When is the derivative of a $2\pi$ periodic function $2\pi$ periodic?

Question

I have that the function $f:\mathbb{R}\rightarrow \mathbb{R}$ is a $2\pi$-periodic, defined on $\mathbb{R}$ with: $$f(x)=(x^2-\pi^2)^2, x\in[-\pi,\pi[$$ I calculated the Fourier coefficients as follows:

$$c_{k}=\cfrac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-ikx}dx=$$ $$c_{k}=\cfrac{1}{2\pi(ik)^{3}}\int_{-\pi}^{\pi}f'''(x)e^{-ikx}dx=$$ $$c_{k}=\cfrac{1}{2\pi(ik)^{3}}\int_{-\pi}^{\pi}24xe^{-ikx}dx=$$ $$c_{k}=-\frac{24}{k^{4}}(-1)^{k}$$ where I use the following:
I know that if $f$ is continously differentiable and $2\pi$-periodic function on $\mathbb{R} $ then $\mathcal{F}(f')_{k}=ik\mathcal{F}(f)_{k}\Leftrightarrow \mathcal{F}(f)_{k}=\cfrac{\mathcal{F}(f')_{k}}{ik}$.

I also know that:

$f(x)=(x^2-\pi^2)^2$
$f'(x)=4x(x^2-\pi^2)$
$f''(x)=4(3x^2-\pi^2)$
$f'''(x)=24x$
are all continously differentiable but i'm not sure if $f',f''$ are $2\pi $-periodic, because $f'(x+2pi)\neq f'(x)$. Can anyone help me with this?

Answer 1

This has nothing to do with fourier series. If a function is derivable at $a$ and periodic with period $T$ it will be derivable at $a+T$ with the same derivate as at $a$:

$$f'(a+nT) = \lim_{h\to0}{f(a+nT+h)-f(a+nT)\over h} = \lim_{h\to 0}{f(a+h)-f(a)\over h} = f'(a)$$

Here only the periodicity is used. We see from this that the derivate is periodic with the same period.

Answer 2

Let's define $f_0(x)=(x^2-\pi^2)^2$ for $x\in[-\pi,\pi]$ and complete by periodicity.

Let's call $f_k(x)$ the expression of $f$ on interval $[(2k-1)\pi,(2k+1)\pi]$.

For $y\in[(2k-1)\pi,(2k+1)\pi]\quad f_k(y)=f_0(x)\quad$ for $x=y-2k\pi\in[-\pi,\pi]$

$$f_k(y)=((y-2k\pi)^2-\pi^2)^2$$

$$f_k'(y)=4(y-2k\pi)((y-2k\pi)^2-\pi^2)$$

So $f_k'$ has not the same expression on each of the intervals than $f'_0$.

Your mistake is comparing $f_0'(x+2\pi)$ and $f_0'(x)$ instead of $f_1'(x+2\pi)$ and $f_0'(x)$.

The reason $f'$ has the same period than $f$ is explained in skyking post, but it just says $f_k'(x+2\pi)=\left(f_k(x+2\pi)\right)'=\left(f_0(x)\right)'=f_0'(x)$

Another disturbing thing in your post is that you say continuously differentiable (on supposedly whole $\mathbb R$) but never explicitly say that the joins are continuous.

In fact $f^{(n)}(-\pi)=f^{(n)}(\pi)$ for $n=0,1,2$ but not for $n=3$.

So considering periodicity $f$ is $C^2$ on $\mathbb R$ but is only piecewise $C^3$ on $\bigcup\limits_{k\in\mathbb Z}](2k-1)\pi,(2k+1)\pi[$.