Weak Convergence Controls the Size of a Sequence
Weak convergence tests vectors using bounded linear functionals rather than the norm. At first, this seems to give only separate information: for each functional, its values on the sequence converge. The important conclusion is stronger. In a Banach space, those separate bounds combine into one bound on the norms of all the sequence elements. The mechanism is the uniform boundedness principle, whose completeness hypothesis is essential.
Let \(X\) be a real or complex Banach space, with dual space \(X^*\), the space of bounded linear functionals on \(X\). No measure space is needed for the abstract result. In particular, it applies when \(X=L^p(\mu)\) for a measure space \((\Omega,\mathcal{F},\mu)\) and \(1\le p\le\infty\), since these are Banach spaces with their usual norms.
The key step is to regard each vector \(x_n\) as a functional on \(X^*\). The resulting family of functionals is bounded at each fixed point of \(X^*\), because a convergent scalar sequence is bounded. To turn those individual bounds into a common operator-norm bound, we use uniform boundedness.
The Uniform Boundedness Principle
We prove the principle using the Baire category theorem. Recall that a complete metric space cannot be a countable union of closed sets all having empty interior. Here is the needed argument. If closed sets \(F_1,F_2,\ldots\) all had empty interior, choose a closed ball of positive radius avoiding \(F_1\). Inside its interior choose a closed ball of positive radius avoiding \(F_2\), and continue, making the \(n\)-th radius at most \(2^{-n}\). Each choice is possible because the complement of a closed set with empty interior is open and dense. The balls are nested and their radii tend to zero. Their centers form a Cauchy sequence, so completeness gives a limit lying in every ball. That limit avoids every \(F_n\), contradicting that the sets cover the space. (In a one-point space the conclusion is immediate.)
Proof. If \(E=\{0\}\), every \(T\) is zero and the conclusion holds. Otherwise, for each positive integer \(m\), define \(E_m=\{u\in E:\sup_{T\in\mathcal{T}}\|Tu\|_Y\le m\}\). Each \(E_m\) is closed: if \(u_j\to u\) with \(u_j\in E_m\), then for each fixed \(T\), continuity gives \(\|Tu\|_Y=\lim_j\|Tu_j\|_Y\le m\), so the supremum over \(T\) is at most \(m\). Pointwise boundedness says \(E=\bigcup_{m=1}^\infty E_m\). By the Baire category theorem, some \(E_m\) has nonempty interior. Thus there are \(u_0\in E\) and \(r>0\) such that the open ball \(B(u_0,r)\) is contained in \(E_m\); in particular \(u_0\in E_m\).
For any \(h\in E\) with \(\|h\|_E<r\), both \(u_0+h\) and \(u_0\) belong to \(E_m\). For every \(T\in\mathcal{T}\), linearity and the triangle inequality therefore give \(\|Th\|_Y\le\|T(u_0+h)\|_Y+\|Tu_0\|_Y\le2m\). If \(\|v\|_E=1\), take \(h=(r/2)v\). Then \(\|h\|_E<r\), and homogeneity gives \(\|Tv\|_Y\le4m/r\). This bound is independent of both \(T\) and \(v\), so \(\sup_{T\in\mathcal{T}}\|T\|\le4m/r<\infty\). \(\square\)
Applying the Principle to a Weakly Convergent Sequence
Proof. For each \(n\), define the evaluation functional \(T_n:X^*\to\mathbb{R}\) in the real case, or \(T_n:X^*\to\mathbb{C}\) in the complex case, by \(T_n(x^*)=x^*(x_n)\). It is linear and bounded, since \(|T_n(x^*)|\le\|x^*\|\,\|x_n\|_X\). The dual \(X^*\) is Banach: a Cauchy sequence in its norm has a pointwise limit that is linear and bounded, and the Cauchy estimate in operator norm shows convergence to that limit in the dual norm.
For every fixed \(x^*\in X^*\), weak convergence gives \(T_n(x^*)=x^*(x_n)\to x^*(x)\), so \(\sup_n|T_n(x^*)|<\infty\). Apply the Uniform Boundedness Principle with domain \(E=X^*\), target the scalar field, and family \((T_n)\). It follows that \(\sup_n\|T_n\|<\infty\).
For any \(y\in X\), the norm of its evaluation functional on \(X^*\) is \(\sup_{\|x^*\|\le1}|x^*(y)|=\|y\|_X\). The upper bound follows from \(|x^*(y)|\le\|x^*\|\|y\|_X\); equality follows from the Hahn–Banach theorem (and is immediate for \(y=0\)). Consequently \(\|T_n\|=\|x_n\|_X\) for each \(n\). The uniform bound on \(\|T_n\|\) is exactly the claimed bound on \(\|x_n\|_X\). \(\square\)
Worked Example: A Weakly Null Sequence in \(\ell^2\)
In the Banach space \(\ell^2\) with counting measure on \(\mathbb{N}\), let \(e_j\) denote the sequence equal to \(1\) at coordinate \(j\) and \(0\) elsewhere. Set \(x_n=e_{n+1}+\frac1n e_1\). We verify directly that \(x_n\rightharpoonup0\), although the norms do not tend to zero.
