if $f\phi \in L^p(\mathbb{R})$ for every $f \in L^p(\mathbb{R})$, then $\phi \in L^{\infty}(\mathbb{R})$. 
Let $\phi: \mathbb{R} \to \mathbb{C}$ be a measurable function and fix $1 \leq p \leq \infty $. Show if $f\phi \in L^p(\mathbb{R})$ for every $f \in L^p(\mathbb{R})$, then $\phi \in L^{\infty}(\mathbb{R})$.

I have the hint if $p< \infty$ and $\phi \notin L^{\infty}(\mathbb{R})$, then infinitely many $E_k=\{k\leq |\phi | \leq k+1\}$ must have positive measure.
So basically I'm proving the contrapositive of the statement. I've shown the case when $p=\infty$, but the case when $p< \infty$, using the hint, is giving me trouble.
 A: Further hint: $$\int |f \phi|^p \; dx \ge \sum_{k=0}^\infty k^p \int_{E_k} |f|^p \; dx$$
Find $f$ so that $\sum_k \int_{E_k} |f|^p\; dx$ converges but $\sum_k k^p \int_{E_k} |f|^p\; dx$ does not.
A: Replacing $\phi$ by $|\phi|^p$ we can assume that $p=1$ and hence we have $\forall f\in L^1, \ \phi f\in L^1$ . Also define $\phi_n =\phi 1_{|\phi|\leq n}\in L^\infty$. Then $\phi_n\rightarrow \phi$ almost everywhere. Also if you define linear functionals 
$$T_n: L^1\rightarrow \mathbb C$$
$$f\mapsto\int_{\mathbb R}\phi_n f \ d\lambda$$ then each of these $T_n$'s are continuous. 
Further observe 
$$\lim _n \int_{\mathbb R}\phi_n f \ d\lambda=\int_{\mathbb R}lim_n\ \phi_n f \ d\lambda=\int_{\mathbb R}\phi f \ d\lambda$$ by the Dominated Convergence Theorem since $|\phi_f|\leq |\phi f|\in L^1$. Thus $T_n \rightarrow T$ weakly where $T\in L^{1*}$ given by $$f\mapsto \int_{\mathbb R}\phi f \ d\lambda$$. 
Now $L^{1}(\mathbb R)^*=L^\infty(\mathbb R)$ and hence $\phi \in L^{\infty}$
A: Let $k_n$ s.t. $m\{k_{n}\leq |\varphi|<k_{n+1}\}>0$. Suppose there are infinitely many.
$$\int_{[k_n,k_n+1)}|\varphi f|^p\geq |k_n|^p\int_{[k_n,k_{n+1)}}|f|^p.$$
Summing both side will give a contradiction.
