0
$\begingroup$

The following question is based on an argument in the SIAM book Spectral Methods in MATLAB by Trefethen.

For a function $v$ defined on $h\mathbb{Z}$ with value $v_j$ at $x_j=jh$, the semidiscrete Fourier transform is defined by $$ \hat{v}(k) = h\sum_{j=-\infty}^\infty e^{-ikx_j}v_j,\quad k\in[-\pi/h,\pi/h], $$ and the inverse semidiscrete Fourier transform is $$ v_j = \frac{1}{2\pi}\int_{-\pi/h}^{\pi/h}e^{ikx_j}\hat{v}(k)\ dk,\quad j\in\mathbb{Z}. $$

Define the interpolant $p$ by $$ p(x) = \frac{1}{2\pi}\int_{-\pi/h}^{\pi/h}e^{ikx}\hat{v}(k)\ dk,\quad x\in{\mathbb{R}}. $$ It is obvious by definition that $p(x_j)=v_j$. The author claims without a proof that $$ \hat{p}(k) = \begin{cases} \hat{v}(k),&k\in[-\pi/h,\pi/h],\\ 0,&\text{otherwise} \end{cases}\tag{1} $$ where $$ \hat{p}(k) := \int_{-\infty}^\infty e^{-ikx} p(x)\ dx, \quad k\in\mathbb{R}. $$

I think this is a rather standard result but I fail to see why. Neither can I find any other references for this.

Question: how can I get (1)?


I simply tried to follow the definition and got $$ \hat{p}(\xi) = \int_{-\infty}^\infty e^{-i\xi x} \left( \frac{1}{2\pi}\int_{-\pi/h}^{\pi/h}e^{ikx}\hat{v}(k)\ dk \right)\ dx. $$ Exchanging the integral seems not helpful.

$\endgroup$

Your Answer

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