I am wondering about the basis functions for $$e_n(x)=\frac{e^{2\pi i n x/L}}{\sqrt(L)}$$ where $L = b - a$ on the domain of $L^2([a,b])$ when doing fourier series. Basically, we must scale it by $L = b - a$ but I am wondering about the case where we need to shift the domain.
For instance, if we consider $L^2([-1,1])$ which has orthonormal basis functions of $$e_n(x) = \frac{e^{2\pi i n x/2}}{\sqrt(2)}$$
But when changing the domain from $[a,b] = [-.5,1.5]$, why is it that we must subtract $.5$ so that the orthonormal basis function becomes $$e_n(x) = \frac{e^{(2\pi i n (x+.5)/2}}{\sqrt(2)}$$
So I'm wondering why is it that we must shift the domain in the second case but not in the first case and to generalize it in a more general context, must we shift the domain if: $\left| a \right| \neq \left| b \right|$ on the domain of $[a,b]$ ?
Thank you
The short answer is periodicity. If you have for example $\sin x$, or $e^{inx}$, it doesn't really matter too much which $2\pi$ length interval you look at it on, it's still got all of the same information. So if we were doing Fourier series on $[-\pi,\pi)$ or $[0,2\pi)$ we can use the same functions, since translating them all doesn't change them (since they are periodic). However when we change the length of the interval we are doing Fourier series on, we need the basis functions to have period the same length as the interval. Hence $e^{2\pi inx/L}/\sqrt{L}$. The $2\pi/L$ in the exponent changes the period to be $L$, and the $\sqrt{L}$ in the denominator is to normalize the basis, but we don't need to translate since that doesn't change (in any meaningful way) a periodic function (if you translate all functions).
Perhaps an equation will make this more clear.
Let $f$ and $g$ be in $L^2([0,x))$, extend them with period $x$ to $\Bbb{R}$.
Now $$\int_0^x f(t)g(t)\,dt = \int_a^{x+a} f(t)g(t)\,dt$$ since $f$ and $g$ are periodic, so the integral will be the same across any interval of length $x$. Then we can change variables to get $$\int_0^x f(t)g(t)\,dt = \int_0^x f(t+a)g(t+a) \,dt$$
Therefore $\langle f(t),g(t)\rangle=\langle f(t+a),g(t+a)\rangle$ so translating the functions doesn't change the bilinear form, or the theory of Fourier series on this interval.