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In an exercise, I am suppose to study this function: $f(x)= ∑ sin(x/k)/k$. The questions are: where is $f$ defined? continuous? differentiable? twice differentiable?

I find that $f$is defined on $\Bbb R$ since $sin(x/k)$and $k$ are both defined on $\Bbb R$.

But then for the continuity I know that an infinite series is continuous if and only if fn(x) is continuous and ∑∞fn(x) converges uniformly . To show that ∑∞fn(x) converges uniformly, I need to show $ | f_n(x)| ≤M_n(x)$ such that $ M_n(x)$ converges.(Weirstrass M test).

But I have $|(sin(x/k)/k)| ≤ 1/k.$ And I know that $∑ 1/k$ diverges. Does it mean that $∑f_n(x)$ does not converge uniformly and thus $f$ is not continuous on $\Bbb R$?

And for $f$ to be differentiable I need it to be continuous is no first place right? And then I do the same test but with the derivatives?

Thank you in advance for your help.

2 Answers 2

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It is always true that $\;|\sin x|\le |x|\;$ ( hint: check the function $\;f(x)=\sin x -x\;$ is monotone descending in $\;\left[0,\,\frac\pi2\right]\;$) , and then ( $\;k\in\Bbb N\;$ )

$$\left|\frac{\sin\frac xk}k\right|\le\frac{|x|}{k^2}$$

and since the series with general term $\;\frac{|x|}{k^2}\;$ converges for all $\;x\in\Bbb R\;$ your function converges everywhere and the convergence is uniform and absolute on compact subsets of $\;\Bbb R\;$ .

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    Great! Thank you so much! Now for the differentiability, I find that fn'(x)= cos(x/k)/k^2. But to show that the infinite series converges, I can not do the same thing since |cos(x)| is not ≤ |x|... Or am I suppose to say that |cos(x)| ≤ |x| when x is greater than some value (but I don't know how to find it), and then use the same argument than before? Thank you again. – 2017-02-04
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    Clear and concise. (+1) – 2017-02-04
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Actually, I find that cos(x/k)/k^2 is less than 1/k^2. The series with general term 1/k^2 converges so we have uniform convergence of f'(x) and f is differentiable.

Then, -sin(x/k)/k^3 is less than x/k^4 and the series with general term 1/k^4 converges so we have uniform convergence of f''(x) and f is twice differentiable.

Is that right? Thank you for your help!