Inverse fourier transform of kernel density estimator

Question

I have $$f_n(x)={(nh)}^{-1}\sum_{k=1}^n K\Bigg[\frac{x-X_k}{h}\Bigg]$$ I want to show that $$f_n(x)={(2\pi)}^{-1} \int_{-\infty}^\infty e^{-iux}k(hu)\psi_n(u) du$$

where $$\psi_n(u)=\int_{-\infty}^\infty e^{iux} dF_n (x)=n^{-1}\sum_{k=1}^n e^{iuX_k}$$ $$k(u)=\int_{-\infty}^\infty e^{-iuy}K(y) dy $$ $$K(y)={(2\pi)}^{-1}\int_{-\infty}^\infty e^{iuy}k(u) du $$

Answer 1

$$f_n(x)={(nh)}^{-1}\sum_{k=1}^n {(2\pi)}^{-1}\int_{-\infty}^\infty e^{iu(x-X_k)}k(hu) h.du$$ $$f_n(x)={(2\pi)}^{-1} {(n)}^{-1}\sum_{k=1}^n \int_{-\infty}^\infty e^{iux}e^{-iuX_k}k(hu) du$$ $$f_n(x)={(2\pi)}^{-1} {(n)}^{-1}\sum_{k=1}^n e^{-iuX_k} \int_{-\infty}^\infty e^{iux}k(hu) du$$ $$f_n(x)={(2\pi)}^{-1} \int_{-\infty}^\infty e^{-iux}k(hu)\psi_n(u) du$$ where $$\psi_n(u)=\int_{-\infty}^\infty e^{iux} dF_n (x)=n^{-1}\sum_{k=1}^n e^{iuX_k}$$