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Suppose $f\in L^{p}(\mathbb{R}^{n}) \cap L^{q}(\mathbb{R}^{n})$. How can I prove that for any $p \lt r \lt q$, $$ \lVert f \rVert_{r} \leq (\lVert f \rVert_{p})^{(1/r-1/q)/(1/p-1/q)} (\lVert f \rVert_{q})^{(1/r-1/q)/(1/p-1/q)}\:? $$ I tried using Hölder to show that $f$ is in $L^{r}$ but I'm completely lost on how to go about it...

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See also: math.stackexchange.com/q/163042 and math.stackexchange.com/q/31683 –  t.b. Jun 25 '12 at 22:13
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1 Answer

up vote 2 down vote accepted

If $p<r<q$ then $\frac{1}{q}<\frac{1}{r}<\frac{1}{p}$, hence there exists $\theta \in ]0,1[$ s.t. $\frac{1}{r}=\frac{\theta}{q}+\frac{1-\theta}{p}$; therefore:

$|f|^r=|f|^{r\theta}\ |f|^{r(1-\theta)} =|f|^{p\ \frac{r\theta}{p}}\ |f|^{q\ \frac{r(1-\theta)}{q}}$,

and you can apply Hölder's inequality, because $|f|^{p \frac{r\theta}{p}} \in L^{\frac{p}{r\theta}}$ and $|f|^{q\ \frac{r(1-\theta)}{q}} \in L^{\frac{q}{r(1-\theta)}}$ and the exponents $\frac{p}{r\theta}, \frac{q}{r(1-\theta)}$ are conjugate.

Finally, to get your claim it suffices to compute explicitly the value of $\theta$.

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Thanks Pacciu; after seeing your hint I proceeded exactly as you and got to the answer :) I forgot to thank you before, but here it goes. –  squ1d Mar 23 '11 at 12:38
    
@squ1d: I'm glad you succeeded... Finding the right exponents to apply Hölder's inequality can be quite messy sometimes. XD –  Pacciu Mar 23 '11 at 14:44
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