Worked Dual Representations on L1

From a Formula for a Functional to Its Representing Function

A bounded linear functional on \(L^1\) can often be recognized by looking at the coefficient or density multiplying the input. The key distinction is between the functional’s action on \(f\) and its operator norm: the action is an integral pairing, while the norm is determined by the essential size of the representing function. We will use explicit examples to make both computations visible.

We use the complex-linear convention throughout: the functional is written \(T(f)=\int_X f g\,d\mu\), with no conjugate on \(f\) in the integrand. Here \(L^1(\mu)\) denotes the space of measurable complex-valued functions with finite norm \(\|f\|_1=\int_X|f|\,d\mu\), with functions equal \(\mu\)-a.e. identified. The space \(L^\infty(\mu)\) consists of measurable complex-valued functions that are bounded outside a \(\mu\)-null set; \(\|g\|_\infty\) is the least such almost-everywhere bound.

Starting point (the \(L^1\)-dual representation theorem): On a sigma-finite measure space \((X,\mathcal{M},\mu)\), every bounded complex-linear functional \(T\) on \(L^1(\mu)\) has a representing function \(g\in L^\infty(\mu)\) such that \(T(f)=\int_X f g\,d\mu\) for all \(f\in L^1(\mu)\). The representing function is unique up to equality \(\mu\)-a.e. This standard theorem was established in the earlier material on the \(L^1\) dual; here we identify the function and compute the norm in concrete cases.

Theorem (norm of an \(L^1\) pairing): Let \((X,\mathcal{M},\mu)\) be sigma-finite, let \(g\in L^\infty(\mu)\) be complex-valued, and define the complex-linear functional \(T(f)=\int_X f g\,d\mu\) on \(L^1(\mu)\). Then \(T\) is bounded and \(\|T\|=\|g\|_\infty\), where \(\|T\|=\sup_{\|f\|_1\le1}|T(f)|\).

Proof. Put \(M=\|g\|_\infty\). Since \(|g|\le M\) \(\mu\)-a.e., for every \(f\in L^1(\mu)\) the integral is absolutely convergent and \(|T(f)|\le\int_X|f||g|\,d\mu\le M\|f\|_1\). Thus \(T\) is bounded and \(\|T\|\le M\). If \(M=0\), this bound gives \(T=0\) and the equality follows. Suppose \(M>0\), and fix \(c\) with \(0\le c<M\). By the definition of essential supremum, \(E=\{x:|g(x)|>c\}\) has positive measure. Sigma-finiteness gives a countable cover of \(X\) by finite-measure sets, so some such set \(F\) satisfies \(0<\mu(E\cap F)<\infty\). Write \(A=E\cap F\), and define \(f=\chi_A\overline{g}/(|g|\mu(A))\) on \(A\), and \(f=0\) elsewhere. The denominator \(|g|\) is nonzero on \(A\). Therefore \(\|f\|_1=1\), while \(T(f)=\mu(A)^{-1}\int_A|g|\,d\mu>c\). It follows that \(\|T\|\ge c\) for every \(c<M\), hence \(\|T\|\ge M\). Combined with the upper bound, \(\|T\|=M\). \(\square\)

The lower-bound argument explains why the essential supremum, rather than a pointwise maximum, governs the norm. It finds a positive-measure set where the density is nearly as large as its essential supremum, then chooses the phase of \(f\) so that the product \(fg\) is nonnegative real-valued on that set.

Counting Measure: A Representing Sequence

Worked Example: A Functional on \(\ell^1\)

Let \(X=\mathbb{N}=\{1,2,\ldots\}\) with the power-set sigma-algebra and counting measure \(\mu\). This measure space is sigma-finite. Define \(g_n=1+(-1)^n/2\), and for \(x=(x_n)\in\ell^1\) set \(T(x)=\sum_{n=1}^{\infty}x_ng_n\). Find the representing function and the operator norm, then evaluate \(T\) on a finite-support sequence.

