Testing Functionals on Vectors
In the previous tutorial, weak convergence of vectors was defined by testing them against every continuous linear functional. There is a related notion for sequences of continuous linear functionals themselves. Instead of testing a functional against every element of the bidual, weak-star convergence tests it at each fixed vector in the original space. This distinction makes weak-star convergence a useful way to study dual spaces while keeping the tests grounded in \(X\).
Let \(X\) be a real normed space, with continuous dual \(X^*\). For each \(x\in X\), evaluation at \(x\) is a continuous linear functional on \(X^*\): it sends \(F\) to \(F(x)\). Indeed, \(|F(x)|\leq\|F\|\|x\|\). These evaluation maps determine the weak-star topology on \(X^*\).
The quantifiers matter: a single vector \(x\) is fixed first, and the real sequence \(F_n(x)\) must converge. The definition does not ask for convergence uniformly over all vectors in the unit ball. That stronger type of control is closely related to convergence in the dual norm.
More explicitly, a basic neighborhood of \(F\) in the weak-star topology is specified by finitely many vectors \(x_1,\ldots,x_m\in X\) and positive tolerances \(\varepsilon_1,\ldots,\varepsilon_m\). It consists of the \(G\in X^*\) satisfying \(|G(x_j)-F(x_j)|<\varepsilon_j\) for every \(j\). Thus weak-star convergence means that every finite collection of these evaluations eventually lies within its prescribed tolerances.
Three Ways Convergence in the Dual Can Differ
The canonical map \(J:X\to X^{**}\), introduced in the tutorial on dual spaces, satisfies \((Jx)(F)=F(x)\). Consequently, weak-star convergence uses only the functionals in \(J(X)\) to test a sequence in \(X^*\). Weak convergence in \(X^*\), by contrast, tests against every member of \(X^{**}\). Therefore weak convergence in \(X^*\) implies weak-star convergence, but the definitions impose different tests. Norm convergence also implies weak-star convergence, as the next proposition records.
Proof. Fix \(x\in X\). The operator-norm inequality gives $$ |F_n(x)-F(x)|=|(F_n-F)(x)|\leq\|F_n-F\|\,\|x\|. $$ The right-hand side tends to zero, so \(F_n(x)\to F(x)\). Since this holds for every \(x\in X\), the definition gives \(F_n\rightharpoonup^*F\). \(\square\)
Worked Example: A Norm-Convergent Sequence of Functionals
Take \(X=C([0,1])\) with the supremum norm. Define \(\delta_t(f)=f(t)\) for \(t\in[0,1]\). Since \(|f(t)|\leq\|f\|_\infty\), each \(\delta_t\) belongs to \(X^*\), and \(\|\delta_t\|=1\): the upper bound is attained by the constant function \(f=1\).
Set \(F_n=\delta_0+\frac{1}{n}\delta_1\) and \(F=\delta_0\). For every \(f\in X\), $$ F_n(f)-F(f)=\frac{f(1)}{n}, \qquad |F_n(f)-F(f)|\leq\frac{\|f\|_\infty}{n}. $$ Taking the supremum over \(\|f\|_\infty\leq1\) gives \(\|F_n-F\|\leq1/n\). Equality holds by taking the constant function \(f=1\), so \(\|F_n-F\|=1/n\to0\). In particular, \(F_n\rightharpoonup^*F\).
Worked Example: Coordinate Functionals on \(\ell^1\)
Let \(X=\ell^1\), the space of real sequences \(x=(x_k)_{k\geq1}\) with \(\|x\|_1=\sum_{k=1}^{\infty}|x_k|<\infty\). Define \(\phi_n(x)=x_n\). The estimate \(|\phi_n(x)|\leq\|x\|_1\) shows that \(\phi_n\in X^*\) and \(\|\phi_n\|\leq1\). For the sequence \(e_n\) with \(n\)th coordinate \(1\) and all other coordinates \(0\), \(\|e_n\|_1=1\) and \(\phi_n(e_n)=1\), so \(\|\phi_n\|=1\).
For any fixed \(x\in\ell^1\), its coordinates satisfy \(x_n\to0\). Otherwise, some \(\varepsilon>0\) would satisfy \(|x_n|\geq\varepsilon\) for infinitely many \(n\), forcing \(\sum_k|x_k|=\infty\). Hence \(\phi_n(x)=x_n\to0\) for every \(x\), and \(\phi_n\rightharpoonup^*0\). Yet \(\|\phi_n-0\|=1\) for all \(n\), so this convergence is not norm convergence.
Worked Example: Moving Evaluation Points
Again let \(X=C([0,1])\), and consider \(\delta_{1/n}\). For each fixed \(f\in X\), continuity at \(0\) gives \(f(1/n)\to f(0)\). Thus \(\delta_{1/n}\rightharpoonup^*\delta_0\). All these functionals have norm \(1\), so their norms do not converge to \(\|\delta_0\|=1\) in a way that distinguishes the limit; instead, examine their distances from it.
