Convergence Through Continuous Linear Measurements
Norm convergence asks whether the distance \(\|x_n-x\|\) tends to zero. The dual space offers another way to compare vectors: apply continuous linear functionals and examine the resulting real numbers. Weak convergence uses all such measurements at once. It can describe convergence even when the sequence does not approach its limit in norm.
The previous tutorial introduced the continuous dual \(X^*\), its operator norm, and the canonical map from a normed space into its bidual. We will use those ideas to define weak convergence and establish two important facts: weakly convergent sequences are bounded, and the norm cannot jump upward at a weak limit.
Definition and First Consequences
Thus, weak convergence is ordinary real convergence after every continuous linear measurement. The requirement is universal: it is not enough to check a few convenient functionals unless a separate argument shows that those functionals determine all the others. Since every \(F\in X^*\) is continuous and linear, this definition also says that \(F(x_n-x)\to0\) for every \(F\in X^*\).
Proof. Fix \(F\in X^*\). By the operator-norm bound, $$ |F(x_n)-F(x)|=|F(x_n-x)|\leq\|F\|\,\|x_n-x\|. $$ The right-hand side tends to zero because \(x_n\to x\) in norm. Therefore \(F(x_n)\to F(x)\). This holds for every \(F\in X^*\), so \(x_n\rightharpoonup x\). \(\square\)
The converse does not hold in general. Weak convergence tests each fixed functional separately; it does not require the vectors themselves to become close in norm. The following examples make both the definition and this distinction concrete.
Worked Example: Weak Convergence in the Euclidean Plane
Let \(X=\mathbb{R}^2\) with the Euclidean norm, and write \(x_n=(s_n,t_n)\) and \(x=(s,t)\). Every linear functional on \(\mathbb{R}^2\) has the form $$ F_{a,b}(u,v)=au+bv $$ for some \(a,b\in\mathbb{R}\). It is continuous because the Cauchy–Schwarz inequality gives $$ |F_{a,b}(u,v)|\leq\sqrt{a^2+b^2}\sqrt{u^2+v^2}. $$
If \(x_n\rightharpoonup x\), apply the definition first to \(F_{1,0}\) and then to \(F_{0,1}\). These give \(s_n\to s\) and \(t_n\to t\). Conversely, if these two coordinate sequences converge, then for every \(a,b\in\mathbb{R}\), $$ F_{a,b}(x_n)=as_n+bt_n\longrightarrow as+bt=F_{a,b}(x). $$ Hence \(x_n\rightharpoonup x\). In this space, coordinate convergence also gives $$ \|x_n-x\|=\sqrt{(s_n-s)^2+(t_n-t)^2}\longrightarrow0. $$ So weak convergence and norm convergence agree in \(\mathbb{R}^2\). This finite-dimensional conclusion should not be assumed in arbitrary normed spaces.
Worked Example: A Norm-Convergent Sequence of Continuous Functions
Let \(X=C([0,1])\) with the supremum norm, and define \(f_n(t)=t/(n+1)\). For \(0\leq t\leq1\), $$ |f_n(t)|\leq\frac{1}{n+1}, $$ with equality at \(t=1\). Thus \(\|f_n\|_\infty=1/(n+1)\to0\). For any \(F\in X^*\), the operator-norm bound yields $$ |F(f_n)-F(0)|=|F(f_n)|\leq\|F\|\,\|f_n\|_\infty =\frac{\|F\|}{n+1}\longrightarrow0. $$ Therefore \(f_n\rightharpoonup0\), as also follows from the proposition. The calculation illustrates that norm convergence controls every continuous linear measurement at once.
A Weakly Convergent Sequence That Does Not Converge in Norm
For a contrasting example, consider the real sequence space $$ \ell^2=\left\{x=(x_k)_{k\geq1}:\sum_{k=1}^{\infty}|x_k|^2<\infty\right\}, \qquad \|x\|_2=\left(\sum_{k=1}^{\infty}|x_k|^2\right)^{1/2}. $$ Let \(e_n\) be the sequence whose \(n\)th coordinate is \(1\) and whose other coordinates are \(0\). To test weak convergence, we need to understand what an arbitrary \(F\in(\ell^2)^*\) does to these vectors; checking only a particular list of functionals would not suffice.
Worked Example: The Standard Basis in \(\ell^2\)
Fix \(F\in(\ell^2)^*\) and set \(a_k=F(e_k)\). For a positive integer \(N\), define \(u_N=\sum_{k=1}^N a_k e_k\). By linearity, $$ F(u_N)=\sum_{k=1}^N a_kF(e_k)=\sum_{k=1}^N a_k^2, \qquad \|u_N\|_2=\left(\sum_{k=1}^N a_k^2\right)^{1/2}. $$ Boundedness of \(F\) therefore gives $$ \sum_{k=1}^N a_k^2=|F(u_N)| \leq\|F\|\,\|u_N\|_2 =\|F\|\left(\sum_{k=1}^N a_k^2\right)^{1/2}. $$ If the sum is positive, division by its square root shows that \(\sum_{k=1}^N a_k^2\leq\|F\|^2\); the same inequality holds if the sum is zero. This bound holds for every \(N\), so \(\sum_{k=1}^{\infty}a_k^2<\infty\). In particular, \(a_n\to0\), since the terms of a convergent series of nonnegative numbers tend to zero.
