Weak and Weak* Convergence

A sequence $\{x_n\}$ in a normed space $X$ converges weakly to $x$ (written $x_n \rightharpoonup x$) if $\varphi(x_n) \to \varphi(x)$ for every $\varphi \in X^*$.

Proposition 6.1. If $x_n \to x$ in norm, then $x_n \rightharpoonup x$ (strong convergence implies weak convergence).

Proposition 6.2. If $x_n \rightharpoonup x$ is weakly convergent, then $\sup_n \|x_n\| < \infty$.

Theorem 6.3. In a Hilbert space $H$, $x_n \rightharpoonup x$ if and only if $\langle x_n, y\rangle \to \langle x, y\rangle$ for every $y \in H$.

A sequence $\{\varphi_n\} \subseteq X^*$ converges weak* to $\varphi$ (written $\varphi_n \overset{w^*}{\to} \varphi$) if $\varphi_n(x) \to \varphi(x)$ for every $x \in X$.

Theorem 6.4 (Banach-Alaoglu). The closed unit ball of $X^*$ is weak*-compact.

Theorem 6.5. In $\ell^p$ ($1 < p < \infty$), $x_n \rightharpoonup x$ if and only if $x_n$ is bounded and $x_n(i) \to x(i)$ for each coordinate $i$.

Example. In $\ell^2$, the standard basis vectors $e_n$ converge weakly to $0$ but not in norm: $\|e_n\| = 1$ for all $n$, but $\langle e_n, y\rangle = y_n \to 0$ for every $y \in \ell^2$.

Proposition 6.6 (Weak vs. Weak*). In a normed space $X$:

• If $X$ is reflexive, then weak and weak* convergence on $X^*$ coincide.
• For non-reflexive spaces, weak convergence on $X^*$ implies weak* convergence, but the converse fails.

Proposition 6.7 (Uniqueness of Limits). Weak limits and weak* limits are unique when they exist.

Proposition 6.8 (Weak Convergence in $L^p$). For $1 \leq p < \infty$, a sequence $f_n \rightharpoonup f$ in $L^p(\mu)$ if and only if $\int f_n g\, d\mu \to \int f g\, d\mu$ for every $g \in L^q(\mu)$, where $1/p + 1/q = 1$. For $p = \infty$, weak* convergence is often more useful: $f_n \overset{w^*}{\to} f$ in $L^\infty$ if $\int f_n g \to \int f g$ for every $g \in L^1$.

Lemma 6.9 (Mazur). Let $X$ be a normed space and $x_n \rightharpoonup x$. Then there exists a sequence of convex combinations $y_n = \sum_{k=n}^{N_n} \lambda_k x_k$ (with $\lambda_k \geq 0$, $\sum \lambda_k = 1$) such that $y_n \to x$ in norm.

Corollary 6.10. If $C \subseteq X$ is convex, then $C$ is weakly closed if and only if it is strongly closed.

Corollary 6.11. If $x_n \rightharpoonup x$, then $\|x\| \leq \liminf_{n\to\infty} \|x_n\|$. This is the weak lower semicontinuity of the norm.

Theorem 6.12 (Eberlein-Smulian). In a Banach space, a set is weakly compact if and only if it is weakly sequentially compact. That is, $A \subseteq X$ is weakly compact iff every sequence in $A$ has a weakly convergent subsequence with limit in $A$.

Theorem 6.13 (Kakutani). A Banach space $X$ is reflexive if and only if the closed unit ball $B_X$ is weakly compact (equivalently, weakly sequentially compact).

Corollary 6.14. In a reflexive Banach space, every bounded sequence has a weakly convergent subsequence.

Example. $L^p(\mu)$ for $1 < p < \infty$ is reflexive, so every bounded sequence in $L^p$ has a weakly convergent subsequence. $L^1(\mu)$ is not reflexive: the sequence $f_n = n\chi_{[0,1/n]}$ on $[0,1]$ is bounded in $L^1$ but has no weakly convergent subsequence.

Proposition 6.15. If $T : X \to Y$ is a bounded linear operator and $x_n \rightharpoonup x$ in $X$, then $T x_n \rightharpoonup T x$ in $Y$. That is, bounded linear operators are weakly continuous.

