Proof Involving Fourier Basis Functions 
Given the equations $$u(x,t) = \sum_{k \in \mathbb{Z}} = \hat{u}(k,t) \exp(i \pi k x), \hspace{5mm} \hat{u}(k,t) = \frac{1}{2} \int_{-1}^{1}u(x,t) \exp(-i \pi k x)dx,$$ where $i$ is the imaginary unit. Due to the orthogonality and completeness of the Fourier basis functions, we have $$u_t(x,t) = \sum_{k \in \mathbb{Z}} \hat{u}_t(k,t) \exp(i \pi k x) = \sum_{k \in \mathbb{Z}} \exp(-i \pi k x(1 + \pi^2) t + i \pi^3 k^3t) \hat{u}(k,t) \exp(i \pi k x),$$ and $$ u(x,t) = \sum_{k \in \mathbb{Z}} \hat{u}(k,0) \exp(-i \pi k(1 + \pi^2)t + i \pi^3 k^3 t) \exp(i \pi k x) $$ Prove that for any time $ t > 0$, $$\lVert u(x,t) \rVert := \sqrt{\frac{1}{2} \int_{-1}^1 |u(x,t)|^2 dx} = \sqrt{\frac{1}{2} \int_{-1}^1 |u(x,0)|^2 dx} $$

My attempt: I begin by trying to show that $|u(x,t)|^2 = |u(x,0)|^2$ by using that $$|u(x,t)|^2 = u(x,t)^2 = \bigg ( \sum_{k \in \mathbb{Z}} \hat{u}(k,0) \exp(-i \pi k(1 + \pi^2)t + i \pi^3 k^3 t) \exp(i \pi k x) \bigg )^2$$
Then inserting for $\hat{u}(k,0)$ gives $$u(x,t)^2 = \bigg ( \sum_{k \in \mathbb{Z}} \frac{1}{2} \int_{-1}^1 u(x,0) \exp(-i \pi k(1 + \pi^2)t + i \pi^3 k^3 t) \exp(i \pi k x) \exp(-i \pi k x) \bigg ) ^2$$
$$= \bigg ( \sum_{k \in \mathbb{Z}} \frac{1}{2} \int_{-1}^1 u(x,0) \exp(-i \pi k(1 + \pi^2)t + i \pi^3 k^3 t) \bigg )^2$$
But here Im stuck and can't really progress further. Any help would be appreciated.
 A: We will make use of the complex variables identity that $|f(x)|^2=f(x)\overline{f(x)}$. If you don't want to cite Parseval's Theorem/identity, we have the following calculations.
First note that:
\begin{equation}
\begin{split}
&\int_{-1}^1|u(x,t)|^2dx \\
&=\int_{-1}^1 u(x,t)\overline{u(x,t)}dx
\\&=\int_{-1}^1\Big{(}\sum_{k\in \mathbb{Z}}\hat{u}(k,t)\exp(i\pi kx)\Big{)}\overline{\Big{(}\sum_{m\in \mathbb{Z}}\hat{u}(m,t)\exp(i\pi mx)\Big{)}}dx\\
&=\int_{-1}^1\Big{(}\sum_{k\in \mathbb{Z}}\hat{u}(k,t)\exp(i\pi kx)\Big{)}\Big{(}\sum_{m\in \mathbb{Z}}\overline{\hat{u}(m,t)\exp(i\pi mx)}\Big{)}dx\\
&=\int_{-1}^1\Big{(}\sum_{k\in \mathbb{Z}}\hat{u}(k,t)\exp(i\pi kx)\Big{)}\Big{(}\sum_{m\in \mathbb{Z}}\overline{\hat{u}(m,t)}\exp(-i\pi mx)\Big{)}dx\\
&=\int_{-1}^1\Big{(}\sum_{k=m, k\in \mathbb{Z}}\hat{u}(k,t)\overline{\hat{u}(k,t)}\exp(i\pi k x)\exp(-i\pi m x)+\sum_{k\neq m;\\ m,k\in \mathbb{Z}}\hat{u}(k,t)\overline{\hat{u}(k,t)}\exp(i\pi k x)\exp(-i\pi m x)\Big{)}dx\\
&\text{By orthogonality, the summation on the right hand side vanishes}\Rightarrow\\
&=\int_{-1}^1\sum_{k=m, k\in \mathbb{Z}}\hat{u}(k,t)\overline{\hat{u}(k,t)}\exp(i\pi k x)\exp(-i\pi m x)dx\,\,\text{Since $k=m$, this simplifies}\\
&=\int_{-1}^1\sum_{k=m, k\in \mathbb{Z}}|\hat{u}(k,t)|^2dx=2\sum_{k=m, k\in \mathbb{Z}}|\hat{u}(k,t)|^2
\end{split}
\end{equation}
Conversely from the formula you provided, we have that:
\begin{equation}
\begin{split}
&\int_{-1}^1|u(x,t)|^2dx\\
&=\int_{-1}^1 u(x,t)\overline{u(x,t)}dx\\
&=\int_{-1}^1\Big{(}\sum_{k\in \mathbb{Z}}\hat{u}(k,0)\exp(-i\pi k(1+\pi^2)t+i\pi^3k^3t)\exp(i\pi k x)\Big{)}\overline{\Big{(}\sum_{m\in \mathbb{Z}}\hat{u}(m,0)\exp(-i\pi m(1+\pi^2)t+i\pi^3m^3t)\exp(i\pi m x)\Big{)}}dx\\
& \text{Following the same steps as the above by using orthogonality, we arrive at:}\\
&=\int_{-1}^1\sum_{k=m, k\in \mathbb{Z}}|\hat{u}(k,0)|^2dx=2\sum_{k=m, k\in \mathbb{Z}}|\hat{u}(k,0)|^2
\end{split}
\end{equation}
Therefore, since the first equation holds for all t, we get that $\int_{-1}^1|u(x,t)|^2dx=\int_{-1}^1|u(x,0)|^2dx$ as desired.
A: The assumption that
$$ u(x,t) = \sum_{k \in \mathbb{Z}} \hat{u}(k,0) \exp(-i \pi k(1 + \pi^2)t + i \pi^3 k^3 t) \exp(i \pi k x) $$
implies that for all times $t$ and integer $k$,
$$|\hat{u}(k,t)|=|\hat{u}(k,0)| \,.$$
The desired conclusion now follows from
https://en.wikipedia.org/wiki/Parseval%27s_theorem
