# Which of the following statements about f is true?

For the function $$f (x)$$ on the real line $$\mathbb{R}$$ defined below, which of the following statements about $$f$$ is true?Choose all the correct options: $$f (x) :=\sum_{n\ge 1}\frac{\text{sin}(x/n)}{n}$$

(a) f is continuous but not uniformly continuous on $$\mathbb{R}$$.

(b) f is uniformly continuous on $$\mathbb{R}$$.

(c) f is differentiable on $$\mathbb{R}$$.

(d) f is an increasing function on $$\mathbb{R}$$.

My attempt:

Option (c) is correct since $$f(x)$$ is $$C^\infty$$ function.

Option (d) is incorrect since sine function is periodic, f(x)is neither decreasing nor increasing function.

I applied $$M-test$$ but reached to a conclusion that $$f(x)\le \frac{x}{n^2}$$. I did this by finding $$f'(x)=0$$ and $$f''(0)<0$$. The other inequality that seems obvious is $$f(x)\le 1/n$$ and $$\sum \frac{1}{n}$$ is divergent sequence.

Am I doing something wrong here while doing $$M-test$$ ? Please help me to solve this question. Thankyou.

• Does $n \in \Bbb{Z}$? Mar 31, 2020 at 8:08
• Since $n\ge 1$, I believe $n \in \mathbb{N}$ Mar 31, 2020 at 8:10
• Your argument about (d) seems incorrect/very weak. $\sin(x)$ being periodic doesn't say anything about the sum, which has $\frac{x}n$ as argument of the sine-function. Also how do you justify your argument about (c)? You are just claiming a very specific thing, without any kind of argument for that. Mar 31, 2020 at 9:14
• @Ingix you're right about my argument (d). It seems incorrect to me now. Mar 31, 2020 at 10:54

User ModCon had a (now deleted) solution to some parts of the problem. There was a small mistake in it that made a few, but not all conclusions incorrect. I'll try to repeat his contributions, and add a few of my own.

We have $$\left|\frac{\sin(x/n)}n\right| \le \frac{|x/n|}n = \frac{|x|}{n^2}.$$

That means for each $$x$$ the series $$\sum_{n=1}^{\infty}\frac{\sin(x/n)}{n}$$ has a convergent majorant

$$\sum_{n=1}^{\infty}\frac{|x|}{n^2} = |x| \sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}6|x|,$$

which means it converges pointwise. This also means that we can use the Weierstrass M-Test on an interval $$[-a,a]$$ for some $$a>0$$ and choose $$M_n=\frac{a}{n^2}$$. That means the series converges absolutely and uniformly on $$[a,-a]$$, which means $$f(x)$$ is continuous on that inverval. Since we can chose any real $$a$$, this means $$f(x)$$ is continuous on the whole $$\mathbb R$$.

If we differentiate the series term-wise, we get another function:

$$g(x)=\sum_{n=1}^{\infty}\frac{\cos(x/n)}{n^2}$$

Using $$|\cos(x/n)| \le 1$$ we can directly apply the M-test on whole real line for $$g(x)$$, when setting $$M_n=\frac1{n^2}$$. This shows that g(x) is actually well defined and the defining series converges uniformly on $$\mathbb R$$ to $$g(x)$$.

So is now $$f'(x)=g(x)$$? By Theorem 1 on page 2 in this university script, the answer is yes. The term-wise derivate series (for $$g(x)$$) converges uniformly on the whole $$\mathbb R$$, the original series (for $$f(x)$$) converges on a point (we already know it converges anywhere), so we now know that

$$f'(x)=g(x),\; \forall x \in \mathbb R$$

This means $$f(x)$$ is differentible, which answers (c) in the affirmative.

We also know that

$$|g(x)| \le \sum_{n=1}^{\infty}\frac{|\cos(x/n)|}{n^2} \le \sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}6,$$

so the derivate of $$f(x)$$ is bounded. That means $$f(x)$$ is uniformly continuous: Given any $$\epsilon > 0$$, we can choose $$\delta=\frac{6\epsilon}{\pi^2}$$ and if we assume $$x_0 < x_1 < x_0+\delta$$, we have $$f(x_1)=f(x_0)+(x_1-x_0)f'(\xi)$$ with $$\xi \in [x_0,x_1]$$ (Mean Value Theorem), so

$$|f(x_1)-f(x_0)| = |f(x_0)+(x_1-x_0)f'(\xi) - f(x_0)| = |x_1-x_0||f'(\xi)| < \delta\frac{\pi^2}6 = \epsilon.$$

This ansers (b) in the affirmative and (a) in the negative.

For (d), because we know $$f(x)$$ is differentiable, it is enough to find a point where $$f'(x)=g(x)$$ is negative, to prove that (d) is not true.

Consider $$x=\pi$$. Then the first element of the series is $$\cos(\pi)=-1$$. We know that all the other elements of the series can sum up to at most $$\sum_{n=\color{red}{2}}^{\infty}\frac{1}{n^2} = \frac{\pi^2}6 - 1 < 1$$, so $$g(\pi) < 0$$ and (d) is not true.

If you don't want to use the non-elementary exact sum of that series, you can can also just say

$$\sum_{n=2}^{\infty}\frac{1}{n^2} < \sum_{n=2}^{\infty}\frac{1}{n(n-1)} = \sum_{n=2}^{\infty}\left(\frac{1}{n-1}-\frac{1}{n}\right) = (1-\frac12)+(\frac12-\frac13)+(\frac13-\frac14)\ldots=1.$$

Finally, a plot of the function (hopefully good enough truncated) by Wolfram Alpha:

• Thanks for such a detailed answer. I just wanted to ask: (we already know it converges anywhere)- does this statement mean we can make it converge anywhere OR it means we know it converges to some point? Mar 31, 2020 at 13:38
• The question was much more difficult than I thought it would be. @ModCon and Ingix cheers to you. Mar 31, 2020 at 13:40
• @s1mple For the theorm I mentioned, you basically have to do "all the work" for the termwise derivative, and for the function itself you only need to check convergence at 1 point. In preparation, we already proofed much more than we needed, so I made that remark. It means we proved it convergens for any $x$, which is much more than we need to apply that theorem. Mar 31, 2020 at 13:44
• +1 here. The way you have dealt with uniform convergence is really nice. One can show the uniform convergence in any finite interval by Weierstrass M test, but for whole of $\mathbb {R}$ we need something different. Luckily the function turned out to have a bounded derivative which made the task easy. Apr 1, 2020 at 2:27