Proposition 6.16. If $T : X \to Y$ is a compact operator and $x_n \rightharpoonup x$ in $X$, then $T x_n \to T x$ in norm. Compact operators map weakly convergent sequences to strongly convergent sequences.

Example. In $L^2([0,1])$, the integral operator $(Tf)(x) = \int_0^1 K(x, y) f(y)\, dy$ with $K \in L^2([0,1]^2)$ is compact. If $f_n \rightharpoonup f$, then $T f_n \to T f$ in $L^2$.

#Example. In $L^2([0,1])$, the integral operator $(Tf)(x) = \int_0^1 K(x, y) f(y)\, dy$ with $K \in L^2([0,1]^2)$ is compact. If $f_n \rightharpoonup f$, then $T f_n \to T f$ in $L^2$.

The four results this page needs proved are the ones its applications rest on: the lower semicontinuity of the norm, Mazur’s lemma, the $\ell^p$ criterion and the compact-operator upgrade. Banach–Alaoglu’s proof is structured but uses Tychonoff, and the page says so rather than hiding it.

Proposition 6.2 (weakly convergent sequences are bounded). If $x_n \rightharpoonup x$, then $\sup_n \|x_n\| < \infty$.

Proof. For each $n$, $\varphi(x_n) \to \varphi(x)$ for every $\varphi \in X^*$, so $\sup_n |\varphi(x_n)| < \infty$ for each fixed $\varphi$. The family $\{x_n\} \subseteq X \subseteq X^{**}$ is therefore pointwise bounded as a family of functionals on $X^*$, and the uniform boundedness principle: page 4’s Theorem 4.11, with $X^*$ in the role of the complete space – gives $\sup_n \|x_n\| = \sup_n \|(x_n)^{**}\| < \infty$. $\blacksquare$

Note the proof’s direction: weak convergence is convergence in the weak* topology of the canonical image of $X$ in $X^{**}$, and Banach–Steinhaus needs the domain complete; which $X^*$ is. The application of the theorem to $X$ itself is the interesting twist, and it is why completeness of $X$ is not needed for this proposition.

Corollary 6.11 (weak lower semicontinuity of the norm). If $x_n \rightharpoonup x$ then $\|x\| \leq \liminf_n \|x_n\|$.

Proof. By Hahn–Banach’s norm-attaining corollary 4.4, for each $x \neq 0$ there is $\varphi_x \in X^*$ with $\|\varphi_x\| = 1$ and $\varphi_x(x) = \|x\|$. Fix such $\varphi_x$ once; weak convergence gives $\varphi_x(x_n) \to \varphi_x(x) = \|x\|$, and

$\blacksquare$

The inequality is the reason norm is not weakly continuous: weak convergence preserves nothing of the norm except that it cannot drop in the limit. The standard example $e_n \rightharpoonup 0$ in $\ell^2$ shows the drop: norms stay $1$ while the weak limit has norm $0$.

Lemma 6.9 (Mazur). If $x_n \rightharpoonup x$, then some sequence of convex combinations of the $x_n$ converges to $x$ in norm.

Proof. For each $N$, let $C_N = \overline{\operatorname{conv}}\{x_m : m \geq N\}$: closed, convex, non-empty. Claim: $x \in C_N$ for every $N$.

Suppose not: $x \notin C_N$ for some $N$. Since $C_N$ is closed, convex and $x$ is a positive distance outside it, the geometric form of Hahn–Banach; the separation theorem; gives $\varphi \in X^*$ and a real $c$ with

But weak convergence gives $\operatorname{Re}\varphi(x_m) \to \operatorname{Re}\varphi(x)$, so for all large $m \geq N$, $\operatorname{Re}\varphi(x_m) > c$; and since $C_N$ is convex, every convex combination $y$ of those tail elements satisfies $\operatorname{Re}\varphi(y) > c$ too. That contradicts $\operatorname{Re}\varphi(y) \leq c$ on $C_N$. So $x \in C_N$ for every $N$.

Now choose $y_N \in C_N$ with $\|y_N - x\| < 1/N$. Each $y_N$ is a convex combination of $\{x_m : m \geq N\}$; finitely many of them, with non-negative coefficients summing to one; and $y_N \to x$ in norm. $\blacksquare$

Corollary 6.10 (norm-closed convex implies weakly closed). A convex subset of a Banach space is weakly closed iff it is norm-closed.

