Why shift by a constant doesn't affect integral for $f \in L^2[0,2\pi]$? For $f \in L^2[0,2\pi]$, a solution to a problem I've been trying to solve states the following:
For all $x\in \mathbb{R}$:
$$ \frac{1}{2\pi} \int_{0}^{2\pi}|f(x-t)|^2dt=\frac{1}{2\pi} \int_{0}^{2\pi}|f(t)|^2dt$$
Why is that? There is nothing else stated about $f$ (I would expect it to be periodic for this to hold, but it's not). Is there some consensus I can't remember where we just assume it becomes periodic outside the domain?
In any case, would appreciate knowing how to show this is true formally.
Thanks for any advice.
 A: You indeed need to extend $f$ to $\mathbb{R}$, since $x-t$ otherwise would not always be an element in $[0,2\pi]$. Namely for example $x>0$ and $t=0$ you would have $-t\not\in [0,2\pi]$. Thus $f$ is indeed $2\pi$ periodic on $\mathbb{R}$. Can you then show it is true?
A: Let $f(t)$ is a periodic function on $\mathbb{R}$ with period $T=2\pi$. We have $f(x)=f(x+2n\pi)$, where $n\in \mathbb{Z}$. Define: $u=x-t$
$$\frac{1}{2\pi} \int_{0}^{2\pi}|f(x-t)|^2dt=\frac{1}{2\pi}\int_{x-2\pi}^x|f(u)|^2du=\frac{1}{2\pi}\int^{x+2\pi}_x|f(u)|^2du$$
If $x=2n\pi$, where $n\in \mathbb{Z}$, then
$$\frac{1}{2\pi}\int^{x+2\pi}_x|f(u)|^2du=\frac{1}{2\pi}\int^{2\pi}_0|f(u)|^2du$$
If $x\neq2n\pi$, where $n\in \mathbb{Z}$, then let $x=2n\pi+y$, where $y\in (0,2\pi)$.
$$\begin{align}
\frac{1}{2\pi}\int^{x+2\pi}_x|f(u)|^2du&=\frac{1}{2\pi}\int^{y+2\pi}_y|f(u)|^2du\\
\\
&=\frac{1}{2\pi}\int^{2\pi}_0|f(u)|^2du+\frac{1}{2\pi}\int^{y+2\pi}_{2\pi}|f(u)|^2du-\frac{1}{2\pi} \int_0^{y}|f(u)|^2du\\
\\
&=\frac{1}{2\pi}\int^{2\pi}_0|f(u)|^2du+\frac{1}{2\pi}\int^{y}_{0}|f(u)|^2du-\frac{1}{2\pi} \int_0^{y}|f(u)|^2du\\
\\
&=\frac{1}{2\pi}\int^{2\pi}_0|f(u)|^2du\end{align}$$
