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$X,M,\mu$ is a measure space for which $$\mu(A)>0 \to \mu(A) \ge 1.$$ If $1 \le p<q\le\infty $, then $L^p \subset L^q$ , and then $$||f||_\infty \le ||f||_q \le ||f||_p \le ||f||_1$$

This inequality is introduced on my book, and very useful. So I'm trying to prove that inequality. How can I do that?

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up vote 1 down vote accepted

For any measurable $f\ne0$, and for any $\varepsilon$, there exists a measurable subset $A$ with positive measure (and so $\mu(A)\geq1$) such that $|f|\geq\|f\|_\infty-\varepsilon$ on $A$. Then $$ \|f\|_q=\left(\int|f|^q\right)^{1/q}\geq\left(\int_A|f|^q\right)^{1/q} \geq\left(\int_A(|f|_\infty-\varepsilon)^q\right)^{1/q}=(\|f\|_\infty-\varepsilon)\,\mu(A)\geq\|f\|_\infty-\varepsilon. $$ As $\varepsilon$ was arbitrary, the first inequality is proven.

For the second inequality, assume first that $\|f\|_q=1$. This implies, by the first inequality, that $|f|\leq1$ a.e. Then, using $p\leq q$ (and so $|f|^p\geq|f|^q$), $$ \|f\|_p=\left(\int|f|^p\right)^{1/p}\geq\left(\int|f|^q\right)^{1/p}=1^{1/p}=1=\|f\|_q $$ The case $\|f\|_q\ne1$ follows easily.

The third inequality is a particular case of the second one.

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It's done well. Thanks @MartinArgerami –  Collins Jun 13 '12 at 6:34
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