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Let $f \in C[0,1]$. For $N \in \mathbb{N}$, define a function $f_N: \{0, 1, \dotsc, N-1\} \to \mathbb{C}$ by $f_N(k) := f(k/N)$.

To show:$\hat{f}(n) = \lim_{N \to \infty}\widehat{f_N}(n)$.

I tried to solve this but writing out the Fourier coefficient $\widehat{f_N}(n)$, but I can't seem to find the relation between $\hat{f}(n)$ and $\widehat{f_N}(n)$. The Fourier coefficient $\widehat{f_N}(n)$: $$\widehat{f_N}(n) = \frac{1}{N}\sum_{m = 0}^{N-1}f_N(m)e^{-2\pi i mn/N} = \frac{1}{N}\sum_{m = 0}^{N-1}f(m/N)^{-2\pi i mn/N}, \ \ \text{for} \ N \in \mathbb{N}.$$

But $$\hat{f}(n) = \int_0^1f(t)e^{-2\pi i nt}dt.$$ When $N$ is large, the sum becomes integral? Any hints would be appreciated.

Vicky
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    Riemann sums converge to the integral. Just partition $[0,1]$ into $N$ equal parts. – Kavi Rama Murthy Oct 21 '20 at 09:01
  • Right, but I'm not sure if what I'm doing is correct. So, I partition $[0,1]$ into $N$ equal parts. We have for $k \in {0, 1, \dotsc, N-1}$, $$f(k/N)\exp\left(-2\pi i nkt/N \right).$$ Also each length is $1/N$. Summing $$\frac{1}{N}f(k/N)\exp\left(-2\pi i nkt/N \right)$$ over $k$, we get basically a Riemann sums. Is the reasoning correct? – Vicky Oct 21 '20 at 09:15
  • Yes, it is correct. – Kavi Rama Murthy Oct 21 '20 at 09:16

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