I need a proof for this theorem that says :
Assume that $V \in L_{1}^{1},$ and that $V$ has no bound states, then $\tilde{\mathcal{F}}$ is an isometry on $L^{2}$, $$ \|\widetilde{\mathcal{F}} f\|_{L^{2}}=\|f\|_{L^{2}}, \quad \forall f \in L^{2} $$ and $\widetilde{\mathcal{F}}$ is a bijection with $$ \widetilde{\mathcal{F}}^{-1} \phi(x)=\int_{\mathbb{R}} \psi(x, k) \phi(k) d k $$ Moreover, the distorded Fourier transform diagonalizes $-\partial_{x}^{2}+V:$ $$ -\partial_{x}^{2}+V=\widetilde{\mathcal{F}}^{-1} k^{2} \widetilde{\mathcal{F}} $$
where the distorted Fourier transform is defined by $$ \widetilde{\mathcal{F}} \phi(k)=\widetilde{\phi}(k)=\int_{\mathbb{R}} \overline{\psi(x, k)} \phi(x) d x $$ $$ \psi(x, k):=\frac{1}{\sqrt{2 \pi}}\left\{\begin{array}{ll} {T(k) f_{+}(x, k)} & {\text { for } k \geq 0} \\ {T(-k) f_{-}(x,-k)} & {\text { for } k<0} \end{array}\right. $$ and $T(k)$ is the transmission , and $f_{+}(x, k)$ and $f_{-}(x, k)$ defined by the requirements that $$\left(-\partial_{x}^{2}+V\right) f_{\pm}=k^{2} f_{\pm}, \quad \forall x \in \mathbb{R}, \quad$$ and $$\quad\left\{\begin{array}{l}{f_{+}(x, k) \sim e^{i x k} \quad \text { as } x \rightarrow \infty} \\ {f_{-}(x, k) \sim e^{-i x k} \quad \text { as } x \rightarrow-\infty}\end{array}\right.$$
