# Embedded Lp spaces [duplicate]

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 $$L^\infty(Ω,F,P)⊂L^2(Ω,F,P)⊂L^1(Ω,F,P)$$

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

• $P$ 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
It is a consequence of Holder inequality $$E[|XY|]\leq E[|X|^p]^{1/p}E[|Y|^q]^{1/q}$$
• 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? – 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
• 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