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Suppose $\mathcal{M}$ is a multiplier operator on $L^2(\mathbb{R})$, in the sense that, for any $u(x)\in L^2(\mathbb{R})$,
$\widehat{\mathcal{M}u}(k)=m(k)\hat{u}(k),$
where the scalar complex function $m$ and its inverse $m^{-1}$ are bounded, analytic on an infinite strap, that is, on $\{z\in\mathbb{C}\mid \mathrm{Im} z\in[-k_0,k_0], \mathrm{Re}z\in\mathbb{R}\}$ for some $k_0>0$.

Given $\gamma_\pm\in\mathbb{R}$ and denoting the Heaviside function as $H$, we define the anisotropic weighted space
$L^2_{\gamma_-,\gamma_+}(\mathbb{R}):=\{u\in L^2_{loc}(\mathbb{R})\mid (1+x^2)^{\gamma_-/2}H(-x)u(x)\in L^2(\mathbb{R}),(1+x^2)^{\gamma_+/2}H(x)u(x)\in L^2(\mathbb{R})\}.$
Can we show that $\mathcal{M}: L^2_{\gamma_-,\gamma_+}(\mathbb{R})\to L^2_{\gamma_-,\gamma_+}(\mathbb{R})$ is a bounded invertible operator for any given $\gamma_\pm\in\mathbb{R}$?

The result when $\gamma_-=\gamma_+$ is classical and thus true. Any ideas for the anisotropic weight case? I have been trying for days and only get a hunch that the result when $\gamma_-\neq\gamma_+$ is probably true.

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