Prove or disprove: the Fourier series of the function $f:[0,L] \to \mathbb{R}$ extended periodically of period $2L$ is unique.
What does "periodically extended" mean? ($\ast$)
The Fourier series of $f$ is
$$f(x) = \frac{a_{0}}{2} + \sum_{n=1}^{\infty}\left(a_{n}\cos\left(\frac{n\pi x}{L}\right) + b_{n}\sin\left(\frac{n\pi x}{L}\right)\right).$$
The coeficients $a_{0},a_{n},b_{n}$ are given in terms of $f(x)$ (by integration). So, the unicity seems intuitive to me, but I dont know how to prove it, indeed I didnt fully understand the question ($\ast$)
Best Answer
"Periodically extended" in the context of Fourier series, in an intuitive sense, basically is like copying and pasting the graph over an interval to all of $\mathbb{R}$, with a potential slight alteration.
A periodic extension that isn't really said to be anything special is generally just assumed to be straight up copying and pasting. In other words, the function appears the same on $[0,L]$ as on $[L,2L]$ or $[-L,0]$, aside from just being at different x-values.
An "even" periodic extension first extends the function by flipping it horizontally onto an interval of length $2L$ (assuming we start at $[0,L]$ - the idea is similar on other intervals, just the math becomes a little nastier). So in other words, if we have $a \in [0,L]$, then $f(\alpha) = f(- \alpha)$ - because we extend it so it becomes an even function. And then we just "copy and paste" this length $2L$ interval all over the real line.
An "odd" periodic extension has it flip horizontally and vertically (across the y-axis) first and sticks it to either end of the function, onto an interval of length $2L$. So basically it's like the even extension, but you flip it upside down too, to give you an odd function. So if we have $\alpha \in [0,L]$, then $f(\alpha) = -f(- \alpha)$.
A more graphic example, using $f(x) = |x|$ over $[0,2]$:
Insofar as uniqueness is concerned, I believe when this came up in my class, it was insofar as the period of the function was concerned. Having a full copy of the question might be more helpful since this was a few months ago for me, though, just for clarity's sake.
But if so, if I remember correctly, was mostly a matter of considering two functions, $f,g$, which were the same function but you chose a different period for each (say $p$ for one and $kp, k \in \mathbb{Z}$ for another) and then worked through the formulas and definitions to show that the choice of period was irrelevant by both functions ultimately having the same Fourier expansion.