Ngô Quốc Anh

April 29, 2011

Uniformly boundedness implies weak convergence

Filed under: Uncategorized — Ngô Quốc Anh @ 5:01

In the study of L^p spaces, there is a well-known theorem that used frequently in PDE. Its statement is the following

Theorem. Let \{f_n\}_n be a sequence of functions in L^p(\Omega) with 1<p<\infty. If there is a positive constant C and a function f such that

\|f_n\|_p \leqslant C, \quad \text{ for any } n

and

f_n \to f \quad \text{ almost everywhere in } \Omega

then f \in L^p(\Omega) and f_n \rightharpoonup f weakly in L^p(\Omega).

For a proof, we refer the reader to Brezis’s book. The idea is as follows

Step 1. If f_n \rightharpoonup f weakly in L^p(\Omega) and f_n \to \overline f almost everywhere in \Omega then f = \overline f almost everywhere.

The idea of the proof of Step 1 is well-known. If f_n \rightharpoonup f weakly in L^p(\Omega), then we can select a new sequence g_n such that g_n \to f strongly in \Omega. Moreover, g_n \to f almost everywhere.

Step 2. The boundedness and the reflexivity of L^p(\Omega)  imply there is a subsequence of \{f_n\}_n converging to f.

This is a standard property.

Step 3. For any subsequence, if we can find a subsubsequence converging to the same limit, the whole sequence also converges to the limit.

This can be proved by contradiction.

The advantage of the theorem is very clear. If we want to prove something like

\displaystyle \int_\Omega (u_n)^pvd\mu \to \int_\Omega u^pvd\mu

as n\to \infty, u_n \in H^1(\Omega), u_n \rightharpoonup u, and for certain p higher enough that we don’t have any compactness property, the uniformly bounded of \{u_n\}_n in some L^q space, q>p, will help us.

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