0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Math.se has latex support if you enclose the math in dollar signs see this faq. $\endgroup$ – JSchlather Jan 31 '13 at 2:47
4
$\begingroup$

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 $n$.

$\endgroup$
  • $\begingroup$ i make a mistake , sorry. the correct suposition is 1<p<q< infty $\endgroup$ – math student Jan 31 '13 at 2:43
  • $\begingroup$ @Leandro: OK, I adjusted my answer accordingly (it doesn't make a difference as to the outcome). $\endgroup$ – Jonas Meyer Jan 31 '13 at 2:45
  • 1
    $\begingroup$ 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. $\endgroup$ – A Blumenthal Jan 31 '13 at 4:00

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.