How to proove that $x \cos nx$ is equicontinuous?

Question

How to proove that $x\cos(nx)$ is equicontinuous on $[0;1]$? I proved that is is equicontinuous on 0, however I cannot prove either it is equicontinuous on[0;1] or not. I also tried to do something with $1$, like $\cos n - (1-\frac{1}{n})\cos(n(1-\frac{1}{n})))$ = $\cos(n) - \cos ((n-1)) +\frac{1}{n}\cos(n-1)$ = [ with $n\to\ \infty$] = $\cos(n)-\cos(n-1)$. But it seems that it isn't something helpful. What should I do?

Answer 1

Pick $x_n=1-\frac{1}{n}, y_n=1$; since $x_n-y_n \to 0$ if $f_m(x)=x\cos mx$ would be equicontinuous it would follow that for every $\epsilon >0$ there is $n(\epsilon)$ st $|f_m(x_n)-f_m(y_n)| \le \epsilon$ for all $n > n(\epsilon)$ and for all $m$

However $f_n(x_n)-f_n(y_n)=(1-1/n)\cos (n-1)-\cos n=2\sin \frac{1}{2} \sin \frac{2n-1}{2}-\frac{\cos (n-1)}{n}$

Now $(2n-1)/(2 \pi)$ mod $1$ is equidistributed in $[0,1]$ by well known results so there is a sequence $n_k$ st the fractional part of $(2n_k-1)/(2 \pi)$ goes to $1/2$ hence $|\sin \frac{2n_k-1}{2}| \to 1$ so in particular choosing a subsequence we can assume $\sin \frac{2n_k-1}{2} \to a$ where $a=1$ or $a=-1$; in either case $|f_{n_k}(x_{n_k})-f_{n_k}(y_{n_k})| \to 2 \sin \frac{1}{2}$ so choosing $\epsilon =\sin \frac{1}{2}$ in the above and $n_k >n(\epsilon)$ large enough we get a contradiction, so the family is not uniformly equicontinuous.

A similar proof applies on any subinterval of $[0,1]$

Answer 2

Here's a proof that uses different values than you, but doesn't require equidistribution results, in case it may be simpler to use for you. First, let us show that $\cos(nx)$ is not equicontinuous on $[0,1]$. Simply pick $x_n=0$ and $y_n=\frac{\pi}{n}$ for example (with $n\geq4$ so that $0\leq y_n\leq 1$). We thus have $|x_n-y_n|\to0$ and $|\cos(nx_n)-\cos(ny_n)|=|1-(-1)|=2$, and so $\cos(nx)$ isn't equicontinuous.

Now we can do something similar for $x\cos(nx)$. However, you'll have noticed we can't use the same $x_n$ and $y_n$, since $|x_n\cos(nx_n)-y_n\cos(ny_n)|\to0$. No problem, we can just use the fact that $\cos(nx)$ is periodic with period $T_n=\frac{2\pi}{n}$ and apply the same results away from $0$.

For instance, we can set an integer $k_n=\lfloor\frac{1}{2T_n}\rfloor+1=\lfloor\frac{n}{4\pi}\rfloor+1$ so that $\frac{1}{2T_n} \leq k_n \leq \frac{1}{2T_n}+1$, and so $\frac{1}{2} \leq k_nT_n \leq \frac{1}{2} + \frac{2\pi}{n}$.

Now we can pick $x_n = k_nT_n$ and $y_n=k_nT_n+\frac{\pi}{n}$ and making sure $n$ is large enough so all the values are in $[0,1]$, we have from the periodicity: $|x_n\cos(nx_n)-y_n\cos(ny_n)| = |x_n-(-y_n)|\to1$, and $|x_n-y_n|\to0$, therefore $x\cos(nx)$ isn't equicontinuous.