$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\arcsinh}{\operatorname{arcsinh}}$ ###5.1.A. Justification Let $u(x)$ be smooth fast decaying function; let us decompose it as in [Section 4.B.](../Chapter4/S4.B.html) (but now we are in the simpler $1$-dimensional framework and $\Gamma=2\pi \mathbb{Z}$): \begin{equation} u(x)= \int\_{0}^{1} u(k;x)\,dk \label{eq-5.1A.1} \end{equation} with \begin{equation} u(k;x)= \sum\_{m=-\infty}^\infty e^{-2\pi km i} u(x+2\pi m). \label{eq-5.1A.2} \end{equation} Here $u(k;x)$ is *quasiperiodic with quasimomentum $k$* \begin{equation} u(k;x+2\pi n)= e^{2\pi n k i}u(k;x)\qquad \forall n\in \mathbb{Z}\ \forall x\in \mathbb{R}. \label{eq-5.1A.3} \end{equation} Indeed, \begin{multline\*} u(k;x+2\pi n)= \sum\_{m=-\infty}^\infty e^{-2\pi km i} u(x+2\pi (m+n))\overset{m:= m+n}{=}\\\\ \sum\_{m=-\infty}^\infty e^{-2\pi k(m-n) i}u(x+2\pi m)=e^{2\pi n k i}u(k;x). \end{multline\*} Then $e^{-ikx}u(k;x)$ is periodic and one can decompose it into Fourier series \begin{equation} 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} \end{equation} (where we restored $u(k;x)$ multiplying by $e^{ikx}$) with \begin{equation} 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} \end{equation} and \begin{equation} 2\pi \sum \_{n=-\infty}^\infty |c\_n(k)|^2 = \int_0^{2\pi} |u(k;x)|^2\, dx. \label{eq-5.1A.6} \end{equation} 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} \,d\omega = \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. --------------- [$\Leftarrow$](../Chapter5/S5.1.html) [$\Uparrow$](../contents.html) [$\Rightarrow$](./S5.2.html)