Proof. Weakly closed implies norm-closed always: the weak topology is weaker, so it has fewer closed sets; the content is the converse for convex $C$. Let $x_n \in C$ with $x_n \rightharpoonup x$; Mazur’s lemma gives convex combinations $y_N \in C$ with $y_N \to x$ in norm, and $C$ norm-closed forces $x \in C$. So $C$ contains its weak limits. $\blacksquare$

This corollary is the technical heart of the direct method in the calculus of variations, in the applications section below: it is what lets a variational argument, which naturally produces weak limits, still land in the constraint set; provided the constraint set is convex. Non-convex constraint sets are where weak convergence arguments fail, and the failure is structural.

Theorem 6.5 (the $\ell^p$ criterion). For $1 < p < \infty$, $x_n \rightharpoonup x$ in $\ell^p$ iff $\sup_n \|x_n\|_p < \infty$ and $x_n(i) \to x(i)$ for every coordinate $i$.

Proof. ($\Rightarrow$) Boundedness is Proposition 6.2. Coordinate functionals $\eta_i(y) = y(i)$ are bounded on $\ell^p$ with $\|\eta_i\| = 1$, so weak convergence gives $x_n(i) = \eta_i(x_n) \to \eta_i(x) = x(i)$.

($\Leftarrow$) Let $y \in \ell^q$ with $1/p + 1/q = 1$; by Hölder,

The tail $\sum_{i>N}|y(i)|^q \to 0$ as $N \to \infty$ because $y \in \ell^q$, so choosing $N$ large makes the second term $\leq \epsilon \cdot M$ with $M = \sup_n\|x_n - x\|_p < \infty$ by Proposition 6.2. On the finite head, the convergence $x_n(i) \to x(i)$ coordinate by coordinate is a finite sum, so it tends to zero. Given $\epsilon$, choose $N$ for the tail, then $n_0$ for the head. $\blacksquare$

The proof’s two halves are the general template for every weak-convergence criterion on a sequence space: a density argument on the finite head and a tail estimate controlled by boundedness. The same split proves the $L^p$ criterion of Proposition 6.8 with dense simple functions in place of finitely supported sequences.

Theorem 6.4 (Banach–Alaoglu). The closed unit ball of $X^*$ is weak* compact.

Proof structure. Map $\varphi \in \{ \varphi : \|\varphi\| \leq 1\}$ to the tuple $\big(\varphi(x)\big)_{x \in X}$ in the product $\prod_{x \in X} \bar D_x$, where $\bar D_x = \{z \in \mathbb{C} : |z| \leq \|x\|\}$ is a closed disc. Each disc is compact; Tychonoff makes the product compact; and the unit ball’s image is closed in the product topology because linearity $\varphi(x + y) = \varphi(x) + \varphi(y)$ and the bound $|\varphi(x)| \leq \|x\|$ are conditions on finitely many coordinates, each closed. A closed subset of a compact product is compact. The product topology restricted to this image is exactly the weak* topology, and the map is injective (functionals are their coordinates). $\blacksquare$

Where Tychonoff is used, and what it costs: the product of compact spaces is compact is equivalent to the ultrafilter lemma, strictly weaker than the full axiom of choice but not provable in ZF. For separable $X$ there is a diagonal-sequence proof that avoids Tychonoff entirely; which is why Banach–Alaoglu for separable spaces is elementary while the general statement is not.

Application 1: Calculus of Variations. Weak convergence is central to the direct method in the calculus of variations. To minimize a functional $I$ over a space $X$, one takes a minimizing sequence $x_n$ with $I(x_n) \to \inf I$. If $I$ is weakly lower semicontinuous and the sequence is bounded, weak compactness gives a convergent subsequence whose limit is the minimizer.

Application 2: PDE Theory. Weak solutions of PDEs are often obtained by constructing approximate solutions and extracting a weakly convergent subsequence. The existence theory for elliptic PDEs via the Lax-Milgram theorem relies on weak convergence in Hilbert spaces.

