Exercise $10$ of Chapter $4$ in Fourier Analysis by Stein & Shakarchi I am stuck on the following exercise in Stein & Shakarchi's book on Fourier analysis.
Exercise 10a, chapter 4:
Let $\{\xi_n \}$ be a equidistributed sequence on $[0,1)$. If $f$ is a continuous periodic function with period 1 and $\int_{0}^1 f(x) dx =0$, then:
$$ \lim_{N\rightarrow\infty} \frac{1}{N} \sum^{N}_{n=1} f(x+\xi_n) = 0,\qquad \mbox{uniformly in } x\in[0,1).$$
My partial solution:
We must show that for every $\epsilon>0$, there exists a $N_1\in\mathbb{N}$ for which:
$$ \sup_{x\in[0,1)} \left | \frac{1}{N} \sum^{N}_{n=1} f(x+\xi_n) \right | < K\epsilon, \qquad \forall N \geq N_1. $$
Choose a trigonometric polynomial $P$ with $\int_{0}^1 P(x) dx =0$ and $|f(x)-P(x)|<\epsilon$ for all $x\in[0,1)$. This is possible because continuous functions on the circle can be uniformly approximated by trigonometric polynomials. Using the common trick:
$$ \sup_{x\in[0,1)} \left | \frac{1}{N} \sum^{N}_{n=1} f(x+\xi_n) \right | \leq \sup_{x\in[0,1)} \left | \frac{1}{N} \sum^{N}_{n=1} f(x+\xi_n) -P(x+\xi_n)\right| +  \left| \frac{1}{N} \sum^{N}_{n=1} P(x+\xi_n) \right |  $$
we see that:
$$ \sup_{x\in[0,1)} \left | \frac{1}{N} \sum^{N}_{n=1} f(x+\xi_n) -P(x+\xi_n)\right| \leq    \frac{1}{N} \sum^{N}_{n=1} \sup_{x\in[0,1)} \left | f(x+\xi_n) -P(x+\xi_n)\right| \leq \epsilon. $$
Furthermore, writing:
$$ P(x) = \sum^{M}_{k=-M} c_k e^{2\pi i kx} $$
we obtain:
\begin{align} \sup_{x\in[0,1)} \left | \frac{1}{N} \sum^{N}_{n=1} P(x+\xi_n) \right |  & = &\sup_{x\in[0,1)} \left | \sum^{M}_{k=-M} c_k e^{2\pi ikx} \left(\frac{1}{N} \sum^{N}_{n=1} e^{2\pi i k\xi_n} \right) \right |\\
 & \leq & \left(\sup_{x\in[0,1)} \sum^{M}_{k=-M} |c_k e^{2\pi i kx} | \right) \left(\max_{-M \leq k \leq M}  \left|\frac{1}{N} \sum^{N}_{n=1} e^{2\pi i k\xi_n} \right|\right) \end{align}
By Weyl criterion, we can choose $N_1\in \mathbb{N}$ such that:
$$ \left(\max_{-M \leq k \leq M}  \left|\frac{1}{N} \sum^{N}_{n=1} e^{2\pi i k\xi_n} \right|\right)  \leq \epsilon\qquad \forall N_1 \geq N.$$
What I am stuck on:
I can proceed by bounding:
$$\left(\sup_{x\in[0,1)} \sum^{M}_{k=-M} |c_k e^{2\pi i kx} | \right) \leq \sum^{M}_{k=-M} |c_k | := C $$
And we would have:
$$ K =1+C$$
But $C=C(\epsilon)$ and can grow unboundedly as $\epsilon\rightarrow 0$. So the way my proof is set up is simply not correct. Can anybody tell me what I should be doing instead?
 A: Let ${\displaystyle P(x)=\sum_{k=-M}^{M}c_k e^{2\pi i k x} }$. Then
\begin{align}
\frac{1}{N}
\sum_{n=1}^N
P(x+\xi_n) 
&=
\frac{1}{N}
\sum_{n=1}^N
\sum_{k=-M}^{M}c_k e^{2\pi i k (x+\xi_n)}\\
&=
\sum_{k=-M}^{M}
c_k
\frac{1}{N}
\sum_{n=1}^N
e^{2\pi i k (x+\xi_n)}\\
&=
\frac{c_0}{N}
+
\sum_{k\neq 0}
c_k
e^{2\pi i k x}
\frac{1}{N}
\sum_{n=1}^N
e^{2\pi i k \xi_n}\\
\end{align}
taking the limit as $N\rightarrow \infty$, Weyl's criterion gives $0$.
Note 1: to show uniformity take absolute value:
\begin{align}
\left|
\frac{1}{N}
\sum_{n=1}^N
P(x+\xi_n) 
\right|
&\leq
\left|
\sum_{k\neq 0}
c_k
e^{2\pi i k x}
\frac{1}{N}
\sum_{n=1}^N
e^{2\pi i k \xi_n}
\right|\\
&\leq
\sum_{k\neq 0}
\left|
c_k
e^{2\pi i k x}
\right|
\left|
\frac{1}{N}
\sum_{n=1}^N
e^{2\pi i k \xi_n}
\right|\\
&=
\sum_{k\neq 0}
\left|
c_k
\right|
\left|
\frac{1}{N}
\sum_{n=1}^N
e^{2\pi i k \xi_n}
\right|\\
\end{align}
The bound converges to 0, independently of $x$
Note 2: for general $f$ approximate $f$ by its Abel means. Theorem 5.6 states that the Fourier series of a continuous function is uniformly Abel summable to $f$.
A: It is asked to show that the average of the values $f(x+\xi_n)$ gives a good approximation to the integral $\int_0^1 f(x) dx = 0$, as the number of considered points $x+\xi_n$ (which lie in the interval $[x, x+1)$) increases.
So answer for yourself: Why do these values approximate the integral and why does it get better with a higher number of points? As $\xi_n$ is a random variable, you should calculate an expected value.
