In the Torus (circle). Let $[0,2\pi]\to\mathbb ]0,\infty[\colon \theta\mapsto w(\theta)$ a weight function, i.e. nonnegative and integrable on $[0,2\pi]$. If $\mathbb{Z}\to\mathbb{R}\colon k\mapsto m(k)$ is a bounded function. Is true that
\begin{align} \left\|{w}^{1/2}\mathcal{F}^{-1}\left(m\widehat{u}\right)\right\|_{L^2([0,2\pi])}\leq C\left\|w^{1/2}u\right\|_{L^2([0,2\pi])}? \end{align}
My attempt: (I used (principally) convolution theorem) \begin{align} \left\|{w}^{1/2}\mathcal{F}^{-1}\left(m\widehat{u}\right)\right\|_{L^2([0,2\pi])}^2 &=\left\|\mathcal{F}^{-1}\left( {w}^{1/2}\right)^\wedge \mathcal{F}^{-1}\left(m\widehat{u}\right)\right\|_{L^2([0,2\pi])}^2\\ &=\left\|\mathcal{F}^{-1}\left((w^{1/2})^{\wedge}*(m\widehat{u})\right)\right\|_{L^2([0,2\pi])}^2\\ &=\left\|(w^{1/2})^{\wedge}*(m\widehat{u})\right\|_{l^2(\mathbb{Z})}^2\\ &=\sum_{l\in\mathbb{Z}}\left|\sum_{k\in\mathbb{Z}}(w^{1/2})^{\wedge}(l-k)m(k)\widehat{u}(k)\right|^2\\ &=\left\|m\right\|_{L^\infty(\mathbb{Z})}^2\sum_{l\in\mathbb{Z}}\left|\sum_{k\in\mathbb{Z}}(w^{1/2})^{\wedge}(l-k)\widehat{u}(k)\right|^2\\ &=\left\|m\right\|_{L^\infty(\mathbb{Z})}^2\left\|(w^{1/2})^{\wedge}*\widehat{u} \right\|_{l^2(\mathbb{Z})}\\ &=\left\|m\right\|_{L^\infty(\mathbb{Z})}^2\left\| w^{1/2}u\right\|_{L^2([0,2\pi])} \end{align}