Application 3: Ergodic Theory. Von Neumann’s mean ergodic theorem states that if $T$ is a unitary operator on a Hilbert space $H$, then $\frac{1}{n}\sum_{k=0}^{n-1} T^k x$ converges weakly to the projection of $x$ onto the subspace of $T$-invariant vectors.

Problem 1. Show that $e_n \rightharpoonup 0$ in $\ell^2$ but not in $\ell^1$.

Solution. For $\ell^2$: $\langle e_n, y\rangle = y_n \to 0$ for any $y \in \ell^2$ since $\sum y_n^2 < \infty$ implies $y_n \to 0$. For $\ell^1$: $\ell^1$ has dual $\ell^\infty$. Take $\varphi \in (\ell^1)^*$ corresponding to $(1, 1, 1, \ldots) \in \ell^\infty$. Then $\varphi(e_n) = 1 \not\to 0$, so $e_n$ does not converge weakly to $0$ in $\ell^1$. $\blacksquare$

Problem 2. Let $f_n(x) = \sin(nx)$ in $L^2([0, 2\pi])$. Show $f_n \rightharpoonup 0$.

Solution. For any $g \in L^2$, the Riemann-Lebesgue lemma gives $\int_0^{2\pi} \sin(nx) g(x)\, dx \to 0$. Hence $\langle f_n, g\rangle \to 0$, so $f_n \rightharpoonup 0$. Note that $\|f_n\|_2 = \sqrt{\pi}$, so $f_n$ does not converge strongly. $\blacksquare$

• Normed Spaces and Banach Spaces: Defines the dual spaces and weak topologies that underpin the notions of weak and weak* convergence.

• Bounded Linear Operators: Bounded operators are weakly continuous, and compact operators upgrade weak convergence to strong convergence.

• Compact Operators: Compact operators map weakly convergent sequences to strongly convergent sequences, connecting the two convergence modes.

• Quantum Mechanics

• Graph Theory

• Classical Mechanics

• Electromagnetism

Weak convergence generalises the idea of convergence beyond the strong topology. In infinite dimensions, bounded sequences need not converge in norm, but they may converge weakly: a sequence converges weakly if every continuous linear functional applied to it produces a convergent sequence of numbers. This is a weaker requirement, like asking whether all measurements stabilise rather than whether the object itself stabilises. The Banach-Alaoglu theorem guarantees that bounded sets in dual spaces are weak-star compact, providing the existence of convergent subsequences. Weak convergence is the natural mode of convergence in the calculus of variations and in PDE theory, where direct methods extract weakly convergent subsequences from minimising sequences.

Mistake 1: Assuming weak convergence implies norm convergence Weak convergence is strictly weaker than norm convergence. The standard basis vectors $e_n$ in $\ell^2$ converge weakly to $0$ but $\|e_n\| = 1$ for all $n$. Students often try to pass limits inside norms or integrals under weak convergence, which is invalid without additional compactness or weak lower semicontinuity arguments.

Mistake 2: Confusing weak convergence with weak* convergence Weak convergence $x_n \rightharpoonup x$ requires testing against all functionals in $X^*$, while weak* convergence $\varphi_n \xrightarrow{w^*} \varphi$ only tests against elements of $X \subseteq X^{**}$. In reflexive spaces these coincide, but in non-reflexive spaces like $L^1$ or $\ell^1$ they differ. The Banach-Alaoglu theorem gives weak* compactness, not weak compactness.

Mistake 3: Forgetting that weakly convergent sequences are bounded If $x_n \rightharpoonup x$, then $\sup_n \|x_n\| < \infty$ (uniform boundedness principle). Students sometimes work with weakly convergent sequences without first verifying boundedness, which is a necessary hypothesis for many theorems including Mazur’s lemma and the Eberlein-Smulian theorem. 2. Show that if $x_n \rightharpoonup x$ and $\varphi \in X^*$, then $\varphi(x_n) \to \varphi(x)$. 3. Prove that in a finite-dimensional space, weak convergence is equivalent to norm convergence. 4. Show that $L^\infty([0,1])$ is not reflexive by finding a bounded sequence with no weakly convergent subsequence. 5. Let $T : X \to Y$ be compact. Prove that if $x_n \rightharpoonup 0$, then $T x_n \to 0$.

• Applications
• Bounded Linear Operators
• Compact Operators
• Historical Context