I am studying the proof of a theorem and in a part of the proof I have the following situation:

Let $u : \Omega \rightarrow R$ a nonnegative measurable function, with $\Omega$ open and bounded. Consider $c>0$ a constant. Write $v_{n} = log( u + 1/n )$. Supose that $log (u + 1/n) \in H^{1,p}_{loc} (\Omega)$.

I need to show that $lim_{n \rightarrow \infty} \int_{B} e^{-c_1 v_n} = \int_{B} u^{-c_1}$, where B is a open ball with $2B \subset \Omega$.

I think the dominated convergence theorem can help.. Someone can give a hint to prove what i said?


  • $\begingroup$ in the set $\{ u < 1\}$ , we have $e^{-c_1 log (u + 1/n)} \leq e^{c_1 log(1/u)}$ . I dont know if this help... I said this because may can help separate the integral where u > 1 and in u< 1 $\endgroup$ – math student May 12 '13 at 2:38

I don't think we need anything about Sobolev spaces here. The function $u^{-c_1}$ takes values in $[0,+\infty]$, so $\int_B u^{-c_1}$ is defined as a number in $[0,+\infty]$. The functions $f_n:=\exp(-c_1v_n)$ are nonnegative and form an increasing sequence: $f_{n+1}\ge f_n$. Also, $f_n\to u^{-c_1}$ pointwise. By the Lebesgue monotone convergence theorem we have $\lim \int_B f_n=\int \lim f_n$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.