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This question already has an answer here:

Let $U \in \mathbb{R^n}$ be a closed space and $p > q \geq1$. Show that $L^p(U) \subseteq L^q(U)$.

I need some hint to start with this!

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marked as duplicate by Did, user147263, Tim Raczkowski, graydad, Claude Leibovici Nov 21 '15 at 6:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ @MikePierce I haven't any idea so I needed some thing to start with. $\endgroup$ – Melina Nov 20 '15 at 19:30
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    $\begingroup$ Are you certain of the question? For $n=1$, $U=\mathbb{R}$, $p=2$ and $q=1$, I am skeptical. $\endgroup$ – Clement C. Nov 20 '15 at 19:32
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    $\begingroup$ Then $f$ defined by $f(x)= \mathbb{1}_{[1,\infty)}(x)\frac{1}{x}$ may be an issue: $f\in L^2(\mathbb{R})$, but $f\notin L^1(\mathbb{R})$. $\endgroup$ – Clement C. Nov 20 '15 at 19:37
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    $\begingroup$ Two words, Holder's inequality. $\endgroup$ – IAmNoOne Nov 20 '15 at 20:09
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    $\begingroup$ Melina: When people ask for explanations, you ought to be more specific than "Yes, I am sure!" Here the trouble is that $U\subseteq\mathbb R^n$, not $U\in\mathbb R^n$, and that "closed space" probably means "bounded subset" (a quite different concept). Then, and only then, a hint "to start with this" (more than a hint, actually) is the pointwise inequality $$|f|^q\leqslant1+|f|^p.$$ $\endgroup$ – Did Nov 20 '15 at 20:13
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So we know that: $$\int_U|f|^pd\mu (x)<\infty $$ where $\mu $ is our measure.

Then, by Holder's inequality:

$$\int_U|f|^qd\mu (x)\leq \left(\int_U|f|^{q\frac{p}{q}}d\mu (x)\right)^{\dfrac{q}{p}}\left(\int_U 1^{a}d\mu (x)\right)^{\frac{1}{a}}=\left(\int_U |f|^pd\mu (x)\right)^{\frac{q}{p}}(\mu (U))^{\frac{1}{a}}<\infty $$

Where $a$ is such that $\frac{q}{p}+\frac{1}{a}=1.$

And $U$ have to be such that $\mu (U)<\infty.$

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  • $\begingroup$ Or our measure $\mu $ should be $\sigma$-finite. $\endgroup$ – Mesmerized student Nov 20 '15 at 20:09
  • $\begingroup$ It is little confusing to me since in case of $d\mu(x)$ we use dx $\endgroup$ – Melina Nov 20 '15 at 20:12
  • $\begingroup$ Then just interpret that $\mu(x)=x$ and it all will be fine again. But then you need your $U$ to be bounded. $\endgroup$ – Mesmerized student Nov 20 '15 at 20:31

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