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

Question

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?

Answer 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 $f$.

Answer 2

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.