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Let $E$ be a measurable set of finite measure and $1\leq p_1 < p_2 \leq \infty$ . Then $L^{p_2} (E) \subseteq L^{p_1} (E)$ Furthermore $||f||_{p_1} \leq c \cdot ||f||_{p_2}$ for all $f$ in $L^{p_2}(E)$ where $c =[m(E)]^{\frac{p_2-p_1}{p_1p_2}} $ if $p_2<\infty$ and $c=[m(E)]^{\frac{1}{p1}}$ if $p_2 =\infty$

I need help with the second part ($p_2= \infty$)

Here is the proof of the first part: $p_2 < \infty$

Define $p = \frac{p_2}{p_1}> 1 $ and $q$ be the conjugate of $p$. Let $f \in L^{p_2} (E) $ then we can prove that $f^{p_1} \in L^p(E)$ and let $g=\chi_E \in L^q(E)$ ( since $m(E)<\infty$) By Holder inequality we have: $\int_E |f|^{p_1} = \int_E |f|^{p_1} g =||f^{p_1}||_p ||g||_q \leq \Big[\int_E (|f|^{p_1})^{\frac{p_2}{p_1}} \Big]^{\frac{p_1}{p_2}} \cdot \Big[\int_E |g|^q \Big]^{\frac{1}{q}}= ||f||_{p_2}^{p_1} \Big[ m(E)\Big]^{\frac{1}{q}}$

$\Rightarrow \text{ by power to }\frac{1}{p_1}$ we get $||f||_{p_1} \leq \Big[m(E)\Big]^{\frac{p_2-p_1}{p_1p_2}} ||f||_{p_2}$

Can anyone help me with the second part: $p_2= \infty$?

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  • $\begingroup$ Please write your email or complete your profile. $\endgroup$ – Hamed Baghal Ghaffari Nov 30 '15 at 16:46
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It's easy. If $f \in L^\infty(E)$, then $ \int_E |f|^{p} \le \int_E \|f\|_\infty^p = \|f\|_\infty^p m(E)$.

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  • $\begingroup$ Ya, you are right, I forget about $|f| \leq ||f||_{\infty} $ Thanks $\endgroup$ – Lucas Nov 30 '15 at 16:39
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$L^{p}(E)$s are not comparable unless you have finite measure space for whole space.

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  • $\begingroup$ My bad, $E$ is measurable set of finite measure, I made the correction, good point, thanks $\endgroup$ – Lucas Nov 30 '15 at 16:50

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