Because \(F(e_n)=a_n\to0=F(0)\), the definition now gives \(e_n\rightharpoonup0\). But \(\|e_n\|_2=1\) for every \(n\), so \(\|e_n-0\|_2\) does not tend to zero. This is weak convergence without norm convergence. The argument works by controlling the values of an arbitrary continuous linear functional, not merely the coordinate functionals.
Weakly Convergent Sequences Are Bounded
Although weak convergence need not force the distances to the limit to vanish, it does impose a uniform bound on the sequence. The proof connects weak convergence to the Uniform Boundedness Principle from earlier in this course. Its application here is to functionals on \(X^*\), which is a Banach space by the Completeness of the Continuous Dual theorem from the previous tutorial.
Proof. For each \(n\), define \(T_n:X^*\to\mathbb{R}\) by $$ T_n(F)=F(x_n). $$ Each \(T_n\) is linear. It is bounded because $$ |T_n(F)|=|F(x_n)|\leq\|F\|\,\|x_n\|. $$ Moreover, \(T_n=Jx_n\), where \(J:X\to X^{**}\) is the canonical map. The Canonical Map Is an Isometry theorem from the previous tutorial therefore gives \(\|T_n\|=\|x_n\|\).
For every fixed \(F\in X^*\), weak convergence implies \(T_n(F)=F(x_n)\to F(x)\). A convergent real sequence is bounded, so \(\sup_n|T_n(F)|<\infty\) for every \(F\in X^*\). The family \((T_n)\) is pointwise bounded on the Banach space \(X^*\). The Uniform Boundedness Principle gives \(\sup_n\|T_n\|<\infty\). Since \(\|T_n\|=\|x_n\|\), it follows that \(\sup_n\|x_n\|<\infty\), as required. \(\square\)
The use of the dual norm here is important: it converts a bound on the functionals \(T_n\) into a bound on the original vectors. The theorem does not say that \(\|x_n-x\|\to0\); the sequence \((e_n)\) in \(\ell^2\) is already a counterexample to that stronger claim.
The Norm Is Lower Semicontinuous Under Weak Convergence
Weak convergence does not generally preserve norms, but it does constrain how small the norms can be near the limit. Specifically, the norm of the weak limit is at most the lower limit of the norms in the sequence. This is called lower semicontinuity of the norm with respect to weak convergence.
Proof. If \(x=0\), the conclusion follows from \(\|x_n\|\geq0\) for every \(n\). Suppose \(x\neq0\). By the Existence of a Norming Functional theorem from the previous tutorial, there is an \(F\in X^*\) such that \(\|F\|=1\) and \(F(x)=\|x\|\). Weak convergence gives \(F(x_n)\to F(x)=\|x\|\). Continuity of absolute value then implies \(|F(x_n)|\to\|x\|\). For every \(n\), the operator-norm bound gives $$ |F(x_n)|\leq\|F\|\,\|x_n\|=\|x_n\|. $$ Taking lower limits in this inequality yields $$ \|x\|=\liminf_{n\to\infty}|F(x_n)| \leq\liminf_{n\to\infty}\|x_n\|. $$ This proves the claim. \(\square\)
For instance, if a weakly convergent sequence satisfies \(\|x_n\|\leq C\) for every \(n\), the theorem implies \(\|x\|\leq C\). The conclusion is one-sided: weak convergence alone does not imply that \(\|x_n\|\to\|x\|\). In the \(\ell^2\) example, \(\|e_n\|_2=1\) while \(\|0\|_2=0\).
How to Use Weak Convergence
Weak convergence is useful when the quantities of interest are continuous linear measurements, even if direct control of the norm is unavailable. Its definition supplies a clear test: fix an arbitrary \(F\in X^*\), calculate \(F(x_n)\), and prove that the resulting real sequence converges to \(F(x)\). In some spaces, a representation theorem or a known description of the dual can make that test especially explicit. Without such a result, verifying convergence for selected functionals alone does not establish weak convergence.
Two reliable consequences help check arguments. First, every weakly convergent sequence is bounded, by the Uniform Boundedness Principle. Second, the norm of the limit cannot exceed the lower limit of the sequence's norms. A common pitfall is to infer norm convergence from convergence under each functional. The standard basis example in \(\ell^2\) shows why this inference fails: every continuous linear functional sends \(e_n\) to a number tending to zero, but each vector remains at norm distance \(1\) from zero.
Check Your Understanding
Use the definition and results in this tutorial to answer the following questions.
- What must be true of \(F(x_n)\) for every \(F\in X^*\) when \(x_n\rightharpoonup x\)?
- Why does norm convergence imply weak convergence?
- In the \(\ell^2\) example, what does boundedness of \(F\) imply about the sequence of numbers \(a_k=F(e_k)\)?
- Where is the Uniform Boundedness Principle applied in the proof that weakly convergent sequences are bounded?
- What inequality does weak lower semicontinuity give for \(\|x\|\) and \(\liminf_n\|x_n\|\)?
- Why does \(e_n\rightharpoonup0\) in \(\ell^2\) not imply that \(e_n\to0\) in norm?