Functionals Tested by Vectors
Compactness of a dual unit ball is not usually visible from the norm: the ball may be infinite-dimensional and need not be norm compact. The weak-star topology provides a different way to compare functionals. It records the values of each functional on vectors in the original space, and Banach–Alaoglu says that these tests make the dual unit ball compact.
Let \((X,\mathcal{M},\mu)\) be any measure space, and let \(1\le p\le\infty\). The space \(L^p(\mu)\), over either the real or complex scalars, consists of equivalence classes of measurable functions with finite \(L^p\) norm; functions equal \(\mu\)-a.e. represent the same element. It is a Banach space. The result below applies to any Banach space \(X\), so we will state it in that form and then take \(X=L^p(\mu)\). Write \(X^*\) for the space of bounded linear functionals on \(X\), and \(X^{**}=(X^*)^*\) for its bidual.
In the case \(X=L^p(\mu)\), the dual ball is \(B_{(L^p)^*}=\{\ell\in(L^p)^*:\|\ell\|\le1\}\). If a duality theorem identifies a functional with integration against a function \(g\), the evaluation can be written as an integral. For example, on \(([0,1],\mathcal{L},m)\), the earlier duality result for \(L^1\) identifies \((L^1)^*\) with \(L^\infty\), with \(\ell_g(f)=\int_0^1 fg\,dm\) and \(\|\ell_g\|=\|g\|_\infty\). The weak-star tests in this case are \(L^1\) functions.
Banach–Alaoglu Compactness
Proof. For each \(x\in X\), let \(D_x=\{z:|z|\le\|x\|\}\) in the scalar field, and form the product \(P=\prod_{x\in X}D_x\), with its product topology. Each disk is compact, including the singleton disk \(D_0=\{0\}\). By the Tychonoff theorem, the product \(P\) is compact. Tychonoff is the standard product-compactness theorem; its proof is not needed here.
A point \(a\in P\) assigns a scalar \(a_x\) to each \(x\in X\). Consider the subset \(K\subseteq P\) consisting of points satisfying, for all \(x,y\in X\) and all scalars \(\alpha\), the equations \(a_{x+y}=a_x+a_y\) and \(a_{\alpha x}=\alpha a_x\). Each equation specifies a closed subset of \(P\): it is the inverse image of \(\{0\}\) under a continuous expression in finitely many coordinates. Thus \(K\), their intersection, is closed in \(P\). A closed subset of a compact space is compact.
For any \(a\in K\), define \(\ell_a(x)=a_x\). The two equations give linearity, and membership in \(P\) gives \(|\ell_a(x)|\le\|x\|\) for every \(x\). Hence \(\ell_a\) is bounded and \(\|\ell_a\|\le1\). Conversely, if \(\ell\in B_{X^*}\), then \(|\ell(x)|\le\|x\|\) for every \(x\), and its coordinates satisfy both equations, so \((\ell(x))_{x\in X}\in K\). Therefore \(K\) is exactly the image of \(B_{X^*}\) under the map \(\ell\mapsto(\ell(x))_{x\in X}\). The product topology on this image is precisely pointwise convergence on \(X\), which is the weak-star topology by definition. Thus the dual unit ball is weak-star compact. If \(X=\{0\}\), the product and the dual ball are both singletons, and the conclusion holds as well. \(\square\)
The proof explains the mechanism: the norm bound puts each evaluation in a compact scalar disk, and linearity is expressed by closed coordinate conditions. Compactness here is topological compactness, not compactness in the dual norm.
Worked Example: The Dual Ball of a Two-Dimensional Space
Take \(X=\mathbb{R}^2\) with \(\|(x_1,x_2)\|_\infty=\max(|x_1|,|x_2|)\). Every linear functional has the form \(\ell_a(x)=a_1x_1+a_2x_2\). Its norm is \(|a_1|+|a_2|\): the triangle inequality gives \(|\ell_a(x)|\le(|a_1|+|a_2|)\|x\|_\infty\); choosing \(x_i=1\) when \(a_i\ge0\), and \(x_i=-1\) when \(a_i<0\), attains that bound. Thus \(B_{X^*}\) is the diamond \(|a_1|+|a_2|\le1\).
