Difference between Cesàro mean of a function and the function itself I'm reading through my professor's class notes for Fourier Analysis and there is a remark that I can't verify.
He notes that:
$$[\sigma_n (f)](x) - f(x) = \frac1{2\pi}\int_{-\pi}^\pi [f(x+t) - f(x)]K_n(t)\;\mathrm dt.$$
Note that here $K_n$ is the Fejer kernel of $f$, and the factor of $2\pi$ comes from the fact that we scaled the $L_1$ norm by this factor which I have noticed is not always done.
I tried using the fact that $\sigma_n (f)$ is equal to $f*K_n$, which let me obtain:
$$[\sigma_n (f)](x) - f(x) = \frac1{2\pi}\int_{-\pi}^\pi f(x+t)K_n (t)\;\mathrm dt - f(x)$$
I can't see how to get the $K_{n}$ inside the integral, let alone attached to the factor of $K_{n}$.  Thanks for the patience, I just am learning how to use this site. :)
Note that the definition of $\sigma_n (f)$ is $\frac1{n}\sum\limits_{j=0}^{n-1}S_j (f)$, where $S_j (f) = \sum\limits_{m=-k}^k\widehat{f}(m) e^{imt}$  is the $j$-th partial sum of the Fourier series of $f$.

You simply use the fact that $\frac{1}{2 \pi} \int_{-\pi}^{\pi} K_n(t) \mathrm{d} t=1$. To see this, note the closed form expression for the Fejer kernel:
$$
   K_n(t) = \frac{1}{n} \sum_{k=0}^{n-1} D_k(t)
$$
where $D_k(t) = \sum_{n=-k}^{k} \exp(i t n)$. Now:
$$
  \frac{1}{2\pi}\int_{-\pi}^\pi K_n(t) \mathrm{d} t = \frac{1}{2 \pi n} \int_{-\pi}^\pi \sum_{k=0}^{n-1} D_k (t) \mathrm{d} t =  \frac{1}{2 \pi n} \sum_{k=0}^{n-1} \int_{-\pi}^\pi  D_k (t) \mathrm{d} t = \frac{1}{2 \pi n} \sum_{k=0}^{n-1}  (2 \pi) = 1
$$
where we used $\int_{-\pi}^\pi  D_k (t) \mathrm{d} t = \sum_{n=-k}^{k} \int_{-\pi}^\pi {\rm e}^{i t n} \mathrm{d} t = \sum_{n=-k}^{k} ( 2\pi \delta_{n,0} ) = 2 \pi$.
