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It is mentioned here that:

Let $R$ be the line parameterized by $x$. Let $f$ be a complex function on $R$ that is integrable. The Fourier transform $\hat f = Ff$ is

$$\hat f(k) = \int_{-\infty}^{+\infty} e^{-ikx} f(x) dx$$

It is a function on the dual real line parameterized by $k$.

I understand from the discussions here: Can someone clearly explain the discrete Fourier transform (DFT)? and here: Fourier Transform: Understanding change of basis property with ideas from linear algebra that $\hat f(k)$ is the projection of $f(x)$ onto the basis of exponential functions $e^{-ikx}$, i.e., $(f, e_k)$ and $f = (f, e_k)e_k$.

However, it is not clear to me why $k$ itself should belong to a dual real line, "dual" to the real line parameterized by $x$. Can someone explain?

Edit: There's a similar claim on Wikipedia:

That is to say, there are two versions of the real line: one which is the range of t and measured in units of t, and the other which is the range of ξ and measured in inverse units to the units of t. These two distinct versions of the real line cannot be equated with each other. Therefore, the Fourier transform goes from one space of functions to a different space of functions: functions which have a different domain of definition.

S.D.
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  • The simplest incarnation of this is to look at what happens when you scale $x$ by a factor. You’ll find that you have to scale the Fourier variable by the inverse factor. – Qiaochu Yuan Sep 25 '22 at 22:35
  • @QiaochuYuan Interesting, that makes sense. Can this idea also be generalized to (say) an $n$-dimensional lattice as in https://en.wikipedia.org/wiki/Reciprocal_lattice#Mathematical_description? I mean it's not mathematically clear to me why the Fourier variable should lie in the dual vector space of the original variable, although I understand your scaling up $x$ by a factor $c$ idea (in that case, the Fourier variable $k$ must be scaled down by $c$). – S.D. Sep 25 '22 at 22:42
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    Well, how would you even define the Fourier transform for a function on a finite-dimensional vector space $V$? What replaces that $kx$ in the $e^{ikx}$ factor in the Fourier transform? The answer is that in order to write a meaningful operation down without choosing any additional structure you must take $e^{i \langle k, x \rangle}$ where $\langle k, x \rangle$ is the dual pairing between $k \in V^{\ast}$ and $x \in V$. You'll end up getting a generalization of what I wrote above if you then try applying a linear transformation in $V$ and seeing its effect on $k$. – Qiaochu Yuan Sep 26 '22 at 01:17

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