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In the accepted answer here, the claim is made that trigonometric polynomials are dense in $1$ periodic continuous functions.

I take this to mean for example, that if we consider

$$f(x) = x-\lfloor x \rfloor$$

we can approximate it to an arbitrary degree with an expression of the form:

$$f(x) = \operatorname{Re}\left( \sum_m c_m e^{2\pi i m} \right)$$

Is my understanding of the statement correct? And how would I go about proving it?

EDIT: It was pointed out that the fractional part function isn't continuous. However, I've seen others recommending approximating it with trigonometric polynomials as well. See here for example. The question then becomes, can we prove that the fractional part function can be approximated with trigonometric polynomials?

vitamin d
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Rohit Pandey
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    I don't think your function is continuous... – Jose27 Jul 05 '21 at 22:12
  • You're right. In that case, I guess the proof at the link provided doesn't hold? – Rohit Pandey Jul 05 '21 at 22:44
  • Since Fourier series can be generated, the partial sums will be a good approximation, except at the jump. (Gibbs's phenomenon). – herb steinberg Jul 05 '21 at 23:55
  • I'm not sure why you want to approximate the fractional part function. It seems to me that the periodicity of the $g$ in the answer you cite is taking care of this $\text{mod}, 1$ operation, while the fact that convergence takes place for every continuous $g$ is equivalent to the convergence in distribution that they claim. Maybe I'm misunderstanding something here. – Jose27 Jul 05 '21 at 23:58
  • @Jose27 - the claim in the proof was made of trigonometric polynomials. But it was used for the mod 1 function. Doesn't this mean we should be able to approximate the mod 1 function arbitrarily with trigonometric polynomials for it to work? – Rohit Pandey Jul 06 '21 at 00:01
  • @herbsteinberg - does this jump pose a problem in the linked proof? – Rohit Pandey Jul 06 '21 at 00:05
  • If they were using it for what you say, the only thing they'd be able to conclude is that $\mathbb{E}(S_n-\floor{S_n})$ converges to the right thing, which is not enough to prove convergence in distribution. I'd suggest going over the Portmanteau lemma. – Jose27 Jul 06 '21 at 01:03
  • @RohitPandey Sorry - I haven't tried to examine the linked proof (and I wont - too long). Fourier series converges to midpoint at jump. – herb steinberg Jul 06 '21 at 17:04

1 Answers1

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This is more a comment or observation for the OP, but it fits better in this section.

The answer in the link you are referring to shows that, under a non degeneracy condition $|\gamma_m|<1$, $m\in\mathbb{N}$, (which fails for instance when $X_1$ is supported in $\mathbb{Z}$), for any continuous function $f$ on the circle $\mathbb{S}^1$ $$E[f(e^{2\pi iS_n})]\xrightarrow{n\rightarrow\infty}\int^1_0f(e^{2\pi ix})\,dx$$ This is equivalent to saying that the distribution of $e^{2\pi iS_n}$ (which is a random variable in $\mathbb{S}^1$) converges weakly to the uniform distribution on $\mathbb{S}^1$, that is, the normalized arc-length measure on the circle $\mathbb{S}^1$.

The identification of $\mathbb{S}^1$ with $\mathbb{R}/\mathbb{Z}$ allows us to be a little loose and consider functions on $\mathbb{S}^1$ as $1$ periodic functions on $\mathbb{R}$, or as functions on $\mathbb{R}\operatorname{mod}1$. The geometric picture is the of wrapping the real line around the circle $\mathbb{S}^1$, but of course this can be done rigorously with simple topological arguments.

To conclude rigorously that the law of $S_n-\lfloor S_n\rfloor$ converges weakly to the uniform distribution $U(0,1)$, one may use the following observations:

  1. $e^{2\pi ix}=e^{2\pi i\{x\}}$ for all $x\in\mathbb{R}$, where $\{x\}:=x-\lfloor x\rfloor$.
  2. The function $\phi:[0,1)\rightarrow\mathbb{S}^1$ given by $x\mapsto e^{2\pi ix}$ is continuous and bijective; the inverse $\phi^{-1}:\mathbb{S}^1\rightarrow[0,1)$ which maps $z=e^{2\pi i\theta}\mapsto\{\theta\}$ is measurable and continuous in $\mathbb{S}^1\setminus\{1\}$. The point of discontinuity has arc-length measure $0$.

Recall the following result from Probability (Billingsley, P., Convergence of Probability measures, Wiley 1968)

Theorem: Suppose $(S,d)$, $(S',d')$ are metric spaces, and that $\{\mu_n,\mu:n\in\mathbb{N}\}$ are Borel measures on $(S,d)$ such that $\mu_n\stackrel{n\rightarrow\infty}{\Longrightarrow}\mu$. If $f:S\rightarrow S'$ is a measurable function that is continuous $\mu$-almost surely (i.e. the set of discontinuities $D_f$ of $f$ has measure $\mu(D_f)=0$), then $\mu_n\circ f^{-1}\stackrel{n\rightarrow\infty}{\Longrightarrow}\mu\circ f^{-1}$, that is, for any $g\in\mathcal{C}_b(S')$, $$\lim_n\int_Sg(f(x))\,\mu_n(dx)\xrightarrow{n\rightarrow\infty}\int_S g(f(x))\,\mu(dx)$$

In terms of random variables, the result says that of $X_n$, $X$ are random variables with values in $S$ and $X_n\stackrel{n\rightarrow\infty}{\Longrightarrow} X$, then $f(X_n)\stackrel{n\rightarrow\infty}{\Longrightarrow}f(X)$

An application of the theorem above implies that $\phi^{-1}(e^{2\pi iS_n})=S_n-\lfloor S_n\rfloor \stackrel{n\rightarrow\infty}{\Longrightarrow} U(0,1)$, where $U(0,1)$ is the uniform distribution on the interval $[0,1)$.

Mittens
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