# Link between Fourier transform of $\Bbb Z$-valued function and $\Bbb T$

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.

• Are you trying to understand why the equation in the second paragraph holds? – Hans Oct 1 '18 at 5:54
• 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. – Hans Oct 6 '18 at 19:16

$$\operatorname{supp}\hat{f} \subset [-\Omega,\Omega] \implies \hat f(\omega)=\sum_k a_ke^{-ik\frac\omega\Omega\pi},\,\forall \omega\in[-\Omega,\Omega]$$ 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.