# Properties of semidiscrete Fourier transform

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.