Convergence in $L^p$ implies convergence in weighted $L^p$ Let $V \in C^\infty(\Bbb{R}^N)$ be a function such that $V \geq 0$ and $\partial^\alpha V(x) \in L^\infty(\Bbb{R}^N)$ for all $|\alpha| \geq 2$. Let $(v_n)$ be a sequence in $L^2(\Bbb{R}^N, Vdx)$ such that $v_n \to v$ in $H^1(\Bbb{R}^N, \Bbb{C})$. Is it true that $v_n$ converges to $v$ in the weighted space $L^2(\Bbb{R}^N, Vdx)$, that is, does
$$
\int_{\Bbb{R}^N}V(x)|v_n - v|^2 \ dx \to 0
$$
hold? If so, how to prove it?
EDIT 
I will add a little context. I am trying to show the claims below, in a paper I am reading:
Screenshot

Here, we define a function space $X$ by
  $$
X:=\left\{v \in H^{1}\left(\mathbb{R}^{n}, \mathbb{C}\right): V_{1}(x)|v(x)|^{2} \in L^{1}\left(\mathbb{R}^{n}\right)\right\}
$$
  We regard $X$ as a real Hilbert space with the inner product
  $$
(v, w)_{X}:=\operatorname{Re} \int_{\mathbb{R}^{n}}\left(v(x) \overline{w(x)}+\nabla v(x) \cdot \overline{\nabla w(x)}+V_{1}(x) v(x) \overline{w(x)}\right) d x
$$

I took a Cauchy sequence. It would then be a Cauchy sequence in $H^1(\Bbb{R}^N, \Bbb{C})$, and thus have a limit. I am trying to show that $v_n$ converges to this limit in $X$.
 A: I don't think that this is true. Let's consider $N=1$ and the function $V(x)=x^2$. A counterexample is given by
\begin{align}
v_n(x)=\frac{1}{n^{1/2}}e^{-\frac{(x-n)^2}{2}}.
\end{align}
I claim that $v_n\to 0$ in $H^1(\mathbb{R})$:
\begin{align}
||v_n||_{L^2(\mathbb{R})}^2&=\frac{1}{n}\int_{\mathbb{R}}e^{-{(x-n)^2}}dx\\
&=\frac{1}{n}\int_{\mathbb{R}}e^{-{y^2}}dy\to 0.
\end{align}
Similarly we get
\begin{align}
||v_n'||_{L^2(\mathbb{R})}^2&=\frac{1}{n}\int_{\mathbb{R}}e^{-{(x-n)^2}}(x-n)^2dx\\
&=\frac{1}{n}\int_{\mathbb{R}}e^{-{y^2}}y^2dy\to 0.
\end{align}
However $v_n$ does not converge to $0$ in $L^2(\mathbb{R},Vdx)$ since
\begin{align}
\frac{1}{n}\int_{\mathbb{R}}x^2 e^{-{(x-n)^2}}dx&=\frac{1}{n}\int_{\mathbb{R}}(y+n)^2e^{-{y^2}}dy\\
&=\frac{1}{n}\int_{\mathbb{R}}(y^2+2yn+n^2)e^{-{y^2}}dy\\
&=\frac{1}{n}\int_{\mathbb{R}}y^2e^{-{y^2}}dy+\int_{\mathbb{R}}2ye^{-{y^2}}dy+n\int_{\mathbb{R}}e^{-{y^2}}dy\nrightarrow 0.
\end{align}
EDIT (answer to comment): I didn't do the proof in detail but I would argue as follows: The norm on your space $X$ is given by
\begin{align}
||v||_X=||v||_{H^1(\mathbb{R}^n)}+||v||_{L^2(\mathbb{R}^n,Vdx)}
\end{align}
Therefore if you have a Cauchy-sequence $v_n$ than this sequence will be Cauchy with respect to $H^1(\mathbb{R}^n)$ and $L^2(\mathbb{R}^n,Vdx)$, which are both Banach spaces. So you get a limit $w\in H^1(\mathbb{R}^n)$ and $u\in L^2(\mathbb{R}^n, Vdx)$ and using the fact that a subsequence converges a.e. we conclude that the limits are equal a.e., therefore $u=w$. This shows that $w\in X$ and $v_n\to w$ in $X$.
(Take my answers with care, it might very well be that I overlooked something and they are completely wrong :))
