# 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): Suppose $f_n \to f$ a.e. and $\|f_n\|_p \leq C<\infty$ for all $n$ and for some $0. 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, 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, 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, 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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