Is $f_n(x)=\cos (\pi n!x)^{2n}$ uniformly convergent on $[0,1]$

Question

Is $f_n(x)=\cos (\pi n!x)^{2n}$ uniformly convergent on $[0,1]$

I could see that for $x\in \mathbb{Q} \cap[0,1]$ we have $f_n(x) \to 1$ and for irrational I think same is true!, but how do I check uniform Convergence?

Answer 1

Suppose that $x = a_0 + \sum_{n=2}^\infty \frac{a_n}{n!}$, where $a_0$ is an integer, and $a_n \in\{0,1,\dots,n-1\}$. So in effect, we are writing $x$ in base $n!$.

Then $\cos(\pi n! x) = \cos(\pi (a_{n+1}/(n+1) + O(n^{-1}))$. Thus, by analogy with considering the $n$th digit of a number $y$, you can see that in any small neighbourhood that you can make $\cos(\pi n! x)$ converge to anything you like.

Putting things to the power of $2n$ complicates things a little, but not much. If $a_n \sim \sqrt n$ for $n$ even, and $a_n = 0$ is odd, then $\cos(\pi n! x)^{2n} \sim e^{-\pi^2/3}$ for $n$ even, and $\cos(\pi n! x)^{2n} \sim 1$ if $n$ is odd. The set of $x$ for which this happens is dense in $[0,1]$.

You can simplify the argument a little by considering $a_n \sim n/2$ for $n$ even, and $a_n = 0$ is odd. Then $\cos(\pi n! x)^{2n} \sim 0$ for $n$ even, and $\cos(\pi n! x)^{2n} \sim 1$ if $n$ is odd.

So not only does $\cos(\pi n! x)^{2n}$ not converge uniformly. There also exists a dense subset of $[0,1]$ for which it doesn't even converge.

Answer 2

Let me apply one result related to limit at infinity. $\lim_{x \rightarrow \infty} \cos x$ does not exists. You may apply Cauchy's criterion to prove it.

Now we have the sequence of functions $\{f_n (x)\}$ where $f_n (x) = \cos (n! \pi x)^n$, and look at its characteristic.

$f_n(1) = \cos (n!\pi)^n$, and $\lim_{n \rightarrow \infty}f_n(1)$, does not exists.

Let $a \in (0,1)$. $\lim_{n\rightarrow \infty}f_n(a) = \lim_{n\rightarrow \infty}\cos(n!\pi a)^n$ also does not exists except $a$, is a zero of the function $f_n(x)$. $f_n(x)$ has only finitely many zeros in $[0,1]$ (Thanks to Mr. Fischer)

$\lim_{n \rightarrow \infty} f_n(0) = \lim_{n \rightarrow \infty} \cos(n! \pi 0)^n = \lim_{n \rightarrow \infty} \cos 0 = 1$

So we are getting only a few points where the function may converge pointwies. So there is no question of uniform convergence of the sequence of function in $[0,1]$.

Answer 3

This paper shows that for any subsequence $\{n_j\}$ of $\mathbb{N}$, the sets of points $x\in\mathbb{R}$ such that $\cos(n_j\,x)$ converges as $j\to\infty$ is of measure $0$; the same resul holds for $\sin(n_j\,x)$. Let $$ A=\{x\in\mathbb{R}:\cos(\pi\,n!\,x)\ \text{converges}\}. $$ Then $|A|=0$ (where $|A|$ is the Lebesgue measure of $A$) and $\mathbb{Q}\subset A$. Given $x\in\mathbb{R}$ let $L_x$ be the set of limit points of $\cos(\pi\,n!\,x)$. Then

1. $L_x\subset[-1,1]$
2. $x\in\mathbb{Q}\implies L_x=\{1\}$
3. If $x\not\in A$ then $L_x$ has at least two different points (the $\liminf$ and the $\limsup$.)

Let $$ B=\bigl\{x\in\mathbb{R}:L_x=\{-1,1\}\bigr\}. $$ Then $|B|=0$. In fact, if $x\in B$ then $\lim_{n\to\infty}\cos^2(\pi\,n!\,x)=1$, $\lim_{n\to\infty}\sin^2(\pi\,n!\,x)=0$ and $\lim_{n\to\infty}\sin(\pi\,n!\,x)=0$. Thus $B$ is contained in the set of pints $x$ such that $\sin(\pi\,n!\,x)$ converges, and therefore has measure $0$.

It follows that for almost all $x\in\mathbb{R}$ there exists $z\in L_x$ such that $|z|<1$. Then $\liminf_{n\to\infty}\cos^{2n}(\pi\,n!\,x)=0$.