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Let be $1\leq p<\infty$ and $g$ a measurable funtion defined on $E$. I have to prove that if $fg\in L^p$ for every $f\in L^p(E)$, then $g$ is essentialy bounded, that is $g\in L^\infty (E)$.

I approached to this problem trying to prove the equivalent formulation: if $g\notin L^\infty (E)$, there exists a funcion $f\in L^p (E)$ such that $fg\notin L^p (E)$. I defined $E_n =\{x\in E : |g(x)|<n\}$, then $G_n = E_{n+1}\setminus E_n$. Because hypotetis, either $E_n$ and $G_n$ are well defined. I choose now a cube $Q_n$ such that $m(Q_n \cap G_n)>0$ and I define

$\displaystyle f=\sum\limits_{n=1}^{\infty} f_n \chi_{Q_n\cap G_n}$

wehere $f_n$ are constant funtions. I am in troble showing that $f$ belongs to $L^p$ if I choose, for example,

$\displaystyle f_n= \frac{1}{m(Q_n\cap G_n)n^{p+1}}$

but in particular I'm totally struck proving that $||fg||_p=\infty$, that is, $fg\notin L^p{E}$.

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What is $E$ in this context? –  Davide Giraudo Dec 15 '12 at 14:26
    
E is a measurable subset of $\mathbf{R}^k$. –  Horatio Dec 15 '12 at 14:32

1 Answer 1

Here's one way to do it:

Consider the operator $T : L^p \to L^p$ defined by $Tf = fg$. Show that $T$ is a closed operator. (If $f_n \to f$ and $f_n g \to h$ in $L^p$, try passing to a subsequence $f_{n_k}$ with $f_{n_k} \to f$ almost everywhere to show that $h = fg$). By the closed graph theorem, $T$ is bounded.

On the other hand, if $\mu(|g| \ge a) > 0$, take $f = 1_{\{|g| \ge a\}}$ and note that $\|T f\|_p \ge a \|f\|_p$. Since $T$ is bounded, for sufficiently large $a$ we must have $\mu(|g| \ge a) = 0$, which is to say $g \in L^\infty$.

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