le M\) almost everywhere, so for every \(f\in L^1(\mu)\), \[ |T_g(f)|=\left|\int fg\,d\mu\right|\le\int|f||g|\,d\mu\le M\int|f|\,d\mu=M\|f\|_1. \] Thus \(T_g\) is bounded and \(\|T_g\|\le M\). This also covers \(M=0\): then \(g=0\) almost everywhere and \(T_g\) is the zero functional.
For the reverse inequality, suppose \(M>0\) and fix \(a\) with \(0\le a<M\). By the lemma, \(E=\{|g|>a\}\) has positive measure. Sigma-finiteness gives a countable cover \(X=\bigcup_jX_j\) with each \(X_j\) measurable and of finite measure. Since \(E=\bigcup_j(E\cap X_j)\) has positive measure, at least one set \(A=E\cap X_j\) has positive measure. It has finite measure because \(\mu(A)\le\mu(X_j)<\infty\).
Define \(u(x)=\overline{g(x)}/|g(x)|\) where \(g(x)\ne0\), and \(u(x)=0\) where \(g(x)=0\). This is measurable, and \(|u|=1\) on \(A\), since \(|g|>a\ge0\) there. Set \(f=u\chi_A/\mu(A)\). Then \(\|f\|_1=1\), and \[ T_g(f)=\frac{1}{\mu(A)}\int_A |g|\,d\mu>a. \] The strict inequality holds because \(|g|-a\) is positive at every point of \(A\); a positive measurable function on a set of positive measure has positive integral. Consequently \(\|T_g\|\ge a\) for every \(a<M\), and hence \(\|T_g\|\ge M\). Combined with the upper bound this gives equality. \(\square\)
The phase \(u\) in the proof is essential: it makes \(u g=|g|\), so cancellation cannot reduce the integral. Sigma-finiteness is used to find a subset \(A\) of finite, positive measure, making the normalized indicator an admissible \(L^1\) function.
Worked Pairings
Worked Example: A Complex Pairing on a Finite Space
Let \(X=\{1,2,3\}\), let \(\mathcal{M}\) be its power set, and let \(\mu\) be counting measure. Set \(g(1)=3\), \(g(2)=4i\), and \(g(3)=0\). Compute the norm of \(T_g\) and exhibit a unit vector attaining it.
For \(f=(f(1),f(2),f(3))\), the \(L^1\) norm is \(|f(1)|+|f(2)|+|f(3)|\), while \[ T_g(f)=3f(1)+4i f(2). \] Therefore \[ |T_g(f)|\le3|f(1)|+4|f(2)|\le4\bigl(|f(1)|+|f(2)|+|f(3)|\bigr)=4\|f\|_1. \] The essential supremum of \(|g|\) is \(\max\{3,4,0\}=4\). Take \(f(1)=0,\ f(2)=-i,\ f(3)=0\). Its norm is \(|-i|=1\), and \(T_g(f)=(-i)(4i)=4\). Thus the bound is attained and \(\|T_g\|=4=\|g\|_\infty\). Check: the general upper bound gives \(\|T_g\|\le4\), and this unit vector gives \(\|T_g\|\ge4\).
Worked Example: Approaching the Norm on an Interval
On \(X=(0,1)\) with Lebesgue measure \(m\), let \(g(x)=x\). Its essential supremum is \(1\): \(x\le1\) everywhere on the domain, and for every \(a<1\) with \(a\ge0\), the set \(\{x\in(0,1):x>a\}=(a,1)\) has positive measure.
For each integer \(n\ge2\), let \(A_n=(1-1/n,1)\) and \(f_n=n\chi_{A_n}\). The interval has measure \(m(A_n)=1/n\), so \(\|f_n\|_1=n(1/n)=1\). Direct integration gives \[ T_g(f_n)=n\int_{1-1/n}^{1}x\,dx =\frac{n}{2}\left(1-\left(1-\frac1n\right)^2\right) =1-\frac{1}{2n}. \] For \(n=2\) this is \(3/4\); for \(n=4\) it is \(7/8\). In general these values approach \(1\), so the operator norm is at least \(1\), and the theorem's upper bound makes it exactly \(1\). Check: \(1-1/(2n)<1=\|g\|_\infty\) for each finite \(n\), while the limit of these values is \(1\).