For odd \(n\), \((-1)^n=-1\), so \(g_n=1-\tfrac12=\tfrac12\). For even \(n\), \((-1)^n=1\), so \(g_n=1+\tfrac12=\tfrac32\). Thus \(g=(g_n)\) is bounded, and the counting-measure integral gives \(T(x)=\int_{\mathbb{N}}xg\,d\mu=\sum_{n=1}^{\infty}x_ng_n\). The largest coefficient magnitude is \(\tfrac32\), so the theorem gives \(\|T\|=\tfrac32\). To check the lower bound directly, let \(e_2\) be \(1\) at \(n=2\) and \(0\) elsewhere. Then \(\|e_2\|_1=1\) and \(T(e_2)=g_2=\tfrac32\).

For the requested evaluation, take \(x_1=\tfrac12,\ x_2=-\tfrac14,\ x_3=\tfrac14\), and \(x_n=0\) for \(n\ge4\). Its norm is \(\|x\|_1=\tfrac12+\tfrac14+\tfrac14=1\). The functional value is \(T(x)=\tfrac12\cdot\tfrac12+(-\tfrac14)\cdot\tfrac32+\tfrac14\cdot\tfrac12=\tfrac14-\tfrac38+\tfrac18=0\). Check: the three terms sum to \(\tfrac{2}{8}-\tfrac{3}{8}+\tfrac{1}{8}=0\), as computed.

Lebesgue Measure on a Finite Interval

Worked Example: A Piecewise Constant Density on [0,1]

Take \(X=[0,1]\), the Lebesgue sigma-algebra, and Lebesgue measure \(m\); this is a finite, hence sigma-finite, measure space. Define \(g(x)=3\) for \(0\le x<\tfrac13\), and \(g(x)=-2i\) for \(\tfrac13\le x\le1\). For \(T(f)=\int_0^1 f(x)g(x)\,dx\), determine the norm and verify it with an input of \(L^1\)-norm one.

The density is measurable and complex-valued. Its magnitude is \(3\) on \([0,\tfrac13)\) and \(|-2i|=2\) on \([\tfrac13,1]\), so \(\|g\|_\infty=3\). Consequently the theorem gives \(\|T\|=3\). Set \(f(x)=3\) on \([0,\tfrac13)\) and \(f(x)=0\) elsewhere. Then \(\|f\|_1=3\cdot\tfrac13=1\), and \(T(f)=\int_0^{1/3}3\cdot3\,dx=9\cdot\tfrac13=3\). Thus this input attains the upper bound. Check: the product \(fg\) is \(9\) on an interval of measure \(\tfrac13\) and zero elsewhere, so its integral is \(9\cdot\tfrac13=3\).

Combining Atoms and a Continuous Part

A single measure space may contain both atoms and a continuous part. Let \(X=\{a,b\}\sqcup[0,1]\), where the union is disjoint, and give it the sigma-algebra consisting of arbitrary choices of the two atoms together with Lebesgue-measurable subsets of the interval. Define the finite measure \(\mu=2\delta_a+3\delta_b+m|_{[0,1]}\). Thus \(\mu(\{a\})=2\), \(\mu(\{b\})=3\), and the interval has Lebesgue measure one.

Proposition (essential supremum on this mixed space): For a measurable complex-valued \(g\) on \(X\), \(\|g\|_\infty=\max\{|g(a)|,|g(b)|,\operatorname*{ess\,sup}_{x\in[0,1]}|g(x)|\}\). Here the essential supremum on the interval is taken with respect to Lebesgue measure.

Proof. An almost-everywhere bound \(C\) for \(|g|\) on \(X\) must hold at \(a\): otherwise the exceptional set contains \(\{a\}\), which has measure \(2>0\). It must also hold at \(b\), since \(\mu(\{b\})=3>0\). On the interval, it must hold Lebesgue-a.e., because the restriction of \(\mu\) there is Lebesgue measure. Therefore every almost-everywhere bound on \(X\) is at least each of the three quantities in the maximum. Conversely, any number at least their maximum bounds \(|g(a)|\), \(|g(b)|\), and \(|g(x)|\) for Lebesgue-a.e. \(x\in[0,1]\). The exceptional interval set has \(\mu\)-measure zero, and neither atom is exceptional, so this is an almost-everywhere bound on \(X\). Taking the least such bound proves the formula. \(\square\)

Worked Example: Computing a Mixed-Space Functional

On the mixed measure space just defined, let \(g(a)=4,\ g(b)=-2i\), and \(g(x)=1+x\) for \(x\in[0,1]\). Define \(T(f)=\int_X fg\,d\mu\). Compute its norm and evaluate it for \(f(a)=\tfrac12,\ f(b)=\tfrac{i}{3}\), and \(f(x)=1\) on \([0,\tfrac12]\), with \(f(x)=0\) on \((\tfrac12,1]\).

