Integral problems using Weierstrass Approximation Theorem

Question

The question:

(a) Let $f$ be a continuous periodic complex valued function with period $2\pi $.Then prove that given $\epsilon >0,$ there exists a function $Q(x)=\sum_{k=-M}^{M}c_{k}e^{ikx}$ with $c_k\in \mathbb{C}$ and $M\in \mathbb{N}$ defined on $[-\pi ,\pi ]$ such that $\int_{-\pi }^{\pi }\left | f(x)-Q(x) \right |dx<\epsilon $.

(b)Suppose $f$ is a continuous function on $R^{1}$,$f(x+2\pi )=f(x)$, and $\frac{\alpha }{\pi }$ is irrational. Prove that $\lim_{n\rightarrow \infty }\frac{1}{N}\sum_{n=1}^{N}f(x+n\alpha )=\frac{1}{2\pi }\int_{-\pi }^{\pi }f(t)dt$

for every $x$. Hint: Do it first for $f(x)=e^{ikx}$.

My attempt:

(a) $f$ is continuous on $[-\pi ,\pi ]$, so $f$ is Riemann integrable on $[-\pi ,\pi ]$.

Given $\epsilon >0,$ there exists a continuous function g such that $\int_{-\pi}^{\pi}\left | f(x)-g(x) \right |dx<\epsilon $. I wonder how $Q(x)$ is approximated to $g(x)$ here.

(b) I used the hint and put $f(x)=e^{ikx}$ and found that both sides of the equality are $0$, but I don't know how to expand to the trigonometric polynomials and why the condition "$\frac{\alpha }{\pi }$ is irrational" is needed.

Answer 1

A) is a good exercise in using Stone-Weierstrass (https://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem). As noted in your attempt for any $\epsilon > 0$, there exists a $g$ continuous on $[-\pi,\pi]$ such that $ \int_{-\pi}^{\pi}|f(x)-g(x)| < \epsilon/2 $. Now by Stone-Weierstrass, we can uniformly approximate $g$ by a family of trigonometric polynomials. Indeed, this gives us a trigonometric polynomials on $[-\pi, \pi ]$ such that $\sup_{x}|\sum_{k=-N}^{N} a_ke^{ikx} - g(x)| < \epsilon/4 \pi$. Pulling all of these together with the triangle inequality, one has $ \int_{-\pi}^{\pi}|f(x)-\sum_{k=-N}^{N} a_ke^{ikx}| < \int_{-\pi}^{\pi}|f(x)-g(x)| + \int_{-\pi}^{\pi}|g(x)-\sum_{k=-N}^{N} a_ke^{ikx}| \leq \epsilon/2 + 2\pi*\epsilon/4\pi = \epsilon $ as desired.

B) After proving the hint, it should follow directly from A), as A) allows one to approximate any rieemen integrable function that is periodic with period $2\pi$ by a finite linear combination of $e^{ikx}$.

Answer 2

(a) To get your result, you can use either Stone-Weierstrass approximation theorem or the theory of summation kernels using the fact that the Fejér kernel is a summation kernel and that the convolution between Fejér kernel and an integrable function is a trigonometric polynomial.

(b) Actually, with the instruments you used to prove (a), you get that for all continuous $2\pi$-periodic $f$ and for all $\varepsilon>0$ there exists a trigonometric polynomial $P$ such that $$\sup_{t\in\mathbb{R}}|f(t)-P(t)|<\varepsilon$$ Then, if $\alpha,x\in\mathbb{R}$ and $m\in\mathbb{Z}$ such that $m\alpha\notin2\pi\mathbb{Z}$ you have: $$|\frac{1}{N} ∑_{n=1}^Ne^{im(x+nα)}|=|e^{imx} \frac{1}{N}(\frac{1-e^{im(N+1)α}}{1-e^{imα}}-1)|≤\frac{1}{N}(\frac{2}{|1-e^{imα}|}+1)\rightarrow0, N\rightarrow\infty.$$ So, if $\frac{α}{\pi}$ is irrational, $\forall m\in\mathbb{Z}, ((m\neq0)\implies (m\alpha\notin2\pi\mathbb{Z}))$.

Then, fix a continuous $2\pi$-periodic function $f$ and fix $x\in\mathbb{R}$. For each $k\in\mathbb{N}$, choose a trigonometric polynomial $P_k$, say $$P_k(t)=\sum_{m=-M_k}^{M_k}a_{m,k}e^{imt},$$ such that $$\sup_{t\in\mathbb{R}}|f(t)-P_k(t)|<1/k.$$ Observe that $$a_{0,k}=\int_{-\pi}^{\pi}P_k(t)\frac{dt}{2\pi}\rightarrow\int_{-\pi}^{\pi}f(t)\frac{dt}{2\pi}, k\rightarrow\infty.$$

Then: $$\limsup _{N\rightarrow\infty}{|\frac{1}{N} ∑_{n=1}^Nf(x+nα) -∫_{-π}^πf(t) \frac{dt}{2π}|}=\\ \limsup _{N\rightarrow\infty} {|\frac{1}{N} ∑_{n=1}^N(f(x+nα)-P_k(x+nα))+\frac{1}{N} ∑_{n=1}^N P_k(x+nα) -∫_{-π}^πf(t) \frac{dt}{2π}|}≤\\ {\limsup _{N\rightarrow\infty}{({\frac{1}{N} ∑_{n=1}^N|f(x+nα)-P_k(x+nα)|}) +∑_{m=-M_k, \\m≠0}^{M_k}(|a_{m,k}|\limsup _{N\rightarrow\infty}|\frac{1}{N} ∑_{n=1}^N e^{im(x+nα)}|)+{|a_{0,k}- ∫_{-π}^π f(t) \frac{dt}{2π}|}}} \\ \le {\frac{1}{k}+{|a_{0,k}- ∫_{-π}^π f(t) \frac{dt}{2π}|}}\rightarrow0, k\rightarrow\infty$$

and you have got the thesis.