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Let $f \in L^1(\Bbb R)$ and $\operatorname{supp}\hat{f} \subset [-\Omega,\Omega], h< {\pi \over \Omega}$ and set $\tilde{D}_{h}f(x) := f(hx)$.

Now let $k \in \Bbb Z$, then $$\tilde{D}_{h}f(k) = ((\tilde{D}_{h}f)\hat{})\check{}(k) = {1 \over 2\pi} \int_{\Bbb T} (\tilde{D}_{h}f)\hat{}(\omega) e^{ik\omega} d\omega,$$ where $(\tilde{D}_{h}f)\hat{}(\omega) = \sum_{m \in \Bbb Z} f(hm)e^{-im\omega}$.

I know that $\{e^{ik\omega}: k \in \Bbb Z\}$ forms an orthonormal basis for $L^2([-\pi,\pi])$ which can be identified with the circle group $\Bbb T$ but I don't see how this is related to $\Bbb Z$-valued functions and thus don't understand the last two equations.

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  • $\begingroup$ Are you trying to understand why the equation in the second paragraph holds? $\endgroup$ – Hans Oct 1 '18 at 5:54
  • $\begingroup$ I am glad that you liked my answer and accepted it. However, though not a huge deal, I am a bit disappointed that you did not come to check on your question and my answer and accept the latter a couple of days ago in time to award your full bounty to me. $\endgroup$ – Hans Oct 6 '18 at 19:16
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$\operatorname{supp}\hat{f} \subset [-\Omega,\Omega] \implies \hat f(\omega)=\sum_k a_ke^{-ik\frac\omega\Omega\pi},\,\forall \omega\in[-\Omega,\Omega]$ due to the Fourier series expansion. But $a_k=\big(\int_{-\Omega}^\Omega=\int_{\Bbb R}\big)\hat f(\omega)e^{ik\frac\omega\Omega\pi}d\omega=f(k)$. So $\hat f(\omega)=\sum_k f(k)e^{-ik\frac\omega\Omega\pi},\,\forall \omega\in[-\Omega,\Omega]$. Going one step beyond the question, substitue the last equation into the Fourier transform and get a nice expression $$f(x)=\int_{-\Omega}^\Omega \hat f(\omega)e^{i2\pi\omega x}\,d\omega=2\sum_k f(k)\frac{\sin\big(\pi(2\Omega x-k)\big)}{\pi\big(2x-\frac k\Omega\big)}.$$ So an $L^1(\Bbb R)$ function with compactly supported Fourier transform is an interpolation of its values at integral points.

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