5.1.A. Justification

$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\arcsinh}{\operatorname{arcsinh}}$

5.1.A. Justification

Let $u(x)$ be smooth fast decaying function; let us decompose it as in Section 4.B. (but now we are in the simpler $1$-dimensional framework and $\Gamma=2\pi \mathbb{Z}$): $$u(x)= \int_{0}^{1} u(k;x)\,dk \label{eq-5.1A.1}$$ with $$u(k;x)= \sum_{m=-\infty}^\infty e^{-2\pi ikm } u(x+2\pi m). \label{eq-5.1A.2}$$ Here $u(k;x)$ is quasiperiodic with quasimomentum $k$ $$u(k;x+2\pi n)= e^{iky}u(k;x)\qquad \forall n\in \mathbb{Z}\ \forall x\in \mathbb{R}. \label{eq-5.1A.3}$$ Then $e^{-ikx}u(k;x)$ is periodic and one can decompose it into Fourier series $$u(k;x)= \sum_{n=-\infty}^\infty e^{ in x} e^{ikx} c_n(k)= \sum_{n=-\infty}^\infty e^{i(n+k) x} c_n(k) \label{eq-5.1A.4}$$ (where we restored $u(k;x)$ multiplying by $e^{ikx}$) with $$c_n(k) = \frac{1}{2\pi} \int_0^{2\pi} e^{- i (n+k)x} u(k;x)\, dx \label{eq-5.1A.5}$$ and $$2\pi \sum _{n=-\infty}^\infty |c_n(k)|^2 = \int_0^{2\pi} |u(k;x)|^2\, dx. \label{eq-5.1A.6}$$

Plugging (\ref{eq-5.1A.4}) into (\ref{eq-5.1A.1}) we get \begin{equation*} u(x)= \int_{0}^{1} \sum_{n=-\infty}^{\infty} c_n(k) e^{i(n+k) x} \,dk= \sum_{n=-\infty}^{\infty} \int_{n}^{n+1} C(\omega) e^{i\omega x} = \int_{-\infty}^{\infty} C(\omega) e^{i\omega x} \,d\omega \end{equation*} where $C(\omega):=c_n (k)$ with $n=\lfloor \omega\rfloor$ and $k=\omega- \lfloor \omega\rfloor$ which are respectively integer and fractional parts of $\omega$. So, we got decomposition of $u(x)$ into Fourier integral.

Next, plugging (\ref{eq-5.1A.2}) into (\ref{eq-5.1A.5}) we get \begin{multline*} C(\omega) = \frac{1}{2\pi} \int_0^{2\pi} e^{- i \omega x} \sum_{m=-\infty}^{\infty} e^{-2\pi ikm } u(x+2\pi m) \, dx=\\ \frac{1}{2\pi} \int_{2\pi m}^{2\pi(m+1)} e^{- i \omega y} u(y)\,dy= \int_{-\infty}^{2\pi} e^{- i \omega y} u(y)\,dy \end{multline*} where we set $y=x+2\pi m$. So, we got exactly formula for Fourier transform.

Finally, (\ref{eq-5.1A.6}) implies \begin{equation*} 2\pi \sum _{n=-\infty}^\infty\int_0^1 |c_n(k)|^2\,dk = \int_0^{2\pi} \Bigl( \int_0^1 |u(k;x)|^2\, dk\Bigr)\,dx \end{equation*} where the left hand expression is exactly \begin{equation*} 2\pi \sum_{n=-\infty}^{\infty} \int_{n}^{n+1} |C(\omega)|^2\,d\omega=2\pi \int_{-\infty}^{\infty}|C(\omega)|^2\,d\omega \end{equation*} and the right hand expression is \begin{equation*} \int_0^{2\pi} \Bigl(\int_0^1 \sum_{m=-\infty}^{\infty} \sum_{l=-\infty}^{\infty} e^{2\pi ik(l-m) } u(x+2\pi m)\bar{u}(x+2\pi l)\,dk\Bigr)\,dx \end{equation*} and since $\int_0^1 e^{2\pi ik(l-m)}\,dk=\delta_{lm}$ ($1$ as $l=m$ and $0$ otherwise) it is equal to \begin{equation*} \int_0^{2\pi} \sum_{m=-\infty}^{\infty} |u(x+2\pi m)|^2\,dx= \int_{-\infty}^{\infty} |u(x)|^2\,dx. \end{equation*}

So, we arrive to Plancherel theorem.