Identity in Fourier summation

Question

Suppose the following series: \begin{eqnarray} \sum_{k'}k'f_{k'} \end{eqnarray} where $f_{k'}$ are some Fourier coefficients that result from a periodic function $f(t+T)=f(t)$: \begin{eqnarray} f_{k'}=\frac{1}{T}\int_{0}^{T}dt e^{ik'2\pi t/T}f(t). \end{eqnarray} Is it true then that: \begin{eqnarray} \sum_{k'}k'f_{k'}=\frac{1}{T}\int_{0}^{T}dt e^{ik'2\pi t/T}f(t)\left(\sum_{k'}k'e^{ik'2\pi t/T}\right)=0, \end{eqnarray} that is, given that: \begin{eqnarray} g(t)=\left(\sum_{k'}k'e^{ik'2\pi t/T}\right)=2i\sum_{k'=1}^{+\infty}k'\sin(2\pi k' t/T)=0 \end{eqnarray} Is the above identity correct?

EDIT: Wolfram Alpha gives the wrong answer for the sum $g(t)$. So I would like to know how to determine $g(t)$ in closed form, if possible.

Answer 1

Amplifying a bit @DavidCUllrich's answer: ... by the way, using $k'$ as a summing index is a bit confusing (in the context of standard practice) when there's no $k$ without the prime... Especially dangerous if/when derivatives might be involved.

So, somewhat revising the notation, first, we have $f(x)=\sum_{k\in\mathbb Z} \widehat{f}(k)\,e^{2\pi ikx}$, under various hypotheses on periodic $f$.

In circumstances where it is legitimate to differentiate termwise (e.g., either for fairly smooth $f$, or interpreting the summating as converging in a Sobolev space...) $$ f'(x) \;=\; 2\pi i\cdot \sum_k k\cdot \widehat{f}(k)\cdot e^{2\pi ikx} $$ Then, if this had a pointwise-convergence sense, $$ \sum_k k\cdot \widehat{f}(k) \;=\; {f'(0)\over 2\pi i} $$ much as in @DavidC.Ullrich's answer. Again, the notational choice is considerably distracting... :)

EDIT: I can't speak for any software system's reasoning about why $\sum_k k\cdot e^{2\pi ikx}=0$ (nor in what sense this is claimed), but it is true that $\sum_k 1\cdot e^{2\pi ikx}$ is a Fourier series expansion of the periodic Dirac delta, the Dirac comb. Of course this does not converge pointwise anywhere, but it does converge (meaningfully and usefully) in a Sobolev space.

The distributional derivative is indeed $2\pi i\sum_k k\cdot e^{2\pi ikx}$, and this is absolutely correct in a Sobolev space, and the differentiation is likewise absolutely correct, if interpreted properly.

Since the (distributional) derivative of the Dirac comb is locally $0$ away from integers, we might decide to say that the corresponding Fourier series is $0$ away from integers... and since the (distributional) derivative is locally $0$ away from integers, we could also say that the Fourier series for the derivative is $0$ there (even though it does not converge pointwise at all, only in a Sobolev space).

Answer 2

If $f$ is smooth then modulo some constant $$\sum_{k'}k'f_{k'}=cf'(0).$$

So your main question amounts to asking whether $$f'(0)=0$$holds for every periodic smooth $f$.

Edit: So now we want to know what distribution the series $\sum ke^{ikt}$ converges to. This turns out to be very easy. I'm going to take advantage of the Littlewood Convention, to the effect that $2\pi=1$; you may want to clean up the following, inserting the missing constants:

$\newcommand\D{\mathcal D}$ $\newcommand\T{\Bbb T}$ $\newcommand{\ip}[2]{\langle #1,#2\rangle}$

Lemma The series $\sum_{k\in\Bbb Z}e^{ikt}$ converges to $\delta_0$ in the (weak) topology of $\D'(\T)$.

Proof. Say $s_n=\sum_{|k|\le n}$. If $\phi\in\D(\T)$ then $$\ip{s_n}\phi=\sum_{|k|\le n}\widehat \phi(k)\to \phi(0)=\ip{\delta_0}\phi,\quad(n\to\infty)$$which is exactly what it means to say $s_n\to\delta_0$ in $\D'(\T)$.

And now since differentiation is continuous on $\D'(\T)$ it follows that $$i\sum_{k\in\Bbb Z}ke^{ikt}=\delta_0',$$where by definition $$\ip{\delta_0'}\phi=-\phi'(0).$$