For simplicity let us assume we are considering $\mathbb{R}^3$. Let us define the weighted Sobolev norm $\| u \|^2\_{L^2\_{\alpha}}= \int\_{\mathbb{R}^3} |u|^2 \langle x\rangle^{\alpha}$ where $\langle x \rangle = (1+|x|^2)^{\frac{1}{2}}$


Let $ S = \{x\in \mathbb{R}^3: a<x\_1<b \}$. Note the following easy Poincaré inequality first:
$\| \partial\_1 u \|\_{L^2(S)} > \frac{\sqrt{2}}{b-a} \|u\|\_{L^2(S)}$ for all $ u \in C^{\infty}\_c(S)$. 


My question is whether I can write the above inequality globally using weights. Let $-1<\delta <1 $ and $\delta $ is not equal to zero.
Prove that there exist a constant C such that:


$\| \partial\_1 u \|\_{L^2\_{\delta+1}} > C \|u\|\_{L^2\_{\delta-1}}$ for all $ u \in C^{\infty}\_c (\mathbb{R}^3)$ ?


Also prove that the following stronger inequality also holds at least for values of $\delta$ sufficiently close to $-1$:
$ 4 \| \partial\_1 u \|^2\_{L^2\_{\delta+1}} - (\delta+1)^2 \int x\_1^2 \langle x \rangle^{\delta-3}u^2 \geq C \|u\|^2\_{L^2\_{\delta-1}} $


Thanks,


Comment: I have a way of proving both these hypothesis in a way different to the solution below. I will post the solution later if there was any interest.



Global Poincaré type estimate
$$\int \partial\_1 (\langle x\rangle^{\delta-1} x\_1 u^2) \mathrm{d}x = 0 $$
So
$$ \int [ (\delta - 1) \langle {x}\rangle^{\delta - 3} |x\_1|^2 + \langle{x}\rangle^{\delta - 1}] u^2 ~\mathrm{d}x = -\int 2 \langle x\rangle^{\delta - 1} x\_1 u \partial\_1 u ~\mathrm{d}x $$
so by C-S
$$ | \mathrm{LHS} | \leq 2 \left( \langle x\rangle^{\delta - 3}|x\_1|^2 u^2 ~\mathrm{d}x\right)^\frac12 \left( \langle x\rangle^{\delta + 1} (\partial\_1 u)^2 ~\mathrm{d}x\right)^\frac12 $$


So as long as $\delta > 0$ your desired inequality holds by one further application of Young's inequality to the RHS and absorbing the undifferentiated factor on the left, and noting that $(\delta - 1)|x\_1|^2 + \langle x\rangle^2 \gtrsim \langle x\rangle^2$ when $\delta > 0$. 




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For $\delta = 0$ the inequality is false. Let $u\_\lambda = \phi(\lambda^{-1} x\_1) \psi(x\_2,x\_3)$ where $\psi$ has compact support and $\phi$ is equal to $1$ on $[-1,1]$ and $0$ outside $[-2,2]$. You have that for all $\lambda > 1$, $\|\partial\_1 u\_\lambda\|\_{L^2\_1}$ is bounded. But $\|u\_\lambda\|\_{L^{2}\_{-1}} \gtrsim \sqrt{\log(\lambda)}$ as $\lambda \nearrow \infty$. 


