Return to Answer

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

For $\delta = 0$ the inequality is false. Let $u_\lambda = \phi(\lambda x_1) \psi(x_2,x_3)$ where $\psi$ has compact support and $\phi$ is equal to $1$ on $[-1,1]$ and $0$ outside $[-2,2]$. You have that for all $\lambda$, $\|\partial_1 u_\lambda\|_{L^2_1}$ is bounded. But $\|u_\lambda\|_{L^{2}_{-1}} \gtrsim \sqrt{\log(\lambda)}$ as $\lambda \nearrow \infty$.

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

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 $\lambda \nearrow \infty$.

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

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

For $\delta = 0$ the inequality is false. Let $u_\lambda = \phi(\lambda x_1) \psi(x_2,x_3)$ where $\psi$ has compact support and $\phi$ is equal to $1$ on $[-1,1]$ and $0$ outside $[-2,2]$. You have that for all $\lambda$, $\|\partial_1 u_\lambda\|_{L^2_1}$ is bounded. But $\|u_\lambda\|_{L^{2}_{-1}} \gtrsim \sqrt{\log(\lambda)}$ as $\lambda \nearrow \infty$.

$$\int \partial_1 (\langle x\rangle^{\delta-1} x_1 u^2) \mathrm{d}x = 0 $$ 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 $\delta > 1$ your desired inequality holds by one further application of Young's inequality to the RHS and absorbing the undifferentiated factor on the left.

When $\delta \in (-1,1]$ your desired inequality clearly does not hold since the RHS is non-increasing when translating a function of compact support far away toward infinity, while the LHS increases by the positive weight.

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