Difference between $n$-th partial sums of Fourier series

Question

Let $f\in\mathcal{C}(\left[-\pi, \pi\right])$ and let $$S_n(f, x) = \sum_{h= -n}^n \widehat{f}(h) e^{ihx}.$$ Moreover, let $g(t):= f(x+t) - f(x)$. It is true that $$ S_n(f, x) - S_n(g, 0) = f(x)?$$ Could anyone help me with the proof?

Thank you.

Answer 1

We have

$$S_n(g,t) = -f(x) + \sum_{h=-n}^n \hat{f}(h)e^{ihx}e^{iht}$$

from the Fourier translation property. This immediately leads to

$$S_n(f,x)-S_n(g,0) = f(x)$$

Answer 2

$$\widehat{g}(h)=\frac{1}{2\pi}\int_{-\pi}^{\pi}(f(x+t)-f(t))e^{-i ht}dt=\frac{e^{ihx}}{2\pi}\int_{-\pi+x}^{\pi+x}f(u)e^{-ihu}du-\widehat{f}(h)=(e^{ihx}-1)\widehat{f}(h)$$ because $u\mapsto f(u)e^{-ihu}$ is $2\pi$-peridoc. Thus $$ S_n(g,0)=\sum_{h=-n}^n\widehat{f}(h)(e^{ihx}-1)=S_n(f,x)-S_n(f,0) $$ But we don't have $S_n(f,0)=f(x).$