Function in $L^\infty$ is element of $L^2$? Let $\mu$ positive measure and $f\in L^\infty(\mu)$. 
My question is: $f\in L^2(\mu)$?  
Thank you all.
 A: The costant functions are elements of $L^{\infty}$ in lebesgue measure on $\mathbb{R}$, but they're not $L^2$, unless they're not $0$.
It is true that a function $f\in L^{\infty}(\mu)$ is also in $L^2$ if the domain of the function has finite measure.

Answering your question: if $f\in L^{\infty}$ and $g\in L^2$, then 
$$\|fg\|_2^2=\int f^2g^2\le\|f\|_{\infty}^2\int g^2=\|f\|_{\infty}^2\|g\|_2^2$$
A: It depends.
If the domain has finite measure, then it is true; in general $p > q \implies \mathcal L ^p(\Omega) \subset \mathcal L ^ q(\Omega) $ if $|\Omega| < \infty$
Note that with the lebesgue measure, it means that $\Omega$ is finite, but it need not to be that way; for example if you use a probability measure (or every measure such that $\mu(\mathbb R) = L < \infty$) then the above implication is always true.
A: This need not be true in general. For example, take $\mu$ to be the Lebesgue measure on $\mathbb{R}$, and $f=\chi_{[0,\infty]}$. Then $f \in L^{\infty}$, but $f $ doesn't belong to $L^2$.
However, if $\mu$ is a finite measure, then any $L^{\infty}$ function is a $L^2$ function- this follows from the Caucy-Schwartz inequality.
