Weak Topology and Weak Convergence — Mathematics Knowledge System

Overview

The weak topology on a Banach space $X$ is the coarsest topology that makes every bounded linear functional $f \in X^*$ continuous. It is strictly coarser than the norm topology in infinite dimensions, producing a richer landscape of convergent sequences and compact sets. Weak convergence is fundamental in the calculus of variations, PDE theory, and optimization: minimizing sequences often converge weakly but not in norm, and weak compactness of bounded sets (in reflexive spaces) is the key tool for extracting convergent subsequences.

Intuition

Imagine testing a sequence $(x_n)$ not by measuring its distance to $x$ directly, but by probing it with every continuous linear 'measurement' $f$. If every such measurement gives $f(x_n) \to f(x)$, we say $x_n \rightharpoonup x$ weakly. This is like checking a physical object indirectly through all possible linear detectors instead of with a ruler. Weak limits are unique, weak limits are bounded in norm, and strong convergence implies weak convergence—but not vice versa.

Formal Definition

Definition

The weak topology $\sigma(X, X^*)$ on a Banach space $X$ is generated by the sub-basic open sets $\{x \in X : |f(x) - f(x_0)| < \varepsilon\}$ for $f \in X^*$, $x_0 \in X$, $\varepsilon > 0$. A net (or sequence) $(x_\alpha)$ converges weakly to $x$ if $f(x_\alpha) \to f(x)$ for every $f \in X^*$. The weak* topology $\sigma(X^*, X)$ on $X^*$ is the coarsest topology making each evaluation $f \mapsto f(x)$ (for fixed $x \in X$) continuous.

x_n \rightharpoonup x \iff f(x_n) \to f(x) \quad \forall f \in X^*

Weak convergence

f_n \xrightarrow{w^*} f \iff f_n(x) \to f(x) \quad \forall x \in X

Weak-* convergence

Properties

Weak limits are unique

If x_n \rightharpoonup x and x_n \rightharpoonup y, then x = y.

Weakly convergent sequences are norm-bounded

If x_n \rightharpoonup x then \sup_n \|x_n\| < \infty.

Condition: Follows from the Uniform Boundedness Principle applied to the evaluation functionals.

Strong implies weak

\|x_n - x\| \to 0 \implies x_n \rightharpoonup x.

Weak does not imply strong in infinite dimensions

The standard basis e_n \rightharpoonup 0 weakly in \ell^2 but \|e_n\| = 1 \not\to 0.

Theorems

Theorem 2.1: Banach–Alaoglu Theorem

The closed unit ball B^* = \{f \in X^* : \|f\| \le 1\} is compact in the weak-* topology \sigma(X^*, X).

Theorem 2.2: Eberlein–Šmulian Theorem

A Banach space X is reflexive if and only if every bounded sequence has a weakly convergent subsequence.

Theorem 2.3: Mazur's Lemma

If x_n \rightharpoonup x weakly in a Banach space X, then x belongs to the norm-closure of the convex hull of \{x_n\}. In particular, there exist convex combinations of the x_n that converge strongly to x.

Worked Examples

Any $f \in (\ell^2)^*$ is represented by some $y \in \ell^2$ via $f(x) = \sum_k x_k y_k$ (Riesz). Then $f(e_n) = y_n$.
Since $y \in \ell^2$, we have $\sum_k |y_k|^2 < \infty$, so $y_n \to 0$. Hence $f(e_n) \to 0 = f(0)$ for every $f$.
$f(e_n) = y_n \to 0$
On the other hand, $\|e_n - 0\| = \|e_n\| = 1$ for all $n$, so the sequence does not converge in norm.
$\|e_n\|_{\ell^2} = 1 \quad \forall n$

Answer: $e_n \rightharpoonup 0$ weakly, but $\|e_n\| = 1 \not\to 0$, so there is no norm convergence.

Practice Problems

Difficulty 7/10

Prove that a weakly convergent sequence $(x_n)$ in a Banach space is norm-bounded.

Difficulty 8/10

In $L^1([0,1])$, show that weak convergence is strictly stronger than convergence in measure by providing a sequence that converges in measure to 0 but not weakly.

Difficulty 9/10

Use the Banach–Alaoglu theorem to show that every bounded sequence in a reflexive Banach space has a weakly convergent subsequence.

Common Mistakes

Common Mistake

Weak convergence implies norm convergence

This is false in infinite dimensions. The standard basis of $\ell^2$ converges weakly to 0, but every vector has norm 1.

Common Mistake

The weak topology is metrizable on all Banach spaces

The weak topology is metrizable on bounded sets only when $X^*$ is separable. In general it is not metrizable.

Quiz

Summary

The weak topology on $X$ is the coarsest making all $f \in X^*$ continuous; it is strictly coarser than the norm topology in infinite dimensions.
Weak convergence $x_n \rightharpoonup x$ means $f(x_n) \to f(x)$ for all bounded linear functionals $f$.
Weakly convergent sequences are norm-bounded (Uniform Boundedness Principle) and have unique limits.
The Banach–Alaoglu theorem: the unit ball of $X^*$ is weak-* compact.
Reflexive Banach spaces enjoy the Eberlein–Šmulian property: every bounded sequence has a weakly convergent subsequence.

References

BookBrezis, H. Functional Analysis, Sobolev Spaces and Partial Differential Equations. Springer, 2011.
BookConway, J. B. A Course in Functional Analysis. 2nd ed., Springer, 1990.
WebsiteWikipedia — Weak topology