I'm trying to show that the spectral gap of an operator $L = -\Delta + x\cdot\nabla$ in $L^2\left(\mathbb{R}^n, d\gamma\right)$ is bounded by $1$ from below. $d\gamma$ is standard Gaussian measure and I want to use Poincare inequality: $$\int_{\mathbb{R}^n}f^2\,d\gamma - \left(\int_{\mathbb{R}^n}f\,d\gamma\right)^2 \le \int_{\mathbb{R}^n}\lVert\nabla f\rVert^2 d\gamma.$$
Notice that we can assume that $\int_{\mathbb{R}^n}f\,d\gamma=0$ and (I'm not writing the constant):$$\nabla\left(f\nabla f e^{-\frac{\vert x \vert^2}{2}}\right) = \nabla f\cdot \nabla f e^{-\frac{\vert x \vert^2}{2}} + f\Delta fe^{-\frac{\vert x \vert^2}{2}} - f x\cdot\nabla f e^{-\frac{\vert x \vert^2}{2}}= \lVert\nabla f\rVert^2e^{-\frac{\vert x \vert^2}{2}} - fLfe^{-\frac{\vert x \vert^2}{2}}.$$
I wanted to integrate by parts now and everything would be ok if I knew that $\int_{\partial B\left(0,R\right)}f\nabla f e^{-\frac{\vert x \vert^2}{2}}\cdot \eta dS_x \to 0$ as $R\to\infty$. (boundary measure, $\eta$ is normal vector)
Is it true or do I need more assumptions about $f$? Maybe I can assume $f$ is smooth or/and bounded and use some density argument? I guess it is true, because in books they write like that, but I couldn't find detailed proof.