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.


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