Null Sets and Conjugation in Complex Duality

Pairings Depend on Classes, Not Pointwise Values

In complex duality, two conventions for pairing functions are both useful, but they place conjugation in different locations. Confusing them can change a linear functional into a conjugate-linear one. A separate but related issue is that elements of \(L^1\) and \(L^\infty\) are defined only up to equality almost everywhere, so pointwise values on null sets cannot determine a pairing or its norm.

Throughout, let \((X,\mathcal{M},\mu)\) be a sigma-finite measure space; it need not be complete or finite. All functions in this tutorial are complex-valued and measurable. Write \(f\sim f'\) when \(f=f'\) \(\mu\)-a.e. The space \(L^1(\mu)\) consists of equivalence classes of functions \(f\) for which \(\int_X |f|\,d\mu<\infty\), and \(L^\infty(\mu)\) consists of equivalence classes that are essentially bounded. For a measurable \(g\), its essential supremum norm is \(\|g\|_\infty=\inf\{C\ge0: |g|\le C\ \mu\text{-a.e.}\}\).

If \(f\in L^1(\mu)\) and \(g\in L^\infty(\mu)\), then \(fg\) is integrable: using an essentially bounded representative of \(g\), \(|fg|\le \|g\|_\infty |f|\) a.e. There are two standard pairings: \[ B(f,g)=\int_X fg\,d\mu,\qquad \langle f,g\rangle=\int_X f\overline{g}\,d\mu. \] The first is bilinear over the complex numbers. The second is linear in its first argument and conjugate-linear in its second.

Definition: A complex-linear functional \(\Lambda\) satisfies \(\Lambda(\alpha f+\beta h)=\alpha\Lambda(f)+\beta\Lambda(h)\) for all complex scalars \(\alpha,\beta\). A conjugate-linear functional satisfies \(\Lambda(\alpha f+\beta h)=\overline{\alpha}\Lambda(f)+\overline{\beta}\Lambda(h)\). For fixed \(g\), \(f\mapsto B(f,g)\) and \(f\mapsto\langle f,g\rangle\) are complex-linear; the conjugation in the second pairing is on the fixed function, not on \(f\).

The representation theorem for the dual of \(L^1\), established in the earlier material on \(L^1\) duality, uses a complex-linear functional represented as \(\Lambda(f)=\int_X fg\,d\mu\) for some \(g\in L^\infty(\mu)\). If one prefers the sesquilinear notation, set \(h=\overline{g}\); then the same functional is \(\Lambda(f)=\langle f,h\rangle\). The representing function changes by conjugation, not the functional.

Why Representatives Do Not Matter

Proposition: On \((X,\mathcal{M},\mu)\), if \(f,f'\in L^1(\mu)\) satisfy \(f=f'\) a.e. and \(g,g'\in L^\infty(\mu)\) satisfy \(g=g'\) a.e., then \[ \int_X fg\,d\mu=\int_X f'g'\,d\mu \quad\text{and}\quad \int_X f\overline g\,d\mu=\int_X f'\overline{g'}\,d\mu. \] Thus both pairings are well defined on equivalence classes.

Proof. There are null sets \(N_1,N_2\in\mathcal{M}\) outside which \(f=f'\) and \(g=g'\), respectively. Their union \(N_1\cup N_2\) is null by countable subadditivity. Outside this union, both products \(fg\) and \(f'g'\) agree, and so do \(f\overline g\) and \(f'\overline{g'}\). Each product is integrable because an essentially bounded factor times an \(L^1\) factor is integrable. Integrable functions that agree a.e. have the same Lebesgue integral. This proves both equalities. \(\square\)

Worked Example: Changing Values on a Null Set

Take \(X=[0,1]\), \(\mathcal{M}=\mathcal{L}\), and Lebesgue measure \(m\). Let \(f(x)=1\) everywhere, and let \(f'(x)=1\) except that \(f'(1/2)=100\). Set \(g(x)=2+i\) everywhere and \(g'(x)=2+i\) except that \(g'(1/2)=7\).

The functions differ only at \(\{1/2\}\), which has measure zero. For the first representatives, \(\int_0^1 fg\,dm=\int_0^1(2+i)\,dm=(2+i)\cdot1=2+i\). For the altered representatives, the product is still \(2+i\) outside \(\{1/2\}\), so its integral is also \(2+i\). For the sesquilinear pairing, \(\overline{g(x)}=2-i\) a.e., and hence \(\int_0^1 f\overline g\,dm=(2-i)\cdot1=2-i\); the altered representatives give the same value.

