Fourier coefficients for Steklov function I'm solving problems about Fourier series, but I cannot solve this one, could anyone give me an idea to solve it?

Let $\{c_n\}$ be the Fourier coefficients of a function $f\in L_1(T,dt)$. Find the Fourier coefficients $\{c_n(h)\}$ for the (Steklov) smoothed function $f_h=\frac1{2h}\int_{x-h}^{x+h} f(\xi)\,d\xi$

 A: Assuming $c_k(f)$ are the coefficients with respect to the Hilbert basis (of $L^2$) $\left(\frac{\sqrt T}{T}\exp\left(\frac{2ik\pi}Tx\right)\,:\, k\in\Bbb Z\right)$
\begin{align}&c_k(f_h)=\frac{\sqrt T}{2hT}\int_0^T\exp\left(-\frac{2ki\pi}Tx\right)\int_{x-h}^{x+h}f(\xi)\,d\xi\,dx=\qquad\text{(Fubini-Tonelli)}\\&=\frac{\sqrt T}{2hT}\int_{\{(x,\xi)\in\Bbb R^2\,:\, 0<x<T\wedge \lvert \xi-x\rvert<h\}}f(\xi)\exp\left(-\frac{2ki\pi}Tx\right)\,d\xi dx=\\&=\frac{\sqrt T}{2hT}\int_{\{(x,\xi)\in\Bbb R^2\,:\, -h<\xi<T+h\wedge \max(0,\xi-h)<x<\min(\xi+h,T)\}}f(\xi)\exp\left(-\frac{2ki\pi}Tx\right)\,d\xi dx=\\&=\frac{\sqrt T}{2hT}\int_{-h}^{T+h}f(\xi)\int_{\max(0,\xi-h)}^{\min(T,\xi+h)}\exp\left(-\frac{2ki\pi}Tx\right)\,dx\,d\xi=\qquad(\text{if }k\ne0)\\&=\frac{\sqrt T}{2hT}\int_{-h}^{T+h}\frac{T}{2ki\pi}f(\xi)\left(\exp\left(-\frac{2ki\pi}{T}\max(0,\xi-h)\right)-\exp\left(-\frac{2ki\pi}{T}\min(T,\xi+h)\right)\right)\,d\xi=\\&=\frac{\sqrt T}{4hki\pi}\int_{-h}^{T+h}f(\xi)\left(\exp\left(-\frac{2ki\pi}{T}\max(0,\xi-h)\right)-\exp\left(-\frac{2ki\pi}{T}\min(T,\xi+h)\right)\right)\,d\xi\end{align}
Now, the integral splits into $A-B$ with


*

*$A=\int_{-h}^{T+h} f(\xi)\exp\left(-\frac{2ki\pi}{T}\max(0,\xi-h)\right)\,d\xi=\\=\int_{h}^{T+h} f(\xi)\exp\left(-\frac{2ki\pi}{T}(\xi-h)\right)\,d\xi+\int_{-h}^hf(\xi)\,d\xi=\\=\int_{h}^{T+h}f(\xi)\exp\left(-\frac{2ki\pi}{T}(\xi-h)\right)\,d\xi+f_h(0)=f_h(0)+e^{2\pi i kh/T}\int_{h}^{T+h}f(\xi)e^{-2\pi i k\xi/T}\,d\xi=\\=f_h(0)+\sqrt Te^{2\pi ikh/T}c_k(f)$

*$B=\int_{-h}^{T+h} f(\xi)\exp\left(-\frac{2ki\pi}{T}\min(0,\xi+h)\right)\,d\xi=\\=\int_{-h}^{T-h} f(\xi)\exp\left(-\frac{2ki\pi}{T}(\xi+h)\right)\,d\xi+\int_{T-h}^{T+h}f(\xi)e^{2ik\pi T/T}\,d\xi=\\=\int_{-h}^{T-h}f(\xi)\exp\left(-\frac{2ki\pi}{T}(\xi+h)\right)\,d\xi+f_h(T)=f_h(0)+e^{-2\pi ik h/T}\int_{h}^{T+h}f(\xi)e^{-2\pi i k\xi/T}\,d\xi=\\=f_h(0)+\sqrt Te^{-2\pi ikh/T}c_k(f)$
Thus, for $k\ne 0$, $$c_k(f_h)=\frac{\sqrt{T}}{4hki\pi}\left(f_h(0)+\sqrt Te^{2kih\pi/T}c_k(f)-f_h(0)-\sqrt T e^{-2khi\pi /T}c_k(f)\right)=\\=\frac{T}{2hk\pi}c_k(f)\sin\frac{2kh\pi}T$$
I leave to you the calculation of $c_0(f_h)$. I also advise you to actually do the exercise on your own with my work as a roadmap, because I am very likely to have made some mistake of calculation, despite the idea being substantially correct.
