Here and below, $\Phi: \mathbb{R} \rightarrow \mathbb{R}$ is smooth such that $\int_{\mathbb{R}} e^{-\Phi(x)} \mathrm{d} x=1$ and

$$\lim _{|x| \rightarrow+\infty}\left(\frac{\left|\Phi^{\prime}(x)\right|^{2}}{4}-\frac{\Phi^{\prime \prime}(x)}{2}\right)=\ell \in(0,+\infty)$$

$C_{c}^{1}(\mathbb{R})$ denotes the set of continuously differentiable complex-valued functions with compact support on $\mathbb{R}$.

(a) Prove that there are constants $R_{0}>0, \lambda_{1}>0$ and $K_{1}>0$ so that for any $R \geqslant R_{0}$ and $h \in C_{c}^{1}(\mathbb{R})$ :

$$\int_{\mathbb{R}}\left|h^{\prime}(x)\right|^{2} e^{-\Phi(x)} d x \geqslant \lambda_{1} \int_{\{|x| \geqslant R\}}|h(x)|^{2} e^{-\Phi(x)} d x-K_{1} \int_{\{|x| \leqslant R\}}|h(x)|^{2} e^{-\Phi(x)} d x$$

[Hint: Denote $g:=h e^{-\Phi / 2}$, expand the square and integrate by parts.]

(b) Prove that, given any $R>0$, there is a $C_{R}>0$ so that for any $h \in C^{1}([-R, R])$ with $\int_{-R}^{+R} h(x) e^{-\Phi(x)} d x=0$ :

$$\max _{x \in[-R, R]}|h(x)|+\operatorname{sip}_{\{x, y \in[-R, R], x \neq y\}} \frac{|h(x)-h(y)|}{|x-y|^{1 / 2}} \leqslant C_{R}\left(\int_{-R}^{+R}\left|h^{\prime}(x)\right|^{2} e^{-\Phi(x)} d x\right)^{1 / 2}$$

[Hint: Use the fundamental theorem of calculus to control the second term of the left-hand side, and then compare $h$ to its weighted mean to control the first term of the left-hand side.]

(c) Prove that, given any $R>0$, there is a $\lambda_{R}>0$ so that for any $h \in C^{1}([-R, R])$ :

$$\int_{-R}^{+R}\left|h^{\prime}(x)\right|^{2} e^{-\Phi(x)} d x \geqslant \lambda_{R} \int_{-R}^{+R}\left|h(x)-\frac{\int_{-R}^{+R} h(y) e^{-\Phi(y)} d y}{\int_{-R}^{+R} e^{-\Phi(y)} d y}\right|^{2} e^{-\Phi(x)} d x$$

[Hint: Show first that one can reduce to the case $\int_{-R}^{+R} h e^{-\Phi}=0$. Then argue by contradiction with the help of the Arzelà-Ascoli theorem and part (b).]

(d) Deduce that there is a $\lambda_{0}>0$ so that for any $h \in C_{c}^{1}(\mathbb{R})$ :

$$\int_{\mathbb{R}}\left|h^{\prime}(x)\right|^{2} e^{-\Phi(x)} d x \geqslant \lambda_{0} \int_{\mathbb{R}}\left|h(x)-\left(\int_{\mathbb{R}} h(y) e^{-\Phi(y)} d y\right)\right|^{2} e^{-\Phi(x)} d x$$

[Hint: Show first that one can reduce to the case $\int_{\mathbb{R}} h e^{-\Phi}=0$. Then combine the inequality (a), multiplied by a constant of the form $\epsilon=\epsilon_{0} \lambda_{R}$ (where $\epsilon_{0}>0$ is chosen so that $\epsilon$ be sufficiently small), and the inequality (c).]