How to do it by Dominated Conversgence Theorem? I'm trying to find the limit
$$ I = \lim_{n\to\infty} \int_{\mathbb R^d} \frac1{n} |f(x)|^2 x\cdot\nabla\chi (x/n)dx, $$
where $f \in H^1 (\mathbb R^d, \mathbb C)$, $f \in H^2_{loc}(\mathbb R^d, \mathbb C)$ and $f \in L^\infty (\mathbb R^d, \mathbb C)$, the test function $\chi \in \mathrm C_c^\infty (\mathbb R^d, \mathbb R) $ such that $0\le \chi\le 1$ on $\mathbb R^d$, $\chi = 1$ on $B(0,1)$ and $\chi=0$ outside $B(0,2)$. 
I hope to prove that $I=0$ by Dominated Convergence Theorem, but I don't know how to determine a dominator. Could someone give me a hint?
Edit: Thank to Nate Eldredge's comment, it seems not true without the assumptions $f \in L^\infty (\mathbb R^d, \mathbb C)$ and $f \in H^1 (\mathbb R^d, \mathbb C)$.
 A: Under the new assumption that $f \in H^1(\mathbb{R}^d)$, the statement is true.  (The additional assumptions that $f \in L^\infty \cap H^2_{\mathrm{loc}}$ are not necessary.)
Integrating by parts we have:
$$\int_{\mathbb R^d} \frac1{n} |f(x)|^2 x\cdot\nabla\chi (x/n)\,dx = - \int_{\mathbb{R}^d} \frac{1}{n} \chi(x/n) (2 \operatorname{Re} [f(x) \nabla f(x) \cdot x] + d|f(x)|^2)\,dx.$$
Now let us estimate the integrand (call it $h_n(x)$).  By Cauchy-Schwarz (for the $\mathbb{R}^n$ dot product) we have
$$|h_n(x)| \le \frac{1}{n} \chi(x/n) (2 |f(x)| |\nabla f(x)| |x| + d|f(x)|^2).$$
When $|x| > 2n$ we have $\chi(x/n) = 0$ so we can assume $|x| \le 2n$, giving
$$\begin{align*} |h_n(x)| &\le \chi(x/n) \left(4 |f(x)| |\nabla f(x)| + \frac{d}{n} |f(x)|^2\right) \\ &\le 4 |f(x)| |\nabla f(x)| + d|f(x)|^2\end{align*}$$
since $|\chi| \le 1$ and $n \ge 1$.  Since $f \in H^1(\mathbb{R}^d)$ we have $|f|, |\nabla f| \in L^2(\mathbb{R}^d)$, so $|f|^2, |f||\nabla f| \in L^1(\mathbb{R}^d)$, the latter by Cauchy-Schwarz for the $L^2$ inner product.  Hence the last line above can be used as a dominating function

Under the original assumption, $f \in H^2_{\mathrm{loc}}(\mathbb{R}^d)$, the claim is not true.  For example, take $n=1$ and $$f(x) = \begin{cases} 0, & x < 0 \\ x^{47}, & x \ge 0 \end{cases}$$
which is $C^2$ (indeed, $C^{46}$) and thus in $H^2_{\mathrm{loc}}(\mathbb{R}^1)$.  Now if we integrate by parts we  get
$$\begin{align*}
\int_{\mathbb{R}} \frac{1}{n} |f(x)|^2 x \chi'(x/n)\,dx &= -\int_{\mathbb{R}} \frac{1}{n} \chi(x/n) (2 f(x) f'(x) x + |f(x)|^2)\,dx \\
&\le -\int_{\mathbb{R}} \frac{1}{n} \chi(x/n) |f(x)|^2\,dx 
\end{align*}$$
since $f(x) f'(x) x \ge 0$ everywhere.  Now since $\chi \ge 0$ and $\chi = 1$ for $0 \le x \le n$ we have
$$-\int_{\mathbb{R}} \frac{1}{n} \chi(x/n) |f(x)|^2\,dx \le -\int_0^n \frac{1}{n} |f(x)|^2\,dx = -\frac{n^{94}}{95} \to -\infty$$
by direct computation.
