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I have the following situation : let $1< p < q < \infty$ . Consider in $\mathbb R^n$ the Lebesgue measure. Let $\{\varphi_n\}$ a sequence of functions in $C^{\infty} (D)$ ( $D$ a open subset of $\mathbb R^n$ not necessarialy bounded ) . Suppose $\int_{D} |\varphi_n |^q\rightarrow 0$.

I showed that (using Holder inequality) $\int_{G} |\varphi_n |^p\rightarrow 0$ where $G$ is a arbitrary subset of $D$ with the closure of G compact in R^n. Someone can give me a suggestion how to prove that

$$\int_{D} |\varphi_n |^p\rightarrow 0?$$

Thank you

share|cite|improve this question has latex support if you enclose the math in dollar signs see this faq. – JSchlather Jan 31 '13 at 2:47
up vote 4 down vote accepted

If $p=2$, $q=4$, $D=(1,\infty)$ and $\varphi_n(x)=\dfrac{1}{n\sqrt{x}}$, you will have $\int_D|\varphi_n|^4\to 0$ while $\int_D|\varphi_n|^2=\infty$ for all $n$.

share|cite|improve this answer
i make a mistake , sorry. the correct suposition is 1<p<q< infty – math student Jan 31 '13 at 2:43
@Leandro: OK, I adjusted my answer accordingly (it doesn't make a difference as to the outcome). – Jonas Meyer Jan 31 '13 at 2:45
Yeah, the general principle at play here is that although in the finite-measure-space situation, there is a relation amongst the $L^p$ spaces (via the Lyapunov inequality), whereas in the infinite measure case, one can prove something along the lines of "for any $p < q$ there exists $f \in L^p$ such that $f \notin L^q$, and vice versa. – A Blumenthal Jan 31 '13 at 4:00

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