Weak-Star Convergence in LInfinity

The Pairing That Defines Weak-Star Convergence

In a dual space, weak-star convergence is determined by testing against elements of the predual. For \(L^\infty\), the relevant tests are integrable functions: a sequence can oscillate without settling pointwise or in norm, while its integral against each fixed \(L^1\) function tends to zero. The key is to distinguish these tests from all bounded linear functionals on \(L^\infty\), which define the ordinary weak topology.

We work with real-valued functions and the measure space \(([0,1],\mathcal{L},m)\), where \(m\) is Lebesgue measure. This space is finite, hence sigma-finite and complete. As usual, \(L^1(m)\) and \(L^\infty(m)\) consist of equivalence classes modulo equality \(m\)-almost everywhere; their norms are \(\|g\|_1=\int_0^1|g|\,dm\) and \(\|f\|_\infty=\operatorname*{ess\,sup}_{x\in[0,1]}|f(x)|\). The integral pairing is \(\langle f,g\rangle=\int_0^1fg\,dm\). It is well-defined and satisfies \(|\langle f,g\rangle|\le\|f\|_\infty\|g\|_1\).

The standard duality theorem, under the sigma-finiteness hypothesis, identifies the continuous dual of \(L^1(m)\) isometrically with \(L^\infty(m)\) by this pairing. The sigma-finiteness qualification matters: on an arbitrary measure space, \(L^\infty\) need not identify with \((L^1)^*\) through the integral pairing. Here the hypothesis holds, so \(L^\infty(m)\) is the dual of \(L^1(m)\).

The ordinary weak topology on \(L^\infty(m)\), denoted \(\sigma(L^\infty(m),(L^\infty(m))^*)\), tests against every bounded linear functional on \(L^\infty(m)\). Each \(g\in L^1(m)\) gives one such functional by integration, but these are not all the functionals in \((L^\infty(m))^*\). Consequently, weak convergence implies weak-star convergence, while the converse need not hold.

Testing Convergence by Integrals

Proof. By the definition of the topology, each map \(T_g(f)=\int fg\,dm\) is continuous. Thus convergence in the topology implies \(T_g(f_n)\to T_g(f)\) for every \(g\in L^1(m)\). Conversely, basic neighborhoods of \(f\) in this topology are finite intersections of sets of the form \(\{h:|T_{g_j}(h)-T_{g_j}(f)|<\varepsilon\}\), where \(g_1,\dots,g_k\in L^1(m)\) and \(\varepsilon>0\). If all the scalar integrals converge, then for this finite list each difference is less than \(\varepsilon\) for all sufficiently large \(n\); taking the largest of the finitely many required indices puts \(f_n\) in the neighborhood. This is convergence in the topology. \(\square\)

A Sequence Converging Weak-Star to Zero

We now verify all the required tests for the same sequence. A dyadic step function of level \(m\) is a function constant on each interval \([j2^{-m},(j+1)2^{-m})\), for \(j=0,\dots,2^m-1\). If \(n>m\), then on each level-\(m\) interval, \(r_n\) has equally many positive and negative subintervals of length \(2^{-n}\). Hence its integral over that interval is zero.

Proof. Simple functions are dense in \(L^1(m)\): for any \(g\in L^1(m)\) and any \(\varepsilon>0\), there is a simple function \(s=\sum_{j=1}^k a_j\chi_{E_j}\) with \(\|g-s\|_1<\varepsilon/2\). By regularity of Lebesgue measure on \([0,1]\), each measurable \(E_j\) can be approximated in measure by a finite union \(A_j\) of intervals. Choose these approximations so that \(\sum_{j=1}^k|a_j|m(E_j\triangle A_j)<\varepsilon/4\). The endpoints of the finitely many intervals can then be approximated by dyadic rationals, producing finite unions \(D_j\) of dyadic intervals for which \(\sum_{j=1}^k|a_j|m(A_j\triangle D_j)<\varepsilon/4\). A common sufficiently fine dyadic partition makes every \(D_j\) a union of cells of that partition. The corresponding dyadic step function \(h=\sum_{j=1}^k a_j\chi_{D_j}\) satisfies \(\|s-h\|_1\le\sum_j|a_j|m(E_j\triangle D_j)<\varepsilon/2\). The triangle inequality gives \(\|g-h\|_1<\varepsilon\), proving density. \(\square\)

Proof. Fix \(g\in L^1(m)\) and \(\varepsilon>0\). By the lemma, choose a dyadic step function \(h\) with \(\|g-h\|_1<\varepsilon\). Let its level be \(m\). For every \(n>m\), the integral of \(r_n\) over each level-\(m\) cell is zero; since \(h\) is constant on each such cell, \(\int_0^1r_nh\,dm=0\). Also \(|r_n|=1\) almost everywhere, so

For this fixed \(g\), the estimate holds for every \(n>m\). As \(\varepsilon\) was arbitrary, \(\int r_ng\,dm\to0\). It holds for every \(g\in L^1(m)\), so the convergence criterion proves \(r_n\to0\) weak-star. \(\square\)

Why the Same Sequence Is Not Weakly Convergent

To show that weak-star convergence is genuinely weaker here, we construct a bounded linear functional on \(L^\infty(m)\) that detects the sequence. For every finite linear combination of the \(r_n\), the signs of the first \(N\) functions take all \(2^N\) possible sign patterns on the level-\(N\) dyadic intervals. Therefore, for real coefficients \(a_1,\dots,a_N\),

Indeed, the triangle inequality gives the upper bound. For the reverse bound, choose a sign pattern with the sign of \(a_k\) in coordinate \(k\) whenever \(a_k\ne0\); the corresponding dyadic interval has positive measure, and on it the sum equals \(\sum_k|a_k|\).

A Common Misidentification

A common mistake is to treat weak-star convergence in \(L^\infty\) as though it tested against every element of \((L^\infty)^*\). That is the definition of weak convergence, not weak-star convergence. The distinction matters precisely because the dual of \(L^\infty\) can contain functionals not represented by integration against \(L^1\) functions. In the example, every \(L^1\) test integral tends to zero, while the Hahn–Banach extension \(F\) has \(F(r_n)=1\) for every \(n\).

Another important qualification is the measure-space hypothesis behind the duality notation. On a sigma-finite measure space, the integral pairing identifies \((L^1(\mu))^*\) with \(L^\infty(\mu)\). Without sigma-finiteness this identification may fail, so one should not assert it for an arbitrary measure space. Regardless of that issue, the convergence test stated here is on the specified finite, complete Lebesgue space \(([0,1],\mathcal{L},m)\), where the duality theorem applies.