if $\int_\Bbb{R}\vert v\vert<\infty \rightarrow \int_\Bbb{R}\vert uv\vert<\infty$ then $u$ is bounded. 
Let $u\in\mathcal{C}^0(\Bbb{R};\Bbb{R}).$
Assume that for all $v\in\mathcal{C}^0(\Bbb{R};\Bbb{R})$ such that $\int_\Bbb{R}\vert v\vert$ converges then $\int_\Bbb{R}\vert uv\vert$ converges as well.

How can I prove that $u$ is bounded ?
I tried by contradiction, if I can construct a function such that $\int_\Bbb{R}\vert uv\vert$ diverges but such that $\int_\Bbb{R}\vert v\vert$ converges.
Any idea how can I do that ?

Suppose $u$ is unbounded. Take $\{x_n:n\geq 0\}\subset \Bbb R$ where $x_0=0$ (or anything in $\Bbb R$) and  $|u(x_n)|>\max (1+|u(x_{n-1})|, 1+4^n)$ for  $n\in \Bbb N.$ 
For each $n\in \Bbb N$ take $r_n>0$ such that 
(i). The interval $J_n=[-r_n+x_n,r_n+x_n]$ satisfies $\forall x\in J_n \;(\; |u(x)-u(x_n)|<1).$
(ii).  $\;J_m\cap J_n=\emptyset$ when $ 0<m< n.$ 
Condition (ii) is possible. Otherwise for some  $m\in \Bbb N$  the point $x_m$ would be a limit point of the set $\{x_{n'}:n'>n\}$ for every $n>m,$ which would make $u$ unbounded on every nbhd of $x_m,$ contradicting the continuity of $u.$ 
Let $S=\Bbb R\setminus \cup_{n\in \Bbb N}J_n.$ There exists a continuous $v$ with $\int_S|v|<1$ such that for $x\in J_n$  we have $$v(x)=\frac {2^{-n}\cdot sgn (u(x_n))}{2r_n}$$       (Note: $sgn(y)=y/|y|$ for $y\ne 0.$)  
Then $\int_{\Bbb R}|v|<2$ and $\int_{\Bbb R}|uv|\geq \sum_{n\in \Bbb N} |J_n| \cdot |4^nv(x_n)|=\sum_{n\in \Bbb N}2^n=\infty.$
