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Suppose that I have an integrable function $f$ on the circle $\mathbb{T}$ and $f$ is a $2 \pi $ periodic function on the real line. Using the fact that $\hat{f}(n)=\frac{1}{2 \pi} \int_{\mathbb{T}} f(t) e^{-i nt}dt,$ I want to show the following formula for a fixed $\tau \in \mathbb{T}$:

$\hat{f}(n-\tau)= \hat{f}(n) e^{-in \tau}$

Here is what I have so far:

$\hat{f}(n) e^{-i n \tau}=\frac{1}{2 \pi} \int_{\mathbb{T}} f(t) e^{-i nt} e^{-in \tau}dt =\frac{1}{2 \pi} \int_{\mathbb{T}}e^{-it (n-\tau+\tau)} e^{-in \tau} dt =\frac{1}{2 \pi} \int_{\mathbb{T}} f(t) e^{-it(n-\tau)} e^{-i \tau (t+n)} dt$

Since $\hat{f}(n-\tau)=\int_{\mathbb{T}}\frac{1}{2 \pi} f(t) e^{-it(n-\tau)} dt,$ it means that I need to have $e^{-i \tau (n+t)}=1$, but I do not see how This is possible.

Where is my mistake?

abc
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  • I believe there is a confusion in notation here: $\hat f(n - \tau)$ doesn't really make sense since $\hat f$ is a function on the integers and $n-\tau$ won't be an integer. Perhaps you want to instead take the transform of the function $g(t) = f(t-\tau)$ or something like this. – User8128 Jan 11 '18 at 23:29
  • yes, I have to calculate $\hat{f}{\tau}(n)$, where $f{\tau}(n)= f(n-\tau)$. However, I do not see what difference it makes. @User8128 – abc Jan 11 '18 at 23:31
  • @User8128 how should I solve this then? – abc Jan 11 '18 at 23:34
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    Well there are two different domains: there is a function $f: [0,2\pi) \to \mathbb C$ and a function $\hat f: \mathbb Z \to \mathbb C$. We can define $f_\tau(t) = f(t-\tau)$ since $t$ and $\tau$ live in the same space. It makes a difference: where you put the $\tau$ really matters. When transforming $f_\tau(t) = f(t-\tau)$, we have $$\hat f_\tau (n) = \frac{1}{2\pi} \int_\mathbb T f(t-\tau) e^{-int} dt$$ whence the correct result follows very easily from a change of variables. However, nowhere should there be an $n-\tau$; this doesn't make sense since $n$ and $\tau$ live in different spaces. – User8128 Jan 11 '18 at 23:39
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    You also shouldn't write $f_\tau(n) = f(n-\tau)$. While formally this isn't incorrect, it is very obfuscating because we typically use $n$ to denote integers and the domain of $f$ is not the integers, but the circle. – User8128 Jan 11 '18 at 23:41
  • ahhhhh thank you SO much @User8128! This makes a LOT more sense! – abc Jan 12 '18 at 00:16

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