# Show that $f_n(x) = \frac{\sin(nx)}{\sqrt{n}}$ converges uniformly to $0$

I need to show that $f_n(x) = \frac{\sin(nx)}{\sqrt{n}}$ converges uniformly to $0$. My book has a theorem that says that if $|f_n(x)|<a_n$ and $\sum |a_n|$ converges, then $f_n$ converges uniformly. But $\sum \frac{1}{\sqrt{n}}$ doesn't converge, so I don't know how to do it. Should it be done by definition?

$$\left|\frac{\sin(nx)}{\sqrt{n}}\right|\le \left|\frac{1}{\sqrt{n}}\right|$$

if I take $n_0 = \frac{1}{\epsilon^2}$, then $n>n_0\implies n>\frac{1}{\epsilon^2}\implies \sqrt{n}>\frac{1}{\epsilon}\implies \frac{1}{\sqrt{n}}<\epsilon$

therefore it converges uniformly. Now, I need to also show that $f'_n$ diverges in every point of the interval $[0,1]$. The derivative is:

$$\frac{n\cos nx}{\sqrt{n}}$$

should I just say that this limit goes to infinity because $n$ grows faster than $\sqrt{n}$?

Hint.- It is enough to be Cauchy $$||f_n(x)-f_m(x)||=\text{Sup } |f_n(x)-f_m(x)| \le\text{Sup }\left( |f_n(x)|+|-f_m(x)|\right)\le\text{Sup }\left|\frac{1}{\sqrt n}+\frac{1}{\sqrt m}\right|$$

This tends to $0$ when $m,n$ tend to $\infty$.

• This answer is very simple and clear! Why is the count of votes very low? Feb 13, 2022 at 8:11

If $x=0$ we have $\sqrt n\cos (nx) = \sqrt n \to \infty.$ It's trickier if $x\in (0,x].$ Let $A = \{e^{it}: t \in [0,1]\},$ $B = \{e^{it}: t \in [\pi,\pi +1]\}.$ Claim*: For each $x\in (0,1],$ $e^{inx}\in A$ for infinitely many $n,$ and $e^{inx}\in B$ for infinitely many $n.$ Accepting this (it's a nice exercise), we see that for each $x\in (0,1],$ $\sqrt n\cos (nx) > \cos 1\cdot \sqrt n$ for infinitely many $n,$ $\sqrt n\cos (nx) <- \cos 1\cdot \sqrt n$ for infinitely many $n.$ So for all $x\in (0,1],$ the sequence $\sqrt n\cos (nx)$ oscillates wildly, taking on arbitrarily large positive and negative values.

$\text{*}$Proof: If $A$ is an arc on the unit circle of arc length $1,$ and $x\in (0,1],$ then $e^{inx} \in A$ for infinitely many $n.$ Proof: The sequence $e^{i nx}$ marches around the circle infinitely many times in steps of fixed arc-length $x\le 1.$ Now remember what your momma taught you: You can't step over a puddle that's larger than your stride. Apply this to our sequence: Because the steps are no more than the length of $A,$ we have to land in $A$ at least once every orbit. That's the idea, and you can make it perfectly rigorous.

• This is the only posted solution thus far that addresses the main issue! +1. You might consider adding a reference to the claim. -Mark Nov 30, 2016 at 21:21

Your explanation for uniform continuity that you wrote is solid.

Without using contradiction, you can show that $f_n^{'}(x)$ diverges at every point in $[0,1]$ by simply showing that $$\lim_{n \rightarrow \infty} \frac{ncos(nx)}{\sqrt{n}} = \pm\infty.$$

Notice that algebraically we can reduce our limit as follows: $$\lim_{n \rightarrow \infty} \frac{ncos(nx)}{\sqrt{n}} = \lim_{n \rightarrow \infty} \sqrt{n}cos(nx).$$ The latter will clearly diverge no matter what $x$ you pick because $cos(nx)$ is bounded by $\pm1$, while $\sqrt{n}$ will shoot off to infinity.

• The limit is certainly not $\pm \infty$. Choose $x=\pi/4\in [0,1]$. Then, $\cos(nx)=0$ whenever $n\pi/4$ is a multiple of $\pi/2$. Furthermore, how does this show that the limit fails to exist for any choice of $x$? Nov 30, 2016 at 20:53
• @Dr.MV is right, there are problems with this proof.
– zhw.
Nov 30, 2016 at 21:13
• You are correct that technically the limit does not exist, I guess an appropriate edit would be to say that it's not $\pm \infty$ but that instead it diverges due to the fact that it wouldn't exist (it oscillates). I don't know how to succinctly show that, an edit would be welcome Nov 30, 2016 at 21:24

Hint

Let $x\in[0,1]$. Suppose $$\lim_{n\to+\infty}\sqrt{n}\cos(nx)=L$$

then $\lim_{n\to+\infty}\cos(nx)=0$

and

$$\lim_{n\to+\infty}\cos((n+1)x)=0$$

or $$\lim_{n\to+\infty}\sin(nx)\cos(x)=0$$

$$\cos^2(nx)+\sin^2(nx)=1$$