Worked Example: A Complex Phase Near the Essential Maximum
Again use \((0,1)\) with Lebesgue measure, now with \(g(x)=i(1+x)\). Since \(|g(x)|=1+x<2\) on \((0,1)\), and for each \(a<2\) sufficiently close to \(2\) the set \(\{x:1+x>a\}=(a-1,1)\) has positive measure, \(\|g\|_\infty=2\).
For \(n\ge2\), put \(A_n=(1-1/n,1)\) and \(f_n(x)=-in\chi_{A_n}(x)\). Then \(\|f_n\|_1=n\,m(A_n)=1\). On \(A_n\), \(f_ng=(-in)i(1+x)=n(1+x)\), so \[ T_g(f_n)=n\int_{1-1/n}^{1}(1+x)\,dx =n\cdot\frac1n\left(1+\frac{(1-1/n)+1}{2}\right) =2-\frac{1}{2n}. \] For \(n=2\), the value is \(7/4\); for \(n=4\), it is \(15/8\). These approach \(2\), in agreement with the theorem. Check: the average of \(1+x\) over \(A_n\) is \(2-1/(2n)\), and the chosen phase makes the product real and nonnegative.
Norm Attainment and a Common Pitfall
The operator norm is a supremum over the unit ball; it need not be the value at any one function. In the example \(g(x)=x\) on \((0,1)\), no \(f\in L^1(m)\) with \(\|f\|_1\le1\) satisfies \(|T_g(f)|=1\). Indeed, \[ |T_g(f)|\le\int_0^1|f(x)|x\,dm =\|f\|_1-\int_0^1|f(x)|(1-x)\,dm. \] If \(f\ne0\) as an \(L^1\) equivalence class, then \(|f|(1-x)>0\) on a set of positive measure: \(|f|>0\) on a set of positive measure, and \(1-x>0\) everywhere on \((0,1)\). Its integral is therefore strictly positive. Hence \(|T_g(f)|<\|f\|_1\le1\). For \(f=0\), the pairing is \(0<1\) as well. Thus the norm \(1\) is approached by the displayed sequence but is not attained.
A common mistake is to try to attain the norm by normalizing the indicator of the set where \(|g|=\|g\|_\infty\). For \(g(x)=x\) on \((0,1)\), that set is empty, so this construction cannot work. The correct argument uses the positive-measure level sets \(\{|g|>a\}\) for \(a<\|g\|_\infty\), and then lets \(a\) increase to the essential supremum.
Sigma-finiteness is also a real hypothesis for this version of the result. For example, take \(X\) uncountable, \(\mathcal{M}=\{\varnothing,X\}\), and \(\mu(\varnothing)=0,\ \mu(X)=\infty\). This measure space is not sigma-finite. Every measurable complex-valued function on it is constant: a nonconstant function would have two distinct values separated by an open set, whose preimage would have to be neither \(\varnothing\) nor \(X\). For \(g=1\), the essential supremum is \(1\), but every nonzero constant function has infinite \(L^1\) norm, so \(L^1(\mu)=\{0\}\) and \(T_g\) has norm \(0\). This shows why one cannot omit sigma-finiteness without further assumptions on the measure space.
Check Your Understanding
Use the definition of the pairing and the proof of the norm formula to answer the following.
- For counting measure on \(\{1,2\}\), let \(g(1)=2i\) and \(g(2)=1\). Find \(\|T_g\|\) and a unit \(f\) for which \(|T_g(f)|=\|T_g\|\).
- Explain why \(\int fg\,d\mu\) is unchanged if \(f\) is replaced by an almost-everywhere equal representative.
- For \(g(x)=x\) on \((0,1)\), compute \(\|n\chi_{(1-1/n,1)}\|_1\) and its pairing with \(g\).
- In the lower-bound proof, why must the selected set \(A\) have finite as well as positive measure?
- Give the reason that the essential supremum may fail to be attained by any individual unit \(L^1\) function.