0

Looking at the graph of the functions $\cos(n^a t)$ ($a>0$) it looks obvious that they don't converge to $1$ as $n \rightarrow \infty$. For integer odd $a$ it's easy to prove this directly, as $\cos(n^a \pi/2)=\cos(m \pi/2)=0$ for some odd $m$ in this case. I've tried proving this for any $a$ then, but reached no success in it. There seem to be no theorems covering this case, which would give uniform convergence for example, or anything like this to help.

I've read about a theorem that $\cos(n^a t)$ is uniformly distributed on $[-1,1]$, but it seems much stronger and harder to prove.

Can anyone give an idea of how this could be proven?

learner
  • 6,726
aplavin
  • 591
  • When you say converge, you mean for one $t$, or that the sequence of functions $(f_n)_n$ defined by $f_n =(t\mapsto \cos(n^a t))$ pointwise converge/ converge uniformly (on a domain of $t$ that you didn't define) ? – user37238 Nov 22 '13 at 08:32
  • @user37238: I mean pointwise convergence, for all $t$. So, if it doesn't converge for one $t$ - then what I need is proven. – aplavin Nov 22 '13 at 08:33

1 Answers1

1

Let $u_n:t\mapsto\cos(n^at)$. If $u_n\to1$ pointwise on some interval $(x,y)$, then, by bounded convergence, $\int\limits_x^yu_n\to y-x$. But a primitive of $u_n$ is $t\mapsto n^{-a}\sin(n^at)$, which is uniformly bounded by $n^{-a}$, hence $\int\limits_x^yu_n\to0$.

Hence, for every interval $(x,y)$ with $x\lt y$, $u_n$ does not converge pointwise to $1$ on $(x,y)$. In particular, the interior of the set $\{t\mid u_n(t)\to1\}$ is empty.

Did
  • 279,727