# Fejér's Theorem (Problem in Rudin)

Can you solve Problem 19 from Chapter 8 of Rudin's Principles of Mathematical Analysis, I'm having a lot of difficulty with it

I've proven the first part, namely $$\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^N \exp(ik(x+n\alpha))=\frac{1}{2\pi}\int_{-\pi}^\pi(\cdots) = \begin{cases} 1\text{ if }k=0\\0\text{ otherwise}\end{cases}$$

Now I want to prove that if $f$ is continuous in $\mathbb{R}$ and $f(x+2\pi)=f(x)$ for all $x$ then

$$\lim_{N\to\infty} \sum_{n=1}^{N} \frac{1}{N} f(x+n\alpha)=\frac{1}{2\pi} \int\limits_{-\pi}^{\pi}f(t)\mathrm dt$$

for any $x$, where $\alpha/\pi$ is irrational.

I've tried writing it as

$$\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^N \sum_{k=0}^N\frac{1}{2\pi}\int_{-\pi}^\pi e^{ikt}f(x+n\alpha)$$ but that was not helpful.

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@stephen: What is the exercise no. – anonymous Oct 22 '10 at 8:05
@Stephen: Rudin wrote multiple books, what book do you mean? – Jonas Teuwen Oct 22 '10 at 8:54
Given what you have proved in the first part, have you tried to apply it to the second part? – Matt E Oct 22 '10 at 15:16
Hint: Can you prove the relation you are asking about for trigonometric polynomials? – Akhil Mathew Oct 22 '10 at 16:22
@Stephen. Which Rudin's book (and which edition) is this problem from? I looked thru two of my Rudin's books and couldn't find this problem. – TCL Nov 28 '10 at 15:59

For the sake of completeness, here's a solution. First I'll prove the lemma that the asker already could do. $$\lim_{N\to \infty}\frac{1}{N}\sum_{n=1}^N \exp(ik(x+n\alpha)) = \lim_{N\to \infty} \exp(ikx)\frac{1}{N}\sum_{n=1}^N \exp(ikn\alpha)$$ If $k=0$, the right hand side evaluates to $1\frac{1}{N}N = 1$. If $k\neq 0$, the sum is geometric, so we know how to evaluate it. $$\exp(ikx)\lim_{N\to \infty}\frac{1}{N} \frac{\exp((N+1)ik\alpha)-1}{\exp(ik\alpha)-1} = \frac{\exp(ikx)}{\exp(ik\alpha)-1}\lim_{N\to\infty}\frac{1}{N}(\exp((N+1)ik\alpha) -1)$$ Because $\alpha$ is an irrational multiple of $\pi$, $k\alpha$ is never an integer multiple of $2\pi$, so the denominator is nonzero. $\exp((N+1)ik\alpha)-1$ is bounded, so the limit evaluates to zero. This means that $$\lim_{N\to \infty} \frac{1}{N}\sum_{n=1}^N \exp(ik(x+n\alpha)) = \delta_{k0} = \frac{1}{2\pi}\int_{-\pi}^\pi \exp(ikt)dt$$ Where $\delta$ is the Kronecker delta.

Now the main problem asks us to show that $$\lim_{N\to \infty} \frac{1}{N}\sum_{n=1}^N f(x+n\alpha) = \frac{1}{2\pi} \int_{-\pi}^\pi f(t)dt$$ for any continuous $2\pi$-periodic function $f$ on the reals. If $f$ is a trigonometric polynomial, it follows easily from the result for $\exp(ikx)$. But we know that every continuous $2\pi$-periodic function is a uniform limit of trigonometric polynomials. If $f_1, f_2, \ldots$ is a sequence of trigonometric polynomials that converges uniformly to $f$, we know that for each $i$, $$\lim_{N\to \infty}\frac{1}{N}\sum_{n=1}^N f_i(x+n\alpha) = \frac{1}{2\pi}\int_{-\pi}^\pi f_i(t)dt$$ Standard theorems about uniform convergence then tell us that $$\frac{1}{2\pi}\int_{-\pi}^\pi f(t)dt = \frac{1}{2\pi}\int_{-\pi}^\pi \lim_{i\to \infty}f_i(t)dt = \lim_{i\to \infty} \frac{1}{2\pi}\int_{-\pi}^\pi f_i(t)dt$$ $$= \lim_{i\to \infty} \lim_{N\to \infty} \frac{1}{N}\sum_{n=1}^N f_i(x+n\alpha) = \lim_{N\to \infty} \frac{1}{N}\sum_{n=1}^N \lim_{i\to\infty}f_i(x+n\alpha) = \lim_{N\to\infty} \frac{1}{N}\sum_{n=1}^N f(x+n\alpha)$$

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Let $f:\mathbb{R}\longrightarrow{\mathbb{C}}$ be continuous and have period $2\pi$ , and $\alpha$ such that $\displaystyle\frac{\alpha}{\pi}$ is irrational. Then $\displaystyle\lim_{N \to\infty}{\displaystyle\frac{1}{N}\displaystyle\sum_{n=1}^N{f(x+n\alpha)}}=\displaystyle\frac{1}{2\pi}\displaystyle\int_{-\pi}^{\pi}f(x)dx$ .

