Essentially bounded function If $(X,\Omega,\mu)$ is a finite measure space, $\varphi:X\to \Bbb C$ is an measurable function, $1\leq p\leq\infty$, and $\varphi f\in L^p(\mu),$ whenever $f\in L^p(\mu)$. 
Show that $\varphi \in L^\infty(\mu)$. Unfortunatly, 
I do not have any idea about it. 
Thanks for all help.
 A: Assume that $\varphi$ is not essentially bounded. This means that for every $n>0$, the set
$$
E_n=\{x\in X: |\varphi(x)|>n\},
$$
has positive measure.
In particular,
$$
A_n=\{x\in X: 2^{n}\le |\varphi(x)|<2^{n+1}\},
$$
has positive measure, for infinitely many $n$. 
Set 
$$
f(x)=\sum_{n=1}^\infty a_n\,\chi_{A_n}(x),
$$
with the $a_n$'s to be positive reals to be determined later.
Then
$$
\|f\|_p^p=\sum_{n=1}^\infty a_n^p\,\mu(A_n),
$$
while
$$
\|f\varphi\|_p^p\ge \sum_{n=1}^\infty\int_{A_n}|f|^p\,|\varphi|^p\,d\mu\ge \sum_{n=1}^\infty\int_{A_n} 2^{np}a_n^p=\sum_{n=1}^\infty 2^{np}a_n^p\,\mu(A_n).
$$
Clearly, not all $A_n$'s have necessarily positive measure, but infinitely many of them do.
Let $\mu(A_{n_k})>0$, $k\in\mathbb N$.  We now set
$$
a_{n_k}^p=\frac{1}{k^2\mu(A_{n_k})},
$$
and $a_n=0$, if $n\ne n_k$. Then
$$
\|f\|_p^p=\sum_{n=1}^\infty a_n^p\,\mu(A_n)=\sum_{k=1}^\infty a_{n_k}^p\,\mu(A_{n_k})
=\sum_{k=1}^\infty\frac{1}{k^2\mu(A_{n_k})}\,\mu(A_{n_k})=\sum_{k=1}^\infty\frac{1}{k^2}<\infty,
$$
while
$$
\|f\varphi\|_p^p\ge \sum_{n=1}^\infty 2^{np}a_n^p\,\mu(A_n)=
\sum_{k=1}^\infty 2^{n_kp}a_{n_k}^p\,\mu(A_{n_k})=\sum_{k=1}^\infty \frac{2^{n_kp}}{k^2}\ge
\sum_{k=1}^\infty \frac{2^{kp}}{k^2}=\infty.
$$
