Let $f \in L^1(\mathbb{T})$ and $m \in \mathbb{N}$. Furthermore, define the dilation $f_m$ by $f_m(t) := f(mt)$ for any $t \in \mathbb{T}$. I would like to show that $$ \widehat{f_m}(n) = \begin{cases} \hat{f}(n/m) \ \ \ \ \ \ \text{if} \ m \ \text{divides}\ n\\ 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{otherwise}. \end{cases} $$ Letting $u = mt$, we have \begin{align*} \widehat{f_m}(n) &= \frac{1}{2\pi}\int_{-\pi}^{\pi}f(mt)e^{-int}\ dt\\ &= \frac{1}{2\pi}\frac{1}{m}\int_{-m\pi}^{m\pi}f(u)e^{-i(n/m)u}\ du \end{align*} I am not sure what to do next. It looks similar to $\hat{f}(n/m)$, but not quite there yet. I was wondering if anyone can help me with this problem.
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1Hint: split up the interval $[-m\pi,,m\pi]$ to $m$ equal parts. – Berci Sep 08 '20 at 16:12