Let us first prove this property for functions of the form $f(x)=e^{inkx}$ , where $k$ is an integer. If $k$ is an integer with $k\neq{0}$ , then since $\displaystyle\frac{\alpha}{\pi}$ is irrational we must have that $e^{ik\alpha}\neq{1}$ , thus for any $N\geq{1}$ we have $\displaystyle\sum_{i=1}^N{e^{kn\alpha}}=\displaystyle\frac{e^{k(N+1)\alpha}-1}{e^{k\alpha}-1}$ , then $\displaystyle\lim_{N \to\infty}{\displaystyle\frac{1}{N}\displaystyle\sum_{n=1}^N{e^{ik(x+n\alpha)}}}=\displaystyle\lim_{N \to\infty}{\displaystyle\frac{1}{N}\displaystyle\frac{e^{k(N+1)\alpha}-1}{e^{k\alpha}-1}}$ , but the sequence $\frac{e^{k(N+1)\alpha}-1}{e^{k\alpha}-1}$ is bounded; since $e^{k\alpha}-1\neq{0}$ , thus $\displaystyle\lim_{N \to\infty}{\displaystyle\frac{1}{N}\displaystyle\sum_{n=1}^N{e^{ik(x+n\alpha)}}}=\displaystyle\lim_{N \to\infty}{\displaystyle\frac{1}{N}\displaystyle\frac{e^{k(N+1)\alpha}-1}{e^{k\alpha}-1}}=0$ , for $k\neq{0}$ . If $k=0$ , it is clear that $\displaystyle\lim_{N \to\infty}{\displaystyle\frac{1}{N}\displaystyle\sum_{n=1}^N{e^{ik(x+n\alpha)}}}=1$.

But $\displaystyle\frac{1}{2\pi}\displaystyle\int_{-\pi}^{\pi}f(x)dx=0$ if $k\neq{0}$ , and $\displaystyle\frac{1}{2\pi}\displaystyle\int_{-\pi}^{\pi}f(x)dx=1$ if $k=0$ , so that $\displaystyle\lim_{N \to\infty}{\displaystyle\frac{1}{N}\displaystyle\sum_{n=1}^N{f(x+n\alpha)}}=\displaystyle\frac{1}{2\pi}\displaystyle\int_{-\pi}^{\pi}f(x)dx$ si $f(x)=e^{inkx}$ , for some integer $k$ .

Now let $P(x)$ be a trigonometric polynomial , then there exists $M\in{\mathbb{N}}$ such that $P(x)=\displaystyle\sum_{n=-M}^M{c_ne^{in}}$ for some $c_n\in{\mathbb{C}}$ ,then we will have, because of what it was shown above, that $\displaystyle\lim_{N \to\infty}{\displaystyle\frac{1}{N}\displaystyle\sum_{n=1}^N{P(x+n\alpha)}}=\displaystyle\frac{1}{2\pi}\displaystyle\int_{-\pi}^{\pi}P(x)dx$ , since this holds for all the summands of the polynomial.

Now let $f:\mathbb{R}\longrightarrow{\mathbb{C}}$ be continuous, of period $2\pi$ . Let $\epsilon>0$ . Beacause of the Stone-Weiestrass theorem there exists a trigonometric polynomial $P(x)$ such that $\left |{P(x)-f(x)}\right|<\epsilon$ $\forall{x\in{\mathbb{R}}}$ . But $\displaystyle\lim_{N \to\infty}{\displaystyle\frac{1}{N}\displaystyle\sum_{n=1}^N{P(x+n\alpha)}}=\displaystyle\frac{1}{2\pi}\displaystyle\int_{-\pi}^{\pi}P(x)dx$ , so there is $M\in{\mathbb{N}}$ with $\left |{\displaystyle\frac{1}{N}\displaystyle\sum_{n=1}^N{P(x+n\alpha)}-\frac{1}{2\pi}\displaystyle\int_{-\pi}^{\pi}P(x)dx}\right |<\epsilon$ $\forall{N\geq{M}}$ (1), but $\left |{\displaystyle\frac{1}{2\pi}\displaystyle\int_{-\pi}^{\pi}f(x)dx}-\displaystyle\frac{1}{2\pi}\displaystyle\int_{-\pi}^{\pi}P(x)dx\right |<\epsilon$ (2); since $\left |{P(x)-f(x)}\right |<\epsilon$ $\forall{x\in{\mathbb{R}}}$ , and also $\displaystyle\frac{1}{N}\left |{\displaystyle\sum_{n=1}^N{P(x+n\alpha)}-\displaystyle\sum_{n=1}^N{f(x+n\alpha)}}\right |\leq{\displaystyle\frac{1}{N}\displaystyle\sum_{n=1}^N{\left |{P(x+n\alpha)-f(x+n\alpha)}\right |}<\displaystyle\frac{N}{N}\epsilon=\epsilon}$ (3), thus for any $N\geq{M}$ , we have then $\left |{\displaystyle\frac{1}{N}\displaystyle\sum_{n=1}^N{f(x+n\alpha)}-\displaystyle\frac{1}{2\pi}\displaystyle\int_{-\pi}^{\pi}f(x)dx}\right |<3\epsilon$ , using the triangular inequality and the inequalities(1), (2), (3). Therefore $\displaystyle\lim_{N \to\infty}{\displaystyle\frac{1}{N}\displaystyle\sum_{n=1}^N{f(x+n\alpha)}}=\displaystyle\frac{1}{2\pi}\displaystyle\int_{-\pi}^{\pi}f(x)dx$ , since $\epsilon>0$ was arbitrary.

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I'm looking at this about a year after it was submitted.

Not sure how much this would help but if we write

$\lim_{N->\infty} \sum\limits_{n=1}^{N} f(x+n\alpha) = \lim_{N->\infty} \sum\limits_{n=1}^{N} f(x+n\alpha - 2\pi \lfloor \frac{x+n\alpha}{2 \pi} \rfloor)$. If you can show the values of $x+n\alpha - 2\pi \lfloor \frac{x+n\alpha}{2 \pi} \rfloor$ are uniformly distributed in $[0, 2 \pi]$, then I think you are done.

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