For each \(n\), define \(g_n(t)=1-2nt\) on \([0,1/n]\) and \(g_n(t)=-1\) on \([1/n,1]\). This continuous function has \(\|g_n\|_\infty=1\), \(g_n(0)=1\), and \(g_n(1/n)=-1\). Therefore $$ |(\delta_{1/n}-\delta_0)(g_n)|=|-1-1|=2. $$ The triangle inequality gives \(\|\delta_{1/n}-\delta_0\|\leq\|\delta_{1/n}\|+\|\delta_0\|=2\), so in fact the distance equals \(2\) for every \(n\). This is another instance of weak-star convergence without norm convergence.
Uniform Boundedness and the Dual Norm
Weak-star convergence controls each fixed evaluation, but it has an important consequence for the norms of a sequence when \(X\) is complete. This is a direct application of the Uniform Boundedness Principle from earlier in the course. Completeness is the hypothesis that allows that principle to turn pointwise boundedness into a uniform operator-norm bound.
Proof. Regard each \(F_n\) as a bounded linear operator from \(X\) to \(\mathbb{R}\). For every fixed \(x\in X\), the sequence \(F_n(x)\) converges to \(F(x)\), and hence is bounded. Thus the family \((F_n)\) is pointwise bounded on the Banach space \(X\). The Uniform Boundedness Principle gives \(\sup_n\|F_n\|<\infty\), as required. \(\square\)
This conclusion should not be confused with norm convergence: the examples above have uniformly bounded norms but do not converge in norm. Nor should completeness of \(X\) be dropped without justification; it is the domain on which the Uniform Boundedness Principle is being applied.
A related property is that the dual norm cannot be larger at the weak-star limit than the lower limit of the sequence of norms. This is lower semicontinuity: the inequality is one-sided and does not assert convergence of the norms.
Proof. Write \(L=\liminf_{n\to\infty}\|F_n\|\). If \(L=\infty\), the inequality holds immediately. Suppose \(L<\infty\), and fix \(x\in X\). Weak-star convergence implies \(|F_n(x)|\to|F(x)|\), while $$ |F_n(x)|\leq\|F_n\|\,\|x\|. $$ Taking lower limits gives \(|F(x)|\leq L\|x\|\). This holds for every \(x\in X\), so taking the supremum over \(\|x\|\leq1\) yields \(\|F\|\leq L\). \(\square\)
When a Dense Set of Tests Is Enough
The definition asks for convergence at every vector, but sometimes it is convenient to check only a dense subset. Pointwise convergence on a dense set alone is not enough: values could behave badly away from that set. A uniform bound on the norms prevents this, because it controls how much each functional can change when its input is approximated.
Proof. Fix \(x\in X\) and \(\varepsilon>0\). Since \(D\) is dense, choose \(d\in D\) such that $$ \|x-d\|<\frac{\varepsilon}{2(C+\|F\|+1)}. $$ For every \(n\), linearity and the operator-norm bound give $$ |F_n(x)-F(x)| \leq |F_n(d)-F(d)|+|F_n(x-d)|+|F(x-d)| \leq |F_n(d)-F(d)|+(C+\|F\|)\|x-d\|. $$ The second term is less than \(\varepsilon/2\), by the choice of \(d\). Since \(F_n(d)\to F(d)\), there is an index after which the first term is less than \(\varepsilon/2\). Hence \(|F_n(x)-F(x)|<\varepsilon\) for all sufficiently large \(n\). Because \(x\) was arbitrary, \(F_n\rightharpoonup^*F\). \(\square\)
The theorem is useful when the domain has a manageable dense set and the norms are already uniformly controlled. The dense-set hypothesis handles approximation of inputs; the uniform bound controls the approximation error independently of \(n\). For a weak-star convergent sequence on a Banach space, the preceding boundedness theorem supplies that uniform bound automatically.
Interpreting Weak-Star Convergence Correctly
Weak-star convergence is pointwise convergence of functionals on the original space. It is weaker than norm convergence because it does not control the supremum of \(|F_n(x)-F(x)|\) over the unit ball. The coordinate functionals on \(\ell^1\) and the moving evaluations on \(C([0,1])\) demonstrate this gap directly.
A common pitfall is to check convergence on a convenient collection of vectors and silently treat it as convergence everywhere. The dense-test criterion explains exactly what extra ingredient makes that strategy valid: density together with a uniform bound on the norms. Another useful check is lower semicontinuity: any proposed weak-star limit must satisfy \(\|F\|\leq\liminf_n\|F_n\|\). These facts provide control without turning weak-star convergence into norm convergence.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- Which vectors are used to test weak-star convergence of a sequence in \(X^*\)?
- Why does weak convergence in \(X^*\) imply weak-star convergence?
- Why do the coordinate functionals on \(\ell^1\) converge weak-star to zero while remaining at norm \(1\)?
- Where is completeness used in the proof that a weak-star convergent sequence is norm bounded?
- State the lower-semicontinuity inequality for the dual norm.
- What two hypotheses allow convergence on a dense subset to imply weak-star convergence everywhere?