Eberlein–Šmulian theorem

Remarks

The weak topology of an infinite-dimensional space is not metrizable, so compactness and sequential compactness need not agree a priori. The theorem says they agree for the weak topology of a Banach space. For "compact implies sequentially compact", a sequence lies in a separable closed subspace $Y$. On weakly compact subsets of $Y$ the weak topology is metrizable, because countably many functionals separate the points of $Y$. The converse direction is harder. One proof first shows that $A$ is bounded, then uses the Banach–Alaoglu theorem in $X^{**}$. It then shows that the weak* closure of the image $J(A)$ under the canonical embedding $J$ lies in $J(X)$. The same equivalences hold for relative compactness: the weak closure of $A$ is weakly compact exactly when every sequence in $A$ has a weakly convergent subsequence with limit in $X$.

Example: the closed unit ball of a reflexive space is weakly compact, so every bounded sequence in a reflexive space has a weakly convergent subsequence. The analogue fails for the weak* topology: the unit ball of $(\ell^\infty)^*$ is weak*-compact by Banach–Alaoglu, but the evaluations $\delta_n(x) = x_n$ have no weak* convergent subsequence.

Going deeper

For a student

The theorem is used in analysis and partial differential equations whenever one extracts weakly convergent subsequences from bounded sequences in reflexive spaces such as $L^p$ for $1 < p < \infty$ or Sobolev spaces. It also combines with James's theorem and the Krein–Smulian theorem in the theory of weakly compact sets.