# Ngô Quốc Anh

## October 17, 2009

### The Brezis-Lieb lemma and several applications

Filed under: Các Bài Tập Nhỏ, Giải Tích 6 (MA5205), Nghiên Cứu Khoa Học — Tags: — Ngô Quốc Anh @ 14:34

What we did in this topic was just an $L^p$ version of the Brezis-Lieb lemma. In this topic, we will discuss the generalization of this lemma.

Roughly speaking, what we are going to prove is the following:  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

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

we claim 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 $\{f_n\}$.

To be exact, in addition let $j$ satisfy the following hypothesis:

For every sufficiently small $\varepsilon>0$, there exists two continuous, nonnegative functions $\varphi_\varepsilon$ and $\psi_\varepsilon$ such that

$\displaystyle |j(a+b)-j(a)| \leq \varepsilon \varphi_\varepsilon(a) + \psi_\varepsilon(b)$

for all $a, b \in \mathbb C$.

Theorem. Let $j$ satisfy the above hypothesis and let $f_n = f+g_n$ be a sequence of measurable functions from $\Omega$ to $\mathbb C$ such that

1. $g_n \to 0$ a.e.
2. $j(f) \in L^1$.
3. $\displaystyle\int \varphi_\varepsilon(g_n(x))d\mu(x) \leq C < \infty$ for some constant $C$, independent of $\varepsilon$ and $n$.
4. $\displaystyle\int \psi_\varepsilon(f(x)) d\mu(x) < \infty$ for all $\varepsilon >0$.

Then, as $n \to \infty$,

$\displaystyle\lim\limits_{n \to \infty} \int \left| j(f+g_n) - j(g_n) - j(f) \right| d\mu =0.$

Proof. Fix $\varepsilon >0$ and let

$\displaystyle W_{\varepsilon, n} (x) = \Big[ \big|j(f_n(x)) -j(g_n(x)) - j(f(x))\big| - \varepsilon \varphi_\varepsilon (g_n(x))\Big]_+,$

where $[a]_+ = \max\{a,0\}$. As $n \to \infty$, $W_{\varepsilon, n} (x) \to 0$ a.e. On the other hand,

$\displaystyle \big| j(f_n) - f(g_n) - j(f)\big| \leq |j(f_n) - j(g_n)| + |j(f)| \leq \varepsilon \varphi_\varepsilon(g_n) + \psi_\varepsilon(f) + |j(f)|.$

Therefore, $W_{\varepsilon, n} \leq \psi_\varepsilon(f) + |j(f)| \in L^1$. By the Lebesgue Dominated Convergence theorem, $\displaystyle\int W_{\varepsilon, n} d\mu \to 0$ as $n \to \infty$. However,

$\displaystyle |j(f_n) - j(g_n) - j(f)| \leq W_{\varepsilon, n} +\varepsilon \varphi_\varepsilon(g_n)$

and thus

$\displaystyle I_n \equiv \int \big| j(f_n) - j(g_n) - j(f) \big| d\mu\leq \int \big[ W_{\varepsilon, n} + \varepsilon \varphi_\varepsilon(g_n)\big] d\mu .$

Consequently, $\limsup_{n \to \infty} I_n \leq \varepsilon C$. Now let $\varepsilon \to 0$.

Applications.

• The simplest example is when we choose $j(x)=|x|^p$ where $0< p<\infty$. In this situation, one has

$\displaystyle \int \Big(|f_n|^p - |f_n - f|^p - |f|^p \Big) d\mu \to 0.$

• We now assume $u_n \rightharpoonup u$ in $W^{1, 2}$. As a consequence and up to a subsequence, $u_n \to u$ in $L^\alpha$ for every $1<\alpha<2^\star := \frac{2n}{n-2}$ and $u_n \to u$ a.e. Therefore, for a fixed $q \in (2, 2^\star)$, the fact that $u_n \to u$ in $L^q$ implies, by the Brezis-Lieb lemma, that

$\displaystyle u_n^{q-1} \to u^{q-1}$ in $L^\frac{q}{q-1}$.

This is because $\{u_n^{q-1}\}_n \subset L^\frac{q}{q-1}$ is bounded, $u_n^{q-1} \to u^{q-1}$ a.e. and

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \int {\Big( {\underbrace {{{\left| {u_n^{q - 1}} \right|}^{\frac{q}{{q - 1}}}}}_{{{\left| {{u_n}} \right|}^q}} - {{\left| {u_n^{q - 1} - {u^{q - 1}}} \right|}^{\frac{q}{{q - 1}}}}} \Big)d\mu }=\int {\underbrace {{{\left| {{u^{q - 1}}} \right|}^{\frac{q}{{q - 1}}}}}_{{{\left| u \right|}^q}}d\mu }$.

The fact that $u_n \to u$ strongly in $L^p$ implies that $\lim_{n\to \infty} \int |u_n|^p d\mu = \int |u|^p d\mu$. Therefore,

$\displaystyle \mathop {\lim }\limits_{x \to \infty }\int {{{\left| {u_n^{q - 1} - {u^{q - 1}}} \right|}^{\frac{q}{{q - 1}}}}d\mu } = 0$.

As a consequence, one has the following result

$\displaystyle \mathop {\lim }\limits_{x \to \infty } \int {\left( {u_n^{q - 1} - {u^{q - 1}}} \right)\left( {{u_n} - u} \right)d\mu } = 0$.