The title mostly says it. The question is, whether I have an isomorphism between $L^2(\mathbb{R})$ and $L^2([0, 2\pi))\otimes l^2(\mathbb{Z})$ (or more generally also $L^2(\mathbb{R}^n) \cong L^2([0, 2\pi)^n)\otimes l^2(\mathbb{Z^n})$)). My candidate for the isomorphism would be
$$ \Phi:L^2(\mathbb{R}) \to L^2([0, 2\pi))\otimes l^2(\mathbb{Z}) \\ f(x) \mapsto \sum_{k \in \mathbb{Z}} \chi_{[0, 2\pi)}(x) \cdot f(x+2\pi k) \otimes e_k $$
where $\{e_k | k\in \mathbb{Z}\}$ is the usual schauder basis for $l^2(\mathbb{Z})$ and we can interpret $\chi_{[0, 2\pi)}(x) \cdot f(x+2\pi k) \in L^2([0, 2\pi))$ since it vanishes outside of $[0, 2\pi)$. We can further decompose any vector / function $f$ in $L^2([0, 2\pi))\otimes l^2(\mathbb{Z})$ as
$$ f = \sum_{k \in \mathbb{Z}} f_k \otimes e_k $$ with $f_k \in L^2([0, 2\pi))$ and $e_k$ as before. This implies that the invervse to $\Phi$ is then
$$ \Phi^{-1} : L^2([0, 2\pi))\otimes l^2(\mathbb{Z}) \to L^2(\mathbb{R}) \\ f = \sum_{k \in \mathbb{Z}} f_k \otimes e_k \mapsto \sum_{k \in \mathbb{Z}} f_k(\cdot-2\pi k). $$
This construction correctly preserves the scalar product $$ \left< f, g\right>_{L^2(\mathbb{R})} = ... = \sum_{k \in \mathbb{Z}} \sum_{k \in \mathbb{Z}} \left<e_k, e_k'\right> \cdot \left<f(\cdot + 2\pi k), g(\cdot + 2\pi k')\right>. $$
So this certainly feels like it is an isomorphism. But are there more technical details that one has to take care of in infinite dimensional spaces (Namely continouity I suppose)?