If $\varphi f\in L^1(\mu)$ for every $f\in L^1(\mu),$ then $\varphi \in L^\infty$ Let $\varphi$ be a measurable function for which $\varphi f\in L^1(\mu)$ for every $f\in L^1(\mu).$ Show that $\varphi \in L^\infty(\mu).$
 A: Consider the map
$$
\Phi : L^1 \to L^1, f \mapsto \varphi f. 
$$
By assumption, this is a well-defined linear map. Use the closed graph theorem to show that it is bounded (there are some details to fill in here). 
This shows that the functional
$$
\psi : L^1 \to \Bbb{C}, f \mapsto \int f \varphi d\mu
$$
is bounded (why)?
Now use the characterisation of the dual space of $L^1$ as $L^\infty$. Here, you will have to assume something like $\sigma$ finiteness of the measure (the claim is wrong in general without $\sigma$ finiteness). 
EDIT: I only used $\sigma$-finiteness to be able to invoke the $L^1$-$L^\infty$ duality in the form most widely known. One can also prove the claim without using this duality.
The actual claim is even true for semifinite measures $\mu$. Here, $\mu$ is called semifinite if for each measurable $A$ with $\mu(A) > 0$, there is a measurable $B \subset A$ of finite, positive measure. It is easy to see that each $\sigma$-finite measure and also any counting measure is semifinite.
Now, let $M := \Vert \psi \Vert$. We will show $\Vert \varphi \Vert_\infty \leq M < \infty$. If this is not the case, then the set
$$
L_n := |\varphi|^{-1}\left(\left[M + \frac{1}{n}, \infty\right]\right)
$$
has positive measure for some $n \in \Bbb{N}$ (why?). Hence, there is a subset $L_n ' \subset L_n$ of finite positive measure, because $\mu$ is semifinite.
Let $f := \mathrm{sgn}(\varphi) \cdot \chi_{L_n '}$. This implies
$$
M \cdot \mu(L_n ') = \Vert \psi \Vert \cdot \Vert f \Vert_1 \geq |\psi(f)| \geq \psi(f) = \int_{L_n '} |\varphi| d\mu \geq \left(M + \frac{1}{n}\right) \cdot \mu(L_n '),
$$
a contradiction.