Every bounded linear functional on \(\ell^2\) has the form \(x\mapsto\sum_{j=1}^\infty x_j\overline{y_j}\) for some \(y\in\ell^2\). For such a functional, \(x^*(x_n)=\overline{y_{n+1}}+\frac1n\overline{y_1}\). Since \(\sum_j|y_j|^2<\infty\), its terms satisfy \(y_j\to0\): otherwise some positive threshold would be exceeded at infinitely many indices, forcing the sum of squares to diverge. Also \(y_1/n\to0\). Thus \(x^*(x_n)\to0\) for every \(x^*\), proving weak convergence.
The two nonzero coordinates of \(x_n\) are distinct, so \(\|x_n\|_2^2=1+\frac1{n^2}\), and hence \(\|x_n\|_2=\sqrt{1+\frac1{n^2}}\). Since \(1/n^2\le1\), these norms are at most \(\sqrt2\); since \(1/n^2\to0\), they tend to \(1\), not \(0\). Check: at \(n=1\), the norm is \(\sqrt2\); at \(n=2\), it is \(\sqrt{5/4}=\sqrt5/2\). Both values satisfy the formula and are bounded by \(\sqrt2\).
Worked Example: Norm Convergence Also Gives Weak Convergence
In \(\ell^2\), take \(z_n=(1-\frac1n)e_1\) and \(z=e_1\), for \(n\ge1\). For every \(x^*\in(\ell^2)^*\), boundedness gives \(|x^*(z_n)-x^*(z)|=|x^*(z_n-z)|\le\|x^*\|\|z_n-z\|_2=\frac{\|x^*\|}{n}\). The right side tends to zero for each fixed \(x^*\), so \(z_n\rightharpoonup z\).
Directly, \(\|z_n\|_2=|1-\frac1n|=1-\frac1n\), because \(n\ge1\), and \(\|z_n-z\|_2=1/n\). Thus \(\|z_n\|_2\le1\), and the bound is exact as a supremum since these norms tend to \(1\). Check: \(z_1=0\) has norm \(0\), while \(z_2=\frac12e_1\) has norm \(1/2\); both agree with the expression \(1-1/n\). This example verifies weak convergence by the stronger norm-convergence estimate.
Why Completeness Cannot Be Dropped
Worked Example: Pointwise Bounded Functionals on an Incomplete Space
Let \(E=c_{00}\), the vector space of finitely supported scalar sequences, with the \(\ell^1\) norm \(\|a\|_1=\sum_j|a_j|\). This space is not complete. For \(k\ge1\), define \(S_k:E\to\mathbb{R}\) (or \(\mathbb{C}\)) by \(S_k(a)=\frac{2^k}{k}a_k\). Each is linear and bounded, since \(|a_k|\le\|a\|_1\), so \(\|S_k\|\le2^k/k\). Equality holds by testing \(a=e_k\), which has norm \(1\) and satisfies \(|S_k(e_k)|=2^k/k\). Hence \(\|S_k\|=2^k/k\).
For any fixed \(a\in c_{00}\), only finitely many coordinates \(a_k\) are nonzero. Therefore \(S_k(a)=0\) for all sufficiently large \(k\), and \(\sup_k|S_k(a)|<\infty\). Nevertheless the operator norms are unbounded. Indeed, for \(k\ge3\), the ratio of consecutive values is \(\frac{2^{k+1}/(k+1)}{2^k/k}=\frac{2k}{k+1}\ge\frac32\). The inequality follows from \(4k\ge3(k+1)\), equivalent to \(k\ge3\). Starting from \(2^3/3=8/3\), the values grow by at least a factor of \(3/2\) at each subsequent step, so they are unbounded.
Check: the first ratio covered by the estimate is from \(k=3\) to \(k=4\), and it equals \(2\cdot3/4=3/2\). The example does not contradict the Uniform Boundedness Principle: its domain \(c_{00}\) is not Banach. It shows why pointwise boundedness alone cannot replace the completeness assumption.
A Common Misreading
A common mistake is to conclude that because each scalar sequence \(x^*(x_n)\) is bounded, the vector norms \(\|x_n\|\) are bounded by simply taking a supremum over \(x^*\). The quantifiers matter: weak convergence says that for each fixed \(x^*\), there is a bound that may depend on \(x^*\). It does not directly provide one bound that works for every \(x^*\) in the unit ball. Uniform boundedness is the theorem that supplies this uniform control, and its Banach-space hypothesis is what makes the conclusion valid.
For a weakly convergent sequence, the functional family is precisely \(T_n(x^*)=x^*(x_n)\), and its pointwise boundedness follows from scalar convergence. The operator norms are precisely the vector norms. Thus the theorem converts bounds depending initially on the test functional into a single bound on the whole sequence. In an \(L^p(\mu)\) space with \(1\le p\le\infty\), this applies because \(L^p(\mu)\) is complete; it requires no finiteness or sigma-finiteness assumption on \(\mu\).
Check Your Understanding
For each question, identify where completeness or the dual-space evaluation enters the argument.
- State weak convergence \(x_n\rightharpoonup x\) using the elements of \(X^*\).
- Why is the family \(T_n(x^*)=x^*(x_n)\) pointwise bounded when \(x_n\rightharpoonup x\)?
- In the uniform boundedness proof, explain why finding a ball inside one set \(E_m\) gives a bound on \(\|T\|\) for every \(T\) in the family.
- For \(x_n=e_{n+1}+\frac1n e_1\) in \(\ell^2\), compute \(\|x_n\|_2\) and state its limit.
- In the \(c_{00}\) example, why is \(\sup_k|S_k(a)|\) finite for each fixed \(a\), even though \(\sup_k\|S_k\|\) is infinite?