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Let $f(x)= \displaystyle \sum \limits_{n=1}^\infty \frac{\sin(nx)}{n^3}.$ Show that $f(x)$ is differentiable and that the derivative $f'(x)$ is continuous.

In class we solved a similar problem, and I think we had to show that both $f(x)$ and $f'(x)$ converge uniformly, but I am not really sure why that is what we have to show. I think we also showed that the partial sums are continuous.

I would really appreciate it if someone could solve this problem and maybe explain a little the theory behind it.

Thanks!

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3 Answers 3

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To show uniform convergence of the original series and the series of derivatives you can use the facts that $\sin$ and $\cos$ are bounded and that $\sum_{n=1}^\infty \frac{1}{n^3}$ and $\sum_{n=1}^\infty \frac{1}{n^2}$ converge.

The uniform limit of continuous functions is continuous. To see that $f$ is differentiable with derivative equal to the series of derivatives, you can apply an integral convergence theorem. E.g., uniform convergence allows you to pass limits through integrals, and (using the fundamental theorem of calculus) $f(x)=\sum_{n=1}^\infty \int_0^x \frac{\cos(nt)}{n^2}dt$. You can use the fundamental theorem of calculus again to finish, also using the fact that the series of derivatives represents a continuous function by the first sentence of this paragraph.

(I'm intentionally leaving some details for you to fill in.)

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  • $\begingroup$ Nicely given. Let me just remark that in your final sentence one can drop the condition of continuity of the derivative. The reference is theorem 7.17 of Rudin's Principles of Mathematical Analysis, 3rd Ed., and the proof is a little bit longer. $\endgroup$
    – user1119
    Nov 22, 2010 at 0:08
  • $\begingroup$ Is there a way of proving differentiability without using the integral convergence theorem? We haven't learