I'm studying weighted $L^2$ spaces in the circle $[0,2\pi]$
Definition 1 A weight is a function $w\colon [0,2\pi]\to \mathbb{R}^+$ (non negative)
Definition 2 The weighted $L_w^2([0,2\pi])$ is defined by $L_w^2([0,2\pi])=\left\{u\colon [0,2\pi]\to\mathbb{C}\colon \left\|w^{1/2}u\right\|_{L^2([0,2\pi])}<\infty\right\}$
Question If $w$ is a weight function with $w\in L^1([0,2\pi])$ then $L^2([0,2\pi])\subset L_w^2([0,2\pi])$?
My attempt: Let $u\in L^2([0,2\pi])$ then $u=\mathcal{F}^{-1}(g)$ for some $g\in l^2(\mathbb{Z})$. If $w\in L^1([0,2\pi])$ then $w^{1/2}\in L^2([0,2\pi])$, and $w^{1/2}=\mathcal{F}^{-1}(f)$ for some $f\in l^2(\mathbb{Z})$. Moreover, $f\in l^{\infty}(\mathbb{Z})$ because $w^{1/2}\in L^2([0,2\pi])\subset L^1([0,2\pi])$ implies that $\mathcal{F}(w^{1/2})\in l^{\infty}(\mathbb{Z})$. Then
\begin{align} \left\|w^{1/2}u\right\|_{L^2([0,2\pi])}&=\left\|\mathcal{F}^{-1}(f)\mathcal{F}^{-1}(g)\right\|_{L^2([0,2\pi])}\\ &=\left\|\mathcal{F}^{-1}(f*g)\right\|_{L^2([0,2\pi])}\\ &=\left\|f*g\right\|_{l^2(\mathbb{Z})}\\ &\leq \left\|f\right\|_{l^{\infty}(\mathbb{Z})}\left\|g\right\|_{l^2(\mathbb{Z})}\\ &= \left\|f\right\|_{l^{\infty}(\mathbb{Z})}\left\|u\right\|_{L^2([0,2\pi])} \end{align}
Therefore, $u\in L_w^2([0,2\pi])$?
