Ngô Quốc Anh

October 13, 2009

Strong convergence in L^p implies convergence a.e.

Filed under: Các Bài Tập Nhỏ, Giải Tích 6 (MA5205), Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 21:49

This topic is to show how to prove the following statement:

if $\{u_n\}_n$ converges strongly to some $u$ in $L^p(\Omega)$, then up to a subsequence, $\{u_n\}_n$ converges almost everywhere to $u$ in $\Omega$.

The proof relies on the so-called Tchebyshev’s inequality. To this end, we first observe that $\{u_n\}_n$ converges strongly to $u$ in $L^p(\Omega)$ means $\displaystyle\lim\limits_{n \to \infty } \int_\Omega {{{\left| {{u_n} - u} \right|}^p}dx} = 0.$

We now apply the Tchebyshev’s inequality, indeed, for each $\varepsilon>0$ one has $\displaystyle {\rm meas}\left\{ {x:\left| {{u_n}(x) - u(x)} \right| >\varepsilon } \right\} \leqslant \frac{1}{{{\varepsilon ^p}}}\int_{\left\{ {x:\left| {{u_n}(x) - u(x)} \right| > \varepsilon } \right\}} {{{\left| {{u_n} - u} \right|}^p}dx} .$

The right hand side of the above inequality can be dominated by $\displaystyle\frac{1}{{{\varepsilon ^p}}}\int_\Omega {{{\left| {{u_n} - u} \right|}^p}dx}$

which implies that $\displaystyle 0 \leqslant \mathop {\lim }\limits_{n \to \infty } {\rm meas}\left\{ {x:\left| {{u_n}(x) - u(x)} \right| > \varepsilon } \right\} \leqslant \mathop {\lim }\limits_{n \to \infty } \left( {\frac{1} {{{\varepsilon ^p}}}\int_\Omega {{{\left| {{u_n} - u} \right|}^p}dx} } \right) = 0.$

Thus $u_n$ converges to $u$ in measure. It turns out that up to a subsequence, $u_n$ converges to $u$ almost everywhere.