Exercise $10$ of Chapter $4$ in Fourier Analysis by Stein & Shakarchi

Solution 1:

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 $f$.