Can we get an estimate of the form $\|\chi u \|_{L^2} \le C_\chi \|\nabla u\|_{L^2}$ for $\chi, u \in C_0^\infty(\mathbb{R}^n)$? Let $\chi \in C_0^\infty(\mathbb{R}^n)$ with support contained in the cube $(-c,c)^n$. Suppose that $u \in C_0^\infty(\mathbb{R}^n)$.

I would like to know whether there exists a constant $C_\chi$ which may depend on $\chi$ and the support of the $\chi$, but is independent of the support of $u$, such that
$$ \|\chi u \|_{L^2} \le C_\chi \|\nabla u\|_{L^2}.$$

Here's what I have so far. By modifying the proof of Theorem 1.30 in Sobolev Spaces (by Adams and Fournier), I was able to show that
$$ \|\varphi \|_{L^2} \le \sqrt{2} d\|\nabla \varphi \|_{L^2}, \qquad \forall \varphi \in C_0^\infty(\mathbb{R}^n)\text{ with support of $\varphi$ in $(-d,d)^n$.}$$
So now I have
$$\|\chi u\|_{L^2} \le \sqrt{2} c \|(\nabla \chi) u\|_{L^2} + \sqrt{2} c \|\chi (\nabla u)\|_{L^2} \le \sqrt{2} c\|(\nabla \chi) u\|_{L^2} + \sqrt{2} c \|\chi \|_{\infty} \|\nabla u\|_{L^2}. $$
But I still need to deal with the pesky $\sqrt{2} c\|(\nabla \chi) u\|_{L^2}$ term. One idea is to try a scaling $\chi$ to $\chi(\varepsilon x)$, and then send $\varepsilon$ to $0$, but this did not work because then the support of $\chi$ (and hence the constant $c$) grows like $\varepsilon^{-1}$.
Is the above estimate attainable? Comments or answers are greatly appreciated.
 A: Unless $\chi = 0$, you can't expect such a bound. The function $u$ can be constant on $(-c,c)^n$ and then decay very slowly outside $(-c,c)^n$ resulting in an arbitrary small right hand side which maintaining a constant left-hand side.
For a concrete example, consider the one-dimensional case. Assume $c = 1$ and let $0 \neq \chi \in C^{\infty}(\mathbb{R})$ whose support is compactly contained in $(-1,1)$. Given $\delta > 0$, consider the piecewise linear functions
$$ u_{\delta} = \begin{cases} 0 & -\infty < x \leq -1 - \delta, \\
\frac{1}{\delta} \left( x - (-1 - \delta) \right) & -1 - \delta \leq x \leq -1, \\
1 & -1 \leq x \leq 1 ,\\
-\frac{1}{\delta} \left( x - (1 + \delta) \right) & 1 \leq x \leq 1 + \delta, \\
0 & x > \delta.
\end{cases} $$
Note that $||\chi u_{\delta}||_{L^2} = ||\chi||_{L^2}$ is independent of $\delta$ while $u_{\delta} \in W^{1,2}_0(\mathbb{R}^n)$ with $||\nabla (u_{\delta})||_{L^2} = \frac{\sqrt{2}}{\sqrt{\delta}}.$  Assuming that we have a constant $C_{\chi}$ in your inequality that is independent of $u$, we would get
$$ ||\chi||_{L^2} = ||\chi u_{\delta}||_{L^2} \leq C_{\chi} \frac{\sqrt{2}}{\sqrt{\delta}}$$
which would imply that
$$ \sqrt{\frac{\delta}{2}} ||\chi||_{L^2} \leq C_{\chi} $$
for all $\delta > 0$, a contradiction. While $u_{\delta}$ are not smooth, note that if your equality holds for all smooth $u$ then it will also hold for all $u \in W^{1,2}_0(\mathbb{R}^n) = W^{1,2}(\mathbb{R}^n)$. 
