Global Poincaré type estimate

Question

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.

Answer 1

$$\int \partial_1 (\langle x\rangle^{\delta-1} x_1 u^2) \mathrm{d}x = 0 $$ 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 $\lambda \nearrow \infty$.