Evaluation at \(e_1=(1,0)\) returns \(a_1\), and evaluation at \(e_2=(0,1)\) returns \(a_2\). Every other evaluation is \(a_1x_1+a_2x_2\), a continuous function of these same two coordinates. Consequently the weak-star topology here is the ordinary coordinate topology on the diamond. The diamond is closed and bounded in \(\mathbb{R}^2\), hence compact by Heine–Borel, in agreement with Banach–Alaoglu. As a check, \(a=(3/5,2/5)\) has norm \(3/5+2/5=1\) and lies on its boundary; \(a=(3/5,1/2)\) has norm \(3/5+1/2=11/10>1\) and lies outside.
Worked Example: A Weak-Star Neighborhood in the Dual of L¹
Work on \(([0,1],\mathcal{L},m)\), with real scalars. Let \(g_0(x)=1/2\), so \(\|g_0\|_\infty=1/2\), and let \(\ell_{g_0}(f)=\int_0^1 f g_0\,dm\). Test using \(f_1=\chi_{[0,1/2]}\) and \(f_2=\chi_{(1/2,1]}\), both in \(L^1\). Since each interval has measure \(1/2\), the evaluations at the center are \(\ell_{g_0}(f_1)=(1/2)(1/2)=1/4\) and \(\ell_{g_0}(f_2)=(1/2)(1/2)=1/4\).
Consider the weak-star neighborhood of \(\ell_{g_0}\) requiring both evaluations to differ from these values by less than \(1/8\). The function \(g=\chi_{[0,1/2]}\) defines a functional in the dual unit ball, since \(\|g\|_\infty=1\). But \(\ell_g(f_1)=1/2\) and \(\ell_g(f_2)=0\), so the absolute differences are \(|1/2-1/4|=1/4\) and \(|0-1/4|=1/4\). Both exceed \(1/8\), and \(\ell_g\) is outside that neighborhood. This computes membership using strict inequalities as required by the neighborhood definition.
In general, a basic neighborhood uses any finite list of \(L^1\) tests, not just indicators of these two halves. This example shows exactly what the chosen tests constrain; they do not determine all evaluations by themselves.
Weak Topology and Weak-Star Topology
The weak topology on \(X^*\) is \(\sigma(X^*,X^{**})\): it is the smallest topology making every map \(\ell\mapsto F(\ell)\), for \(F\in X^{**}\), continuous. Each \(x\in X\) defines a member \(Jx\in X^{**}\) by \(Jx(\ell)=\ell(x)\). Thus every weak-star test is a weak test, and the weak-star topology is no stronger than the weak topology. The two topologies need not agree.
Proof. If \(J\) is onto, every \(F\in X^{**}\) is \(Jx\) for some \(x\in X\). Hence the tests defining the weak topology are exactly the weak-star evaluations, and the topologies agree.
Conversely, suppose the topologies agree and take any \(F\in X^{**}\). The map \(F\) is continuous for the weak topology, hence for the weak-star topology. We show that any weak-star continuous linear functional \(G:X^*\to\mathbb{K}\) equals \(Jx\) for some \(x\in X\). Continuity at zero gives finitely many \(x_1,\ldots,x_k\in X\) and \(\varepsilon>0\) such that \(|G(\ell)|<1\) whenever \(|\ell(x_i)|<\varepsilon\) for all \(i\). If \(\ell(x_i)=0\) for every \(i\), then every scalar multiple \(t\ell\) satisfies those inequalities, so \(|tG(\ell)|<1\) for all scalars \(t\). This forces \(G(\ell)=0\). Therefore \(G\) vanishes on the kernel of \(A(\ell)=(\ell(x_1),\ldots,\ell(x_k))\). It factors through the finite-dimensional range of \(A\), and a linear functional on that range extends to \(\mathbb{K}^k\). Consequently \(G(\ell)=\sum_{i=1}^k c_i\ell(x_i)=\ell(\sum_{i=1}^k c_ix_i)=Jx(\ell)\), where \(x=\sum_i c_ix_i\). Applying this to \(G=F\) shows \(F\in J(X)\). Since \(F\) was arbitrary, \(J\) is onto. \(\square\)
For \(X=L^p(\mu)\) with \(1<p<\infty\), reflexivity established earlier in the course therefore makes the weak and weak-star topologies on \((L^p)^*\) agree. At an endpoint, the conclusion cannot be inferred from Banach–Alaoglu: the theorem asserts weak-star compactness, not weak compactness. The tests defining the two topologies must be kept distinct unless reflexivity has been established.
