Ngô Quốc Anh

May 8, 2009

A Relation Between Pointwise Convergence Of Functions And Convergence Of Functionals

Filed under: Các Bài Tập Nhỏ, Giải Tích 6 (MA5205), Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 16:40

Let (\Omega, \Sigma, \mu) be a measure space and let \{f_n\}_{n=1}^\infty be a sequence of complex valued measurable functions which are uniformly bounded in L^p= L^p(\Omega, \Sigma, \mu) for some 0 < p<\infty. Suppose that f_n \to f pointwise almost everywhere (a. e.). What can be said about \| f\|_p?

The simplest tool for estimating \| f\|_p is Fatou’s lemma, which yields

\displaystyle\| f\|_p \leq \liminf_{n \to \infty} \|f_n\|_p.

The purpose of this note is to point out that much more can be said, namely

\displaystyle\lim_{n \to \infty} \left( \|f_n\|_p^p - \| f_n - f\|_p^p \right) = \|f\|_p^p.

More generally, if j : \mathbb C \to \mathbb C is a continuous function such that j(0) = 0, then, when f_n \to f a.e. and

\displaystyle\int |j(f_n(x))| d\mu(x) \leq C < \infty

it follows that

\displaystyle\lim\limits_{n \to \infty} \int \left[ j(f_n) - j(f_n - f)\right] = \int j(f)

under suitable conditions on j and/or \{f_n\}.

Statement. The L^p case (0<p<\infty): Suppose f_n \to f a.e. and \|f_n\|_p \leq C<\infty for all n and for some 0<p<\infty. Then the following  limit exists and the equality holds

\displaystyle\lim_{n \to \infty} \left( \|f_n\|_p^p - \| f_n - f\|_p^p \right) = \|f\|_p^p.

Proof.

(i) By Fatou’s lemma, f \in L^p.

(ii) In case 0<p\leq 1, and if we assume that f \in L^p, then we do not need the hypothesis that \|f_n\|_p is uniformly bounded. [This follows from the inequality

\displaystyle|f_n|^p - |f_n - f|^p \leq |f|^p

and the dominated convergence theorem.] However, when 1<p<\infty, the hypothesis that \|f_n\|_p is uniformly bounded is really necessary (even if we assume that f \in L^p) as a simple counterexample shows.

(iii) When 1<p<\infty, the hypotheses imply that f_n \rightharpoonup f weakly in L^p. [By the Banach-Alaoglu theorem, for some subsequence,f_n, converges weakly to some g; but g = f since f_n \to f a.e.] However, weak convergence in L^p is insufficient to conclude that

\displaystyle\lim_{n \to \infty} \left( \|f_n\|_p^p - \| f_n - f\|_p^p \right) = \|f\|_p^p

holds, except in the case p=2. When p\ne 2 it is easy to construct counterexamples, that is

\displaystyle\lim_{n \to \infty} \left( \|f_n\|_p^p - \| f_n - f\|_p^p \right) \ne \|f\|_p^p

under the assumption only of weak convergence. When p=2 the proof of

\displaystyle\lim_{n \to \infty} \left( \|f_n\|_2^2 - \| f_n - f\|_2^2 \right) = \|f\|_2^2

is trivial under the assumption of weak convergence. Indeed,

\|f_n\|_2^2 - \| f_n - f\|_2^2 = \displaystyle\int \big[ f_n^2 - (f_n-f)^2 \big]= \int \big[ 2f_nf -f^2\big] .

Since f_n \rightharpoonup f in L^2, then \int f_n f \to \int f^2 (note that the dual space of L^2 is itself). Thus

\displaystyle\lim_{n \to \infty} \left( \|f_n\|_2^2 - \| f_n - f\|_2^2 \right) = \|f\|_2^2.

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