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Here is the question: Given function $f$ is bounded on a measurable set $E$. Show that if $f \in L^{p_1}(E)$ then $f \in L^{p_2}(E)$ whenever $p_1<p_2$ .

I know that I need to show $|f|_{p_1}\geq c |f|_{p_2}$ for a constant c

I tried to let $p=\frac{p_1}{p_2}$ and then try to use Holder inequality but it did not get anywhere.

I also try to let $g=f^{p_1}$ and try to look at $g^p$ but it did not work .

iipqa_dk
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  • This is false if $f$ is bounded and $\mu(E) = \infty$. Consider $f(x) = 1$. We have, $f\in L_{p_1}({\mathbb R})$ for $p_1=\infty$, but $f\notin L_{p_2}({\mathbb R})$ for $p_2 = 1<p_1$. – Yury Sep 28 '14 at 16:37
  • For this statement to be true, we should have $p_1 \leq p_2$. – Yury Sep 28 '14 at 16:38
  • I am sorry . I meant p2>p1 – iipqa_dk Sep 28 '14 at 16:39
  • @cameron Williams: This does not seem to be a duplicate of the question you linked. Here $f$ is bounded, whereas in the linked question, the measure space is of finite measure. – PhoemueX Sep 28 '14 at 16:43
  • @PhoemueX Oh you're right. My brain scrambled the words around. I retracted my close vote. – Cameron L. Williams Sep 28 '14 at 16:44
  • Hint: (1) use that $f$ is bounded, (2) prove that for every bounded $f$, there exists $C$ such that $|f(x)|^{p_2} \leq C |f(x)|^{p_1}$ for every $x$. – Yury Sep 28 '14 at 16:44
  • As I already pointed out above, this question is not a duplicate of the one as which it is marked now. – PhoemueX Sep 30 '14 at 08:39

1 Answers1

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Simply note that

$$ |f|^{p_2} = |f|^{p_1} \cdot |f|^{p_2 - p_1} \leq |f|^{p_1} \cdot K^{p_2 - p_1}. $$

Cameron L. Williams
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PhoemueX
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  • I edited your post to reflect what I thought you meant. – Cameron L. Williams Sep 28 '14 at 16:46
  • Thank you very much, writing latex on the mobile phone sucks ;) – PhoemueX Sep 28 '14 at 16:47
  • Definitely. I wrote this post http://math.stackexchange.com/questions/784529/integral-int-01-log-left-gamma-leftx-alpha-right-right-rm-dx-frac/784545#784545 on my phone some time ago. Was a nightmare! – Cameron L. Williams Sep 28 '14 at 16:48
  • Thank you but I still do not get how we can continue after that. Since we do not know if m(E) is finite or not, we can not integrate both sides of the inequality . In case m(E) is infinity, the integral of $K^{p_2-p_1}$ over $m(E)$ will blow up. – iipqa_dk Sep 28 '14 at 16:50
  • Note that $K$ is a constant an we have a product, i.e. something like $\int a \cdot f(x) d\mu(x) = a \cdot \int f(x) d\mu(x)$. – PhoemueX Sep 28 '14 at 17:00