Weak Convergence, Weak-* Topology & Reflexivity

Norm convergence requires the whole object to become close. Weak convergence asks only that every bounded linear measurement agree in the limit. The weaker topology creates compactness unavailable in norm, but it also preserves less information. Weak-* convergence is a different construction on a dual space and must not be used as a synonym.

Weak and weak-* convergence #

The weak topology $\sigma(X,X^*)$ is the coarsest topology making every $f\in X^*$ continuous. A sequence satisfies

$$x_n\rightharpoonup x \quad\Longleftrightarrow\quad f(x_n)\to f(x)\quad\text{for every }f\in X^*.$$

The weak-* topology $\sigma(X^*,X)$ on $X^*$ uses only evaluations at primal vectors:

$$f_n\overset{*}{\rightharpoonup}f \quad\Longleftrightarrow\quad f_n(x)\to f(x)\quad\text{for every }x\in X.$$

Weak convergence in $X^*$ tests against all of $X^{**}$ and is generally stronger than weak-* convergence.

Norm convergence implies weak convergence. The converse fails in infinite dimensions: the unit vectors $e_n$ in $\ell^2$ converge weakly to zero but keep norm one.

What weak convergence preserves #

If $x_n\rightharpoonup x$, uniform boundedness gives

$$\sup_n\|x_n\|<\infty.$$

The norm is weakly lower semicontinuous:

$$\|x\|\le\liminf_{n\to\infty}\|x_n\|.$$

Both follow from dual measurements and $\|x\|=\sup_{\|f\|\le1}|f(x)|$. A bounded linear operator preserves weak convergence. A compact operator improves it: if $T$ is compact and $x_n\rightharpoonup x$, then $Tx_n\to Tx$ in norm.

Weak convergence plus convergence of norms implies norm convergence in Hilbert spaces and, more generally, uniformly convex Banach spaces. It is false in arbitrary Banach spaces.

Banach–Alaoglu compactness #

The closed dual unit ball

$$B_{X^*}=\{f\in X^*:\|f\|\le1\}$$

is compact in the weak-* topology. It embeds into the product of compact scalar disks $\prod_{x\in X}\{z:|z|\le\|x\|\}$, and Tychonoff compactness plus closedness gives the theorem.

This is topological compactness. It does not automatically say that every sequence has a weak-* convergent subsequence. If $X$ is separable, the weak-* topology on bounded subsets of $X^*$ is metrizable, so sequential compactness does follow there.

Reflexivity #

The canonical isometry

$$J:X\to X^{**},\qquad (Jx)(f)=f(x)$$

need not be onto. $X$ is reflexive when it is. Equivalent structural characterization: a Banach space is reflexive iff its closed unit ball is weakly compact.

The spaces $\ell^p$ and $L^p$ are reflexive for $1<p<\infty$ under the standard $L^p$ setup. The spaces $\ell^1$, $c_0$, $\ell^\infty$, $L^1$, and $L^\infty$ are standard nonreflexive examples. Closed subspaces of reflexive spaces are reflexive.

Convexity connects weak and norm closure #

For a convex subset of a normed space, norm closure and weak closure agree. Hahn–Banach separation proves that a point outside a norm-closed convex set can be excluded by a weakly continuous functional. Mazur's lemma gives the sequential mechanism: from $x_n\rightharpoonup x$, convex combinations of tails can be chosen to converge to $x$ in norm.

This is why weakly convergent minimizing sequences become useful in convex variational problems: compactness supplies a weak limit, lower semicontinuity controls the objective, and convex/closed constraints survive the limit.

Diagnostic checklist #

• Say weak in which primal space, or weak-* in which dual pairing.
• Derive boundedness before extracting subsequences.
• Do not expect norms or point values to converge weakly.
• Check reflexivity for weak compactness of primal balls.
• Check separability/metrizability before replacing compactness by a sequence statement.
• Use compactness of an operator, not mere boundedness, when upgrading weak to norm convergence.