I am reading a proof in harmonic analysis, and in the middle of the proof, the author seems to claim that
If $$S_{n}(f):=\sum_{k=-n}^{n}\widehat{f}(n)e^{ikx}=(D_{n}*f)(x),$$ where $D_{n}$ is the Dirichlet Kernel, and $$\sigma_{n}(f):=\dfrac{S_{0}(f)+\cdots+S_{n}(f)}{n+1}.$$ Then, $S_{n}(\sigma_{n}(f))=\sigma_{n}(f).$
He said:
For all $n\in\mathbb{N}$, we have that $\sigma_{n}(f)$ is a trigonometric polynomial of degree at most $n$,so $S_{n}(\sigma_{n}(f))=\sigma_{n}(f)$.
I understand that $\sigma_{n}(f)$ is a trigonometric polynomial, but why does this imply the equality?
Thank you!
Edit 1: [Complete Proof]
After read the answer from Thorogtt, I managed to prove the claim. This proof is exactly what Thorogtt suggested, I just add more details for future users who are interested in this question.
We know that $\sigma_{n}(f)$ is a trigonometric polynomial with degree at most $n$, so $S_{n}(\sigma_{n}(f))=\sigma_{n}(f)$ must be true.
Indeed, consider $$g(x):=\sum_{k=-n}^{n}c_{k}e_{k}(x),\ \text{where}\ e_{k}(x):=e^{ikx}.$$
Then, the Fourier coefficient is defined by \begin{align*} \widehat{g}(m):=\dfrac{1}{2\pi}\int_{-\pi}^{\pi}g(x)e^{-imx}dx&=\dfrac{1}{2\pi}\int_{-\pi}^{\pi}\Big(\sum_{k=-n}^{n}c_{k}e^{ikx}\Big)e^{-imx}dx\\ &=\dfrac{1}{2\pi}\sum_{k=-n}^{n}c_{k}\int_{-\pi}^{\pi}e_{k-m}(x)dx\\ &=\dfrac{1}{2\pi}\cdot 2\pi\cdot c_{m}\ \text{by the orthogonality}\\ &=c_{m}. \end{align*}
Therefore, the partial sum $$S_{n}(g):=\sum_{m=-n}^{n}\widehat{g}(m)e^{imx}=\sum_{m=-n}^{n}c_{m}e^{imx}=g(x),$$ as desired.
The $n$-th Fourier polynomial of a trigonometric polynomial of degree at most $n$ is the trigonometric polynomial itself. Say you have such a trigonometric polynomial given as $$g\colon\mathbb{R}\rightarrow\mathbb{C},\,x\mapsto\sum\limits_{k=-n}^nc_ke^{ikx},\qquad c_{-n},...,c_n\in\mathbb{C}.$$ It's a good exercise to check that $\hat{g}(k)=c_k$ for $k$ from $-n$ to $n$ and $0$ otherwise, so $S_ng=g$.