Problem on convergence of Cesaro mean of Fourier series I'm working through Anton Deitmar's A First Course in Harmonic Analysis. And I got stuck at one of the problems:


Using dominated convergence theorem, prove that $\lim_{n\to \infty} \sigma_n f(x) = f(x) $.


Where I have already proved $$\sigma_n f(x) - f(x)= \int_0^1 \frac 12 [f(x+y) + f(x-y) - 2f(x)] F_n(y) dy$$
Where $F_n = \frac 1{n+1} (D_0 + \dots+ D_n) = \frac 1{n+1} (\frac{\sin((n+1)\pi x)} {\sin \pi x}) ^2$
, and $D_n = \sum_{k=-n} ^{n} e^{2\pi ikx}$
I haven't learned dominated convergence theorem yet, but it seems that I'm supposed to prove the limit by proving that the above integral tends to 0. I already know the sequence of integrand tends to the zero function. But I can't find a single bounding function to the sequence because $F_n(0) = F_n(1) =n$. Thanks in advance. 
 A: Dominated Convergence Theorem: Let $\{f_{k}\}$ be a sequence of measurable functions converging pointwise a.e. to $f$. If $|f_{k}(x)| \le g(x)$ for all $k$, such that $g \in L^{1}(\mathbb{R})$, then $f \in L^1(\mathbb{R})$ and 
$$\int f = lim_{k\to \infty} \int f_{k} =  \int lim_{k\to \infty} f_{k}$$
In essence, dominated convergence tells you when you can and cannot take the limit inside the integral.
Now onto your question. 
$$\sigma_{N}(f)(x) - f(x) = \int_{0}^{1} \frac{1}{2} (f(x+y) + f(x-y) - 2f(x)) F_{N}(y) dy$$
Observe that 
$$\frac{1}{n+1}\Big|(f(x+y) + f(x-y) - 2f(x)) \frac{sin^2(n+1)\pi x}{sin^2(\pi x)}\Big| \le \Big| \frac{4}{n+1} \frac{f(x)}{sin^{2} (\pi x)}\Big| $$ which is in $L^1(\mathbb{R})$.
Therefore 
$$\lim_{N\to\infty} \int_{0}^{1} \frac{1}{2} (f(x+y) + f(x-y) - 2f(x)) F_{N}(y) dy $$
$$ = \int_{0}^{1} \lim_{N\to\infty} \Big[ \frac{1}{2} (f(x+y) + f(x-y) - 2f(x)) F_{N}(y)\Big] dy $$
And use the fact that $lim_{N\to\infty} F_{N}(x) = 0$ (easy to observe). 
This is a good reference.
