# weak L¹-convergence

1. The set $$\mathcal{U}$$ is tight, i.e. for any $$\varepsilon > 0$$ there exists a measurable set $$A$$ with $$|A|<\infty$$ $\forall u\in \mathcal{U}: \quad \int_{\Omega\backslash A} | u | < \varepsilon.$ This condition is trivially true if $$|\Omega| < \infty$$.
2. For any $$\varepsilon>0$$ there exists $$\delta >0$$ such that for every measurable set $$E$$ with $$|E|\leq \delta$$ $\forall u\in\mathcal{U}: \quad \int_E | u | \;\dx{x} < \varepsilon.$
Lemma (Equivalent characterisation of equiintegrability). Let $$\Omega\subset \R^n$$, then $$\mathcal{U}\subset L^1(\Omega)$$ is a family of equiintegrable functions if and only if
1. the family $$\mathcal{U}$$ is tight and
2. there exists an increasing superlinear function $$\Psi: [0,\infty)\to [0,\infty]$$ such that $\sup_{u\in \mathcal{U}} \int_\Omega \Psi(|u|) \; \dx{x} < \infty .$
Theorem (Dunford-Pettis). A sequence $$(u_n)_{n\in \mathbb{N}} \subset L^1(\Omega)$$ converges weakly in $$L^1(\Omega)$$ if and only if
1. the sequence is $$u_n$$ is equibounded in $$L^1(\Omega)$$: $\sup_n \Vert u_n \Vert_{L^1(\Omega)} < \infty .$
2. and the sequence $$u_n$$ is equiintegrable.
Lemma (weak lower semicontinuity of convex functions). If $$F:\R \to \R$$ is convex and $u_n \rightharpoonup u \quad \text{in } L^1(\Omega).$ then $\int F(u) \;\dx{x} \leq \liminf_{n\to \infty} \int F(u_n) \;\dx{x} .$

### References

1. K. H. Karlsen, “Notes on weak convergence (MAT4380 – Spring 2006).” pp. 1–14, 2006
2. L. C. Evans.”Weak convergence methods for nonlinear partial differential equations”, volume 74 of CBMS Regional Conference Series in Mathematics. Published for the Conference Board of the Mathematical Sciences, Washington, DC, 1990.

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