Check: The interval has measure \(1\), and changing a function at one point changes no integral. In particular, the product value at \(1/2\), whether \(100(7)\) or another finite number, contributes zero to the integral.

Completeness of the measure space is not needed for this argument. The functions and exceptional sets here are measurable, and equality a.e. means that the set where they differ is contained in a measurable null set. In computations with equivalence classes, it is therefore legitimate to choose convenient representatives, but not to infer an integral or a norm from exceptional point values.

Conjugation Determines Linearity

Worked Example: Two Pairings Give Different Values

Let \(X=\{1,2\}\) with counting measure, so \(\int_X q\,d\mu=q(1)+q(2)\). Let \(f(1)=1+i,\ f(2)=2\), and \(g(1)=1+i,\ g(2)=2i\). Every subset is measurable, and all functions on this finite space are integrable.

For the bilinear pairing, the first product is \((1+i)(1+i)=1+2i+i^2=2i\), and the second is \(2(2i)=4i\). Thus \(\int_X fg\,d\mu=2i+4i=6i\). For the sesquilinear pairing, \(\overline{g(1)}=1-i\) and \(\overline{g(2)}=-2i\). The two products are \((1+i)(1-i)=1-i+i-i^2=2\) and \(2(-2i)=-4i\). Thus \(\int_X f\overline g\,d\mu=2-4i\).

Both pairings are complex-linear in \(f\): replacing \(f\) by \(\alpha f\) multiplies either integral by \(\alpha\), since \(\overline g\) is fixed. But as functions of the second argument, \(B(f,g)\) is linear in \(g\), whereas \(\langle f,g\rangle\) is conjugate-linear in \(g\). For example, multiplying \(g\) by \(i\) multiplies \(B(f,g)\) by \(i\), but multiplies \(\langle f,g\rangle\) by \(\overline{i}=-i\).

Check: For this \(f,g\), the sesquilinear value is \(2-4i\). Replacing \(g\) by \(ig\) gives \(-i(2-4i)=-4-2i\), exactly the original value multiplied by \(-i\), not by \(i\).

There is another convention that must be kept distinct. For fixed \(h\), the map \(f\mapsto\int_X\overline f\,h\,d\mu\) is conjugate-linear in \(f\), because \(\overline{\alpha f}=\overline\alpha\,\overline f\). If \(\Phi\) is a bounded conjugate-linear functional, then \(f\mapsto\overline{\Phi(f)}\) is complex-linear: applying conjugation to \(\Phi(\alpha f)=\overline\alpha\Phi(f)\) gives \(\overline{\Phi(\alpha f)}=\alpha\overline{\Phi(f)}\). The \(L^1\) representation theorem applied to this linear functional gives \(\overline{\Phi(f)}=\int_X fg\,d\mu\), or equivalently \(\Phi(f)=\int_X\overline f\,\overline g\,d\mu\). Conjugation is how the two versions correspond.

The Essential Supremum Is the Dual Norm

Theorem: Let \((X,\mathcal{M},\mu)\) be sigma-finite and let \(g\in L^\infty(\mu)\). The complex-linear functional \(T_g(f)=\int_X fg\,d\mu\) on \(L^1(\mu)\) has operator norm \[ \|T_g\|=\sup_{\|f\|_1\le1}|T_g(f)|=\|g\|_\infty. \] The same norm formula holds for \(f\mapsto\int_X f\overline g\,d\mu\).

Proof. First, for every \(f\in L^1(\mu)\), the integral triangle inequality and the essential bound on \(g\) give \[ |T_g(f)|\le\int_X|f||g|\,d\mu\le\|g\|_\infty\|f\|_1. \] Hence \(\|T_g\|\le\|g\|_\infty\).

Write \(M=\|g\|_\infty\). If \(M=0\), then \(g=0\) a.e., so \(T_g=0\) and the asserted equality holds. Suppose \(M>0\), and fix any real \(a\) with \(0\le a<M\). By the definition of essential supremum, \(E=\{x:|g(x)|>a\}\) has positive measure; otherwise \(|g|\le a\) a.e., contradicting \(a<M\). Sigma-finiteness gives a countable cover of \(X\) by measurable sets of finite measure. At least one intersection of \(E\) with a set in this cover, call it \(F\), has positive measure; otherwise \(E\) would be a countable union of null sets. Thus \(0<\mu(F)<\infty\).

