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Weak $H^{1}$ convergence plus compactness gives strong $L^{2}$ convergence

Statement

Assume the Axiom of Choice. Let $n \geq 1$ , let $Ω \subseteq R^{n}$ be a bounded extension domain and let $(u_{j})$ be bounded in $H^{1} (Ω) = W^{1, 2} (Ω)$ with $u_{j} ⇀ u$ weakly in $H^{1} (Ω)$ . Then $u_{j} \to u$ in $L^{2} (Ω)$ .

Facts & Assumptions

Given: the Axiom of Choice, a bounded extension domain $Ω \subseteq R^{n}$ , $n \geq 1$ , and a bounded sequence $(u_{j})$ with $u_{j} ⇀ u$ weakly in $H^{1} (Ω)$ .

[F1]

First-order Rellich compactness. Since $H^{1} (Ω) = W^{1, 2} (Ω)$ , every bounded sequence in $H^{1} (Ω)$ has a subsequence converging in $L^{2} (Ω)$ on this bounded extension domain. (Compactness of $W^{1, p} (Ω) ↪ L^{p} (Ω)$ on bounded extension domains, The notation $H^{k}$ and the reserved zero-boundary symbol)

[F2]

Finite-measure indicators test both limits. For each measurable $A \subseteq Ω$ , the functional $v \mapsto \int_{A} v$ is bounded on $H^{1} (Ω)$ and on $L^{2} (Ω)$ by H"older's inequality. Weak $H^{1}$ convergence and strong $L^{2}$ convergence therefore give the same limit for these integrals. Since $Ω$ is bounded, $u - v \in L^{1} (Ω)$ ; if its integral over every measurable set is zero, then its real and imaginary parts vanish almost everywhere. (Weak convergence of nets and sequences, Holder's inequality for integrals, including the endpoint cases, The space $L^{p} (μ)$ as the quotient by null functions)

Proof

technique · every subsequence has a further $L^2$-convergent subsequence; indicator tests identify its limit with the weak limit, forcing convergence of the whole sequence

1.1F1given

Let $(u_{j_{k}})$ be any subsequence. It is bounded in $H^{1} (Ω)$ , so by [F1] it has a further subsequence $(u_{j_{k_{r}}})$ converging in $L^{2} (Ω)$ to some $v$ .

2.1F2step 1.1

For each measurable $A \subseteq Ω$ , [F2] gives $\int_{A} u_{j_{k_{r}}} \to \int_{A} u$ by weak convergence and $\int_{A} u_{j_{k_{r}}} \to \int_{A} v$ by strong $L^{2}$ convergence. Thus $\int_{A} (u - v) = 0$ for every such $A$ . Applying this to the sets where the real or imaginary part of $u - v$ is greater than $1 / m$ or less than $- 1 / m$ , for $m \geq 1$ , shows that each part vanishes almost everywhere; hence $u = v$ in $L^{2} (Ω)$ .

3.1F1step 2.1given∎

Every subsequence of $(u_{j})$ therefore has a further subsequence converging in $L^{2} (Ω)$ to $u$ . If the whole sequence did not converge to $u$ , some $ε > 0$ would admit a subsequence staying at distance at least $ε$ from $u$ , contradicting the further-subsequence conclusion. Thus $u_{j} \to u$ in $L^{2} (Ω)$ . The Axiom of Choice is inherited through [F1].

Depends on

Used by

Strong L² convergence preserves an L²-normalisation constraint Example

Dependency tree · two levels

39 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Richard S. Laugesen, Linear Analysis and Partial Differential Equations (University of Illinois, complete graduate notes) (standard reference, not scraped)

Weak H1H1 convergence plus compactness gives strong L2L2 convergence