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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$P$ 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 $f>1$. – 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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Thks, I got it for $L^2 \subset L^1$ (that is CauchySchwartz as particular case of Holder) but not understood how $L^\infty \subset L^2$. Does $1/p+1/q=1$ holds? – user79133 May 23 '13 at 15:11

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? – guacho May 23 '13 at 15:18


1You don't need Holder to show merely set inclusion (Holder does, however, does give you the inequality between the norms). – David Mitra May 23 '13 at 15:35

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 $f\in L_p$. – David Mitra May 23 '13 at 15:47