Worked Example: Oscillation in the Dual of L¹
On \(([0,1],\mathcal{L},m)\), define \(r_n(x)=(-1)^{\lfloor 2^n x\rfloor}\) for \(0\le x<1\), and set \(r_n(1)=1\). Each \(r_n\) is measurable, takes only the values \(1\) and \(-1\), and has \(\|r_n\|_\infty=1\). Through \((L^1)^*=L^\infty\), it defines a member \(\ell_{r_n}\) of the dual unit ball.
We verify weak-star convergence to zero, meaning \(\int_0^1 f r_n\,dm\to0\) for every \(f\in L^1\). First let \(h\) be constant on each interval of a dyadic partition of level \(k\), whose intervals have length \(2^{-k}\). If \(n>k\), each interval of level \(k\) contains \(2^{n-k}\) intervals of level \(n\). Since \(2^{n-k}\) is even and the signs of \(r_n\) alternate on those smaller intervals, the integral of \(r_n\) over each level-\(k\) interval is zero. As \(h\) is constant on each such interval, \(\int_0^1 h r_n\,dm=0\).
Dyadic step functions are dense in \(L^1([0,1])\), the standard density result for Lebesgue \(L^1\). Given \(f\in L^1\) and \(\varepsilon>0\), choose such an \(h\) with \(\|f-h\|_1<\varepsilon\), and let \(k\) be a partition level on which \(h\) is constant. For every \(n>k\), using \(|r_n|=1\) everywhere, \[ \left|\int_0^1 f r_n\,dm\right| =\left|\int_0^1 (f-h)r_n\,dm\right| \le\int_0^1|f-h|\,dm <\varepsilon. \] Thus \(\ell_{r_n}\to0\) in the weak-star topology. Yet \(\|r_n\|_\infty=1\) for every \(n\), so there is no convergence to zero in the \(L^\infty\) norm. This is a check that the topology is determined by evaluations against \(L^1\) functions, not by norm distance.
A Common Testing Error
A frequent mistake is to check only one evaluation and conclude that a functional is determined, or converges, in the weak-star topology. For a concrete correction on \(([0,1],\mathcal{L},m)\), let \(g=\chi_{[0,1/2]}-\chi_{(1/2,1]}\). Then \(\|g\|_\infty=1\), and testing only against the constant function \(1\) gives \(\int_0^1 g\,dm=1/2-1/2=0\). But for \(f=\chi_{[0,1/2]}\), the evaluation is \(\int_0^1 fg\,dm=\int_{[0,1/2]}1\,dm=1/2\), not zero. Agreement on the constant test does not mean agreement on all \(L^1\) tests. The definition of weak-star convergence requires convergence for every test \(f\in L^p(\mu)\), not merely for a convenient finite selection.
Check Your Understanding
Use the definitions of the two topologies and the compactness proof to answer these questions.
- For a basic weak-star neighborhood in \((L^p)^*\), what data specify the neighborhood, and what is being evaluated?
- In the product proof of Banach–Alaoglu, which coordinate conditions ensure that a point of the compact subset represents a linear functional of norm at most one?
- Why does weak convergence in \(X^*\) imply weak-star convergence, but not necessarily conversely?
- For \(X=L^p(\mu)\) with \(1<p<\infty\), why do the weak and weak-star topologies on \(X^*\) coincide?
- Why does \(\int_0^1 g\,dm=0\) alone fail to establish that the functional defined by \(g\) is zero?