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 ?
 A: 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 $\int_{\Bbb R}|uv|\geq \sum_{n\in \Bbb N} |J_n| \cdot |4^nv(x_n)|=\sum_{n\in \Bbb N}2^n=\infty.$
A: EDIT: The following does not actually work, because $v$ is not fixed. 
Hint: Assume otherwise. Using continuity, show that given $n\in\mathbb{N}$, there exists a non-empty intevral $I$ such that $|u|>n$ on $I$. Now, choose (justify this) $v$ such that $\int_\mathbb{R}|v| = 1$ and $\int_I|v| = 1.$ Then, 
$$\infty> \int_\mathbb{R}|uv|\geq\int_I|uv|>n.$$ Since $n$ is arbitrary, this gives the desired contradiction. 
A: Take the contrapositive: if u is unbounded, then |uv| diverges. For simplicity, I'll take u and v to be positive.
Since u is unbounded:
for all M there exists x$_M$ such that u(x$_M$)>M. 
Since u is continuous: 
for all $\epsilon$ there exists $\delta _{\epsilon}$ such that if |x-x$_M$|<$\delta _{\epsilon}$ then |u(x)-u(x$_M$)|<$\epsilon$. 
So now take $\epsilon$ = M/2, and it follows that if |u(x)-u(x$_M$)|<$\epsilon$ then u(x)>M/2.
So choose x$_n$ such that u(x$_n$) > 2$^n$ and x$_{n} > x_{n-1}+10$
Note that the existence of x$_{n}$ such that  x$_{n} > x_{n-1}+10$ follows from continuity (continuous functions are bounded on finite intervals, so if u is unbounded, then u must be unbounded outside of the finite interval [0,$ x_{n-1}+10$]
Choose $\delta _n$ such that 
if |x$_n$-x| <  $\delta _n$ then u > (2$^n$)/2 
and
$\delta _n$ < 1
For each x$_n$, we have an interval of length 2$\delta _n$ for which u > 2$^{n-1}$.  Let v be equal to (2$^{-n}$)/$\delta _n $ on that interval. Then the integral of uv on that interval is at least 2$\delta _n $2$^{n-1}$ * 2$^{-n}$/$\delta _n $ = 1. 
If we set v to be zero outside of these intervals, the total integral of uv is  the sum over all these intervals, and that sum clearly diverges. (Note that strictly speaking, this violates the continuity of v, but for each n, we can take v as going to zero in an arbitrarily small distance.)
The integral of v is its value over these intervals. For each interval, v is 2$^{-n}$ for an interval of at most 1, so the integral for each interval is at most 2$^{-n}$, which converges.
