# 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?

• Forget general $f$ to start with and prove it for a trig polynomial $P$. Use Weyl's criterion, p.112. Commented Jan 15, 2018 at 22:58
• That is what I was about to do, I've added some extra details now. But I don't see the extension to general f, due to the issues I've pointed out. Commented Jan 15, 2018 at 23:12

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 $$f$$.

• Yeah, but they are asking for uniform convergence. Commented Jan 15, 2018 at 23:15
• This is the idea, and you can show that it is uniform with a small modification Commented Jan 15, 2018 at 23:22
• How? That is the part I am stuck on... Commented Jan 15, 2018 at 23:23
• I mean I know how to show uniform convergence for a fixed polynomial P. But I dont see how to pass it on for general f, by approximating it with a sequence of polynomials P. The number of terms required to approximate f uniformly increases as you make $\epsilon$ smaller Commented Jan 15, 2018 at 23:26
• Why does the number of terms matter? Use the Fourier series of $f$ Commented Jan 15, 2018 at 23:38

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.

• The hint given to me in the book was: Establish this result result first for trigonometric polynomials. showing it for trigonometric polynomials was easy. But extending it to the general case seems a little bit harder than I thought. Commented Jan 15, 2018 at 22:50
• If it was easy for trigonometric polynomials, then you should have $\left| \frac{1}{N} \sum^{N}_{n=1} P(x+\xi_n) \right |<\epsilon$ immediately. Commented Jan 15, 2018 at 22:59