At the atoms the magnitudes are \(|g(a)|=4\) and \(|g(b)|=|-2i|=2\). On the interval \(g(x)=1+x\), whose essential supremum on \([0,1]\) is \(2\): it is at most \(2\), and for every \(c<2\) the set where \(1+x>c\) has positive measure. The proposition gives \(\|g\|_\infty=\max\{4,2,2\}=4\), hence \(\|T\|=4\).

The input norm is \(\|f\|_1=2|\tfrac12|+3|\tfrac{i}{3}|+m([0,\tfrac12])=1+1+\tfrac12=\tfrac52\). Its functional value has separate atomic and continuous contributions: \[ T(f)=2\left(\tfrac12\cdot4\right)+3\left(\tfrac{i}{3}\cdot(-2i)\right)+\int_0^{1/2}(1+x)\,dx. \] The first term is \(2\cdot2=4\). Since \(i(-2i)=2\), the second term is \(3\cdot\tfrac23=2\). The interval term is \(\int_0^{1/2}1\,dx+\int_0^{1/2}x\,dx=\tfrac12+\tfrac18=\tfrac58\). Therefore \(T(f)=4+2+\tfrac58=\tfrac{53}{8}\). Check: \(|T(f)|=\tfrac{53}{8}<10=4\cdot\tfrac52=\|g\|_\infty\|f\|_1\), in agreement with the pairing bound.

A Common Norm Calculation Error

For a functional on \(\ell^1\), the operator norm is not the sum of the absolute values of the representing coefficients. For example, on \(\mathbb{N}\) with counting measure, take \(g_n=1/n\). The sum \(\sum_{n=1}^{\infty}|g_n|=\sum_{n=1}^{\infty}1/n\) diverges, so using that sum as the operator norm would incorrectly claim the functional is unbounded. The correction is to use the essential supremum for counting measure, which is \(\sup_n|g_n|=1\). Indeed, \(|\sum_n x_n/n|\le\sum_n|x_n|=\|x\|_1\), while \(T(e_1)=1\) and \(\|e_1\|_1=1\). Thus the exact operator norm is \(1\).

Likewise, on a Lebesgue space the pointwise supremum can be larger than the essential supremum. If \(g=\chi_{\{1/2\}}\) on \([0,1]\), then its pointwise supremum is \(1\), but \(g=0\) Lebesgue-a.e. and \(T(f)=\int_0^1fg\,dx=0\) for every \(f\in L^1\). Its operator norm is therefore \(0\), not \(1\). In each setting, identify the measure and use the corresponding essential bound; then the pairing theorem gives the operator norm.

Check Your Understanding

For each question, specify the measure space and the representing function before computing the norm.

On \(\mathbb{N}\) with counting measure, let \(T(x)=\sum_{n\ge1}x_n(2+(-1)^n)\). Find its representing sequence and operator norm.
On \([0,1]\) with Lebesgue measure, let \(T(f)=\int_0^1 f(x)(1-2x)\,dx\). Find \(\|T\|\) and exhibit a sequence of inputs of norm one whose functional values approach that norm.
On \(X=\{a\}\sqcup[0,1]\) with \(\mu=5\delta_a+m|_{[0,1]}\), suppose \(g(a)=2i\) and \(g(x)=x\). What is the operator norm of \(f\mapsto\int_Xfg\,d\mu\)?
For \(g_n=1/n\) on counting-measure \(\mathbb{N}\), explain why the divergence of \(\sum_n|g_n|\) does not prevent the associated functional from being bounded.
On Lebesgue \([0,1]\), compare the operator norms associated with \(g=\chi_{\{1/2\}}\) and \(g=1\).