Compact resolvent proof

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I want to prove the following proposition:

We consider the operator

$$A=A_V=-\Delta+\frac{1}{4}|\nabla V|^2-\frac{1}{2}\Delta V$$(where $V$ is a polynomial) with domain $D(A)=C_c^{\infty}(R^n)$.

$A$ is a positive operator that verifies:

$$\|g(x)u\|^2\le \|u\|^2+\langle u,Au\rangle, \ \ \ \ \forall u\in D(A),$$ where $g$ is a function in $C^{\infty}(R^n)$ which goes to $+\infty$ as $|x|$ goes to $+\infty$

Then $A$ has a compact resolvent.

Here is my answer:

We suppose that the resolvent is not compact. Then there exists a sequence $\{u_n\} \in C_c^{\infty}R^n)$ such that :

  1. $u_n$ converges weakly to $0$ in $L^2(\mathbb R^n)$

  2. $||u_n||_{L^2(R^n)}=1$

  3. $\|(A+1)u_n\|_{L^2(R^n)}\le c=$constant

We consider a partition of unity $f_1(x)^2+f_2(x)^2=1$ such that, for $j=1,2$ we have :

  1. $f_j=1 $ in $B(0,1)$

  2. $\mathrm{supp\,} f\subset B(0,2)$

  3. $f_j\in C^{\infty}(\mathbb R^n)$

Then we have, for all $m\in \mathbb N^*$,: $$f_1(\frac{x}{m})^2+f_2(\frac{x}{m})^2=1,$$ $$\tag{1}0\leq \langle u_n,Au_n\rangle \le\langle u_n,(A+1)u_n\rangle \leq \|u_n\|\,\|(A+1)u_n\|\le c$$ and $$\tag{2}Au_n=\sum_{j=1}^{2}f_j(\frac{x}{m})Af_j(\frac{x}{m}) u_n -\sum_{j=1}^{2}|\partial_xf_j(\frac{x}{m})|^2u_n.$$

$\langle u_n,Au_n\rangle\ge \langle f_2(\frac{x}{m})u_n,Af_2(\frac{x}{m})u_n\rangle-\sum_{j=1}^{2}\langle|\partial_xf_j(\frac{x}{m})|^2u_n,u_n\rangle$ $\ge ||g(x)f_2(\frac{x}{m})u_n||^2- ||f_2(\frac{x}{m})u||^2-\sum_{j=1}^{2}\langle|\partial_xf_j(\frac{x}{m})|^2u_n,u_n\rangle$

we have : $|||\partial_xf_j(\frac{x}{m})|u_{n}||_{L^2(R^n)}\le c_1||u_n||=c_1 $ car $f_j$ is bounded.

Hence $ c \le ( inf _{B(0,2m)}g(x)-1)||f_2(\frac{x}{m})u_{n}||_{L^2(R^n)}$

Then as $m$ goes to infinity $inf _{B(0,2m)}g(x)-1$ goes to $+\infty$(by hypothèsis for $g$)

Hence as $m$ goes to infinity , $||f_2(\frac{x}{m})u_{n}||_{L^2(R^n)}$ goes to $0$

Here i need your help to prove that we can extruct from $u_n$ a convergent subsequence (this subsequence converges to $0$ since $u_n$ converge weakly to $0$) and we obtain a contradiction with the fact that $||u_n||=1$

Can you please help me ?Thanks