0
$\begingroup$

Let $\gamma:\mathbb{R}\to\mathbb{C}$ be a continuous periodic curve having a bounded variation.

1) Is it true that one can find a sequence of numbers $(r_n)_{n\in\mathbb{N}}\subset (0,\infty)$ and some function $\varphi:\mathbb{R}\times \mathbb{R}\to\mathbb{R},\ \varphi\in C^{1}$ such that the sequence of curves:

$$\gamma_n:\mathbb{R}\to\mathbb{C},\ \gamma_{n}(t)=\sum_{k=0}^n r_k e^{i\varphi(k,t)}$$

converges uniformly to $\gamma$ on $\mathbb{R}$? Moreover can we choose $(r_n)_{n\in\mathbb{N}}$ to be decreasing or strictly decreasing? Can we choose $\varphi (x,t)=2x\pi t+f(x),\forall (x,t)\in\mathbb{R}\times \mathbb{R}$, where $f$ is a $C^1$ function?

2) Is it true that there is a function $r:\mathbb{R}\to \mathbb{C},\ r\in L^1 (\mathbb{R})$ such that $r(k)=r_k e^{if(k)},\ \forall k\in\mathbb{N}$ and

$$\gamma(t)=\int_{-\infty}^{\infty}r(x)e^{i2\pi x t} dx\ ?$$

I'm trying to prove this theorem: http://www.u.arizona.edu/~aversa/scholastic/Mathematical%20Power%20of%20Epicyclical%20Astronomy%20%28Hanson%29.pdf

See the begining of the last page. You can take a look also, at:https://www.youtube.com/watch?v=QVuU2YCwHjw

$\endgroup$
  • 2
    $\begingroup$ The context confuses me, even if it confuses no one else: a "curve" as posed is exactly a complex-valued function on the circle, so certainly has a (complex-valued) Fourier series, and these things have been studied for a long time. I'd think that imbedding it in that context, or at least talking in a way that allows reference to it, would help people answer usefully... $\endgroup$ – paul garrett Jun 17 '15 at 23:40
0
$\begingroup$

The answer to (1) is yes, due to the Jordan criterion for convergence of the Fourier series.

The answer to (2) is no, since by the Riemann-Lebesgue lemma $\gamma$ cannot be periodic (or even almost periodic) and be the (inverse) Fourier transform of an $L^1$ function.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.