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Show that the series $$\sum_{n=0}^{\infty} 2^n \sin (\frac{1}{3^nx})$$ does not converge uniformly on $(0,\infty)$

Solution: enter image description here

My question is what is the technique being used in the solution to show that the series does not converge uniformly? Can anyone explain it with reference to the relevant definition/theorem?


What I don't understand is that how showing that $$\sum_{n=N}^{\infty} 2^n \sin (\frac{1}{3^nx})$$ cannot be made arbitrarily small proves the result. This answer seems more like its showing that the sequence of partial sums is unbounded. I need clarification.

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  • $\begingroup$ the point is, the function converges at different speed, and as x approaches 0, this speed becomes very slow so that there is no uniform speed for the convergence $\endgroup$ – lion Nov 22 '17 at 19:56
  • $\begingroup$ What book are you reading? $\endgroup$ – Jack Nov 25 '17 at 16:48
  • $\begingroup$ @Jack Spivak's Caculus (3rd Ed) $\endgroup$ – helios321 Nov 26 '17 at 12:10
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what is the technique being used in the solution to show that the series does not converge uniformly

Go back to the definition. I will explain it abstractly here.

Consider a sequence of functions $f_n:E\to{\bf R}$, where $E$ is some subset of ${\bf R}$. There are two basic modes of convergence of the sequence $\{f_n\}$.

  • If for each $x\in E$, $f_n(x)\to f(x)$ as $n\to\infty$, we say that $f_n$ converges to $f$ pointwise. In terms of the $\epsilon$-$N$ language, $$ \forall x\in E\ \forall \epsilon>0\ \exists N\in{\bf N}\ \forall n\geq N\ \ |f_n(x)-f(x)|<\epsilon\tag{1} $$
  • We say $f_n$ converges to $f$ uniformly, if the following is true: $$ \forall \epsilon>0\ \exists N\in{\bf N}\ \forall n\geq N\ \forall x\in E\ \ |f_n(x)-f(x)|<\epsilon\tag{2} $$

Note that (2) is a stronger condition than (1).

Remember that a series is nothing but the limit of a partial sum. Suppose $g_n:(0,\infty)\to{\bf R}$, then $\sum_{n=0}^\infty g_n(x)$ converges uniformly means, the sequence of functions $f_k$, defined with $$ f_k(x):=\sum_{n=0}^k g_n(x), $$ converges uniformly to some function $g:(0,\infty)\to{\bf R}$.


Now, note that $f_k$ does not converge uniformly on $(0,\infty)$ means (2) is not true, namely the following holds: $$ \exists \epsilon>0\ \forall N\in{\bf N}\ \exists n\geq N\ \exists x\in E\ \ |f_n(x)-f(x)|\geq \epsilon\tag{3} $$

It is an instructive exercise to check how your "solution" matches (3).

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By definition: a series $\sum_{ n=0}^{\infty} a_n(x)$ converges uniformly iff: $$ \lim_{N\to \infty}\sup_{x\in\Bbb R} \left|\sum_{ n=N}^{\infty} a_n(x)\right| =0$$

see here: Explain $\lim_{N\to \infty}\sup_{x\in\Bbb R} \left|\sum_{ n=N}^{\infty} a_n(x)\right| =0$

They essentially used the following well known inequality

For all $x\in[0,\frac\pi2]$ We have, $$\frac{2}{\pi}x\le \sin x\le x$$

As follows: for very large $N\in\Bbb N,$ Taking $ \color{blue}{x =\frac{2}{3^N\pi}}$ and using the geometric sum we obtain:

$$\infty =\sup_{x\in(0,\infty)} \left|\sum_{ n=N}^{\infty} 2^n \sin (\frac{1}{3^nx}) \right| \overset{\color{blue}{x =\frac{2}{3^N\pi}} }{\ge} \sum_{ n=N}^{\infty} 2^n \sin (\frac{3^N\pi}{3^n2})\\\ge \sum_{ n=N}^{\infty} 3^N \left(\frac{2}{3}\right)^n= 3.2^N.$$

That is $$\infty = \lim_{N\to \infty}\sup_{x\in(0,\infty)} \left|\sum_{ n=N}^{\infty} 2^n \sin (\frac{1}{3^nx}) \right| \ge \lim_{N\to \infty}2^N =\infty.$$ Which will prove the non-uniformly convergence for your series.

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  • $\begingroup$ My book doesn't seem to present that definition. It seems to be similar to the vanishing condition for a series that converges. Is that right? Also what does the $\sup_{x\in\Bbb R}$ represent? $\endgroup$ – helios321 Nov 23 '17 at 3:52
  • $\begingroup$ what is the definition in your book . it must be equivalent to this one. ch3ck carefully $\endgroup$ – Guy Fsone Nov 23 '17 at 7:06
  • $\begingroup$ @helios321 you seem to be spektical with this definition $\endgroup$ – Guy Fsone Nov 24 '17 at 18:46
  • $\begingroup$ A little bit, I have another post here: math.stackexchange.com/questions/2533476/… $\endgroup$ – helios321 Nov 25 '17 at 12:32
  • $\begingroup$ @heliose321 the answer there is perfectly correct. I upvoted it . you should do the same. unless you have another definition you would like to share with you us $\endgroup$ – Guy Fsone Nov 25 '17 at 12:39

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