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This question already has an answer here:

Let $(E,\mathcal{F},\mu)$ be a measure space such that $\mu(E)=1$ and let $L^p=L^p(E, \mathcal{F},\mu)$. Prove that $L^p \subset L^q\text{ if } 1 \le q \le p$.

I let $f \in L^p$. Then $(\int_E |f|^pd\mu)^{1/p} < \infty$. To prove that $f \in L^q$, I should prove that $(\int_E |f|^q d\mu)^{1/q} < \infty $ but I'm unable to do that. I read somewhere that this proof can be done using Holder's inequality but I couldn't do it.

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marked as duplicate by 23rd, Start wearing purple, azimut, Lord_Farin, Dennis Gulko May 13 '13 at 9:50

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  • $\begingroup$ It was mainly about a counterexample right? $\endgroup$ – Bhavish Suarez May 13 '13 at 9:33
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Since $1 \le q \le p < \infty$, $\varphi(x) = x^{p/q}$ is a convex function. Apply Jensen's inequality to get:

$$ \varphi\left(\int \left|f\right|^q \,d\mu\right) \le \int \varphi\left(\left|f\right|^q\right) \,d\mu $$

Hence: $$ \left(\int |f|^q \,d\mu\right)^{p/q} \le \int \left|f\right|^p \,d\mu $$

It follows that $\lVert f \rVert_q \le \lVert f \rVert_p$ and $L^p \subset L^q$.

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Consider applying Holder's Inequality to $$\int_E |f|^q \cdot 1.$$ You'll have to be a little clever about the choice of conjugate exponents. As a hint: you're right that you'll want a term of $\int |f|^p$. Can you think of an exponent that will give you that?

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    $\begingroup$ Okay! Thanks, I'll try this! Actually I tried with $\int_E |f| \cdot 1$ and couldn't do it. If $p$ and $q$ satisfied $1/p + 1/q =1$ then $q(p-1)=p$. Thats' how we prove Holder's inequality I think. Though I doubt if I can assume the same here! $\endgroup$ – Bhavish Suarez May 13 '13 at 9:23
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    $\begingroup$ Right, you cannot assume $1/p + 1/q = 1$ here. But you do have $p/q + (q-p)/q = q/q = 1$, and so $1/(q/p) + 1/(q/(q-p)) = 1$, meaning that $q/p$ and $q/(q-p)$ are conjugate exponents. $\endgroup$ – Jesse Madnick May 13 '13 at 11:28
  • $\begingroup$ Right, got it! I'll try it this way too! Thanks a lot Jesse! $\endgroup$ – Bhavish Suarez May 14 '13 at 10:28

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