prove that a measurable function is in $L^{\infty}([0,1])$ I am trying to solve this problem but I can not find a good idea:
Let $\phi$ be a measurable function on $[0,1]$ and assume that the linear transformation 
A $\colon f \to f \cdot \phi$ 
maps $L^{2}([0,1])$ into itself .Prove that $\phi \in L^{\infty}([0,1])$.
I tried to prove it by contradiction by choosing properly a function f so that $f\phi$ is no longer in $L^2([0,1])$ but I was no able to find any of it. Any hint for this problem would be appreciated.

Here is a simple way to prove the result by contradiction. 
Let $\phi$ be a measurable function on $[0,1]$. Suppose that  $\phi \notin L^{\infty}([0,1])$. We are going to define a function $f \in L^2([0,1])$, such that $f\phi \notin L^2([0,1])$.
For each $n \in \mathbb{N}$, let 
$E_n= \{ x \in [0,1] : n \leqslant |\phi(n)| < n+1 \}$. 
Note that if $n\neq m$ then $E_n$ and $E_m$ are disjoint.  
Now, let $S = \{n \in \mathbb{N} : n>0 \textrm{ and }\mu(E_n)>0 \}$. 
Since $\phi \notin L^{\infty}([0,1])$, $S$ is an infinite set. 
Define $f$ by 
$$ f = \sum_{n \in S} \frac{1}{n\sqrt{ \mu(E_n) }} \chi_{E_n} $$
where $\chi_{E_n}$ is the characteristic function of $E_n$. 
It is easy to see that 
$$ \int_{[0,1]} |f|^2 = \sum_{n \in S} \frac{1}{n^2} <+\infty $$
So $f \in L^2([0,1])$.
However, 
$$ \int_{[0,1]} |f\phi|^2 = \int_{[0,1]} |f|^2|\phi|^2 \geqslant \sum_{n \in S} \frac{1}{n^2 \mu(E_n) } n^2 \mu(E_n)=\sum_{n \in S} 1 = +\infty$$
So  $f\phi \notin L^2([0,1])$.