On \(F\), \(g\ne0\). Define \(f=\chi_F\overline g/(|g|\mu(F))\), and set \(f=0\) outside \(F\). Then \(|f|=\chi_F/\mu(F)\), so \(\|f\|_1=\mu(F)/\mu(F)=1\). Also \(fg=\chi_F|g|/\mu(F)\), so \(T_g(f)=\mu(F)^{-1}\int_F|g|\,d\mu>a\): indeed \(|g|-a\) is positive at every point of \(F\), and its integral on a set of positive measure is positive. Therefore \(\|T_g\|\ge a\) for every \(a<M\), which implies \(\|T_g\|\ge M\). Together with the upper bound this proves equality. Replacing \(g\) by \(\overline g\), whose essential supremum norm is also \(M\), proves the second formula. \(\square\)

This norm is determined by an almost-everywhere bound, not by the largest value of a chosen representative. The proof also shows uniqueness of the representing class: if \(T_g=T_h\), then \(T_{g-h}=0\), so the norm formula gives \(\|g-h\|_\infty=0\), which means \(g=h\) a.e.

Worked Example: Pointwise Supremum Can Give the Wrong Norm

On \([0,1]\) with Lebesgue measure, let \(g(x)=1\) for \(x\ne1/2\) and \(g(1/2)=3\). The pointwise supremum of \(|g|\) is \(3\), but for every \(C\ge1\), \(|g|\le C\) a.e., and for every \(C<1\), the set \(\{|g|>C\}\) has measure \(1\). Therefore \(\|g\|_\infty=1\).

For \(f\in L^1\), the exceptional value at \(1/2\) does not affect the integral, so \(T_g(f)=\int_0^1 f\,dm\). The integral inequality gives \(|T_g(f)|\le\|f\|_1\). Taking \(f=\chi_{[0,1]}\), for which \(\|f\|_1=1\) and \(T_g(f)=1\), shows \(\|T_g\|=1\). This agrees with the essential supremum and not with the pointwise supremum.

Common mistake and correction: The incorrect calculation is \(\|T_g\|=\sup_{x\in[0,1]}|g(x)|=3\). It uses a value at a null set as if it contributed to an integral. The correct calculation uses the essential supremum: \(\|T_g\|=\|g\|_\infty=1\), as also verified directly by the test function above.

Worked Example: A Test Function Detects the Essential Bound

Let \(X=(0,1)\) with Lebesgue measure and set \(g=-3i\) on \(A=(1/2,3/4)\), while \(g=1\) elsewhere. Since \(m(A)=3/4-1/2=1/4\), and \(|g|=3\) on \(A\) and \(1\) on its complement, \(\|g\|_\infty=3\).

For every \(f\in L^1\), \(|\int fg\,dm|\le3\|f\|_1\), so \(\|T_g\|\le3\). Define \(f=4i\chi_A\). Its norm is \(\|f\|_1=4m(A)=4(1/4)=1\). On \(A\), \(fg=(4i)(-3i)=12\); outside \(A\), \(fg=0\). Consequently \(T_g(f)=12m(A)=12(1/4)=3\), proving \(\|T_g\|\ge3\). Thus \(\|T_g\|=3\).

Check: The test function has unit \(L^1\) norm and its pairing has absolute value \(3\). This matches the upper bound and the essential supremum.

Keep the Three Choices Separate

When reading or writing a complex \(L^1\) duality formula, check these features explicitly:

Representatives: functions equal a.e. are the same element; changing values on a null set does not change the pairing.
Conjugation: \(f\mapsto\int fg\,d\mu\) and \(f\mapsto\int f\overline g\,d\mu\) are both linear in \(f\), but their representing functions differ by conjugation.
Norm: the norm of the functional is the essential supremum of the representing function, not its pointwise supremum.

These checks prevent a common chain of errors: treating an equivalence class as a pointwise function, using an exceptional value to estimate a norm, or moving conjugation across a pairing without tracking which variable becomes conjugate-linear.

Check Your Understanding

For each question, state which equalities are pointwise and which are only required a.e.

If \(g\) is changed at one point of \([0,1]\), under what condition does this change affect \(\int_0^1 fg\,dm\) for \(f\in L^1\)?
For fixed \(g\), test whether \(f\mapsto\int f\overline g\,d\mu\) is complex-linear or conjugate-linear.
Give a function on \([0,1]\) whose pointwise supremum is larger than its essential supremum, and compute both values.
On a sigma-finite measure space, explain why \(\|T_g\|\ge a\) whenever \(0\le a<\|g\|_\infty\).
If \(\Phi\) is conjugate-linear, explain why \(f\mapsto\overline{\Phi(f)}\) is complex-linear.