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Let $L^\infty(Ω,F,P)$ be the vector space of bounded random variables $(X ∈ L^\infty (Ω,F,P)$ means that there exists a constant C such that $|X(ω)|≤C$, a.s.$)$. Show that $$L^\infty(Ω,F,P)⊂L^2(Ω,F,P)⊂L^1(Ω,F,P)$$

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marked as duplicate by Norbert, Start wearing purple, Amzoti, Lord_Farin, Brandon Carter May 23 '13 at 17:01

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  • $\begingroup$ $P$ presumably is a probability measure. You could just use comparison tests. For the second inclusion, given $f\in L_2$, consider the sets where $|f|\le 1$ and where $|f|>1$. $\endgroup$ – David Mitra May 23 '13 at 15:09
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It is a consequence of Holder inequality $$ E[|XY|]\leq E[|X|^p]^{1/p}E[|Y|^q]^{1/q} $$

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  • $\begingroup$ Thks, I got it for $L^2 \subset L^1$ (that is Cauchy-Schwartz as particular case of Holder) but not understood how $L^\infty \subset L^2$. Does $1/p+1/q=1$ holds? $\endgroup$ – user79133 May 23 '13 at 15:11
  • $\begingroup$ Yes for $L^\infty$ you don't need to be 'fancy' :-P. But the same is true for general $q$ and then you need Holder, isn't it? $\endgroup$ – guacho May 23 '13 at 15:18
  • $\begingroup$ Got it! Anyway, is there a way to prove it without using Holder? $\endgroup$ – user79133 May 23 '13 at 15:24
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    $\begingroup$ You don't need Holder to show merely set inclusion (Holder does, however, does give you the inequality between the norms). $\endgroup$ – David Mitra May 23 '13 at 15:35
  • $\begingroup$ To wit: for $\infty\ne q>p$, we have $\int |f|^p=\int_{[\,|f|\le 1\,]} |f|^p +\int_{[\,|f|> 1\,]} |f|^p\ \le\ \int_{[\,|f|\le 1\,]} 1 +\int_{[\,|f|> 1\,]} |f|^q$. So if the measure space is finite, then $f\in L_q$ implies $f\in L_p$. $\endgroup$ – David Mitra May 23 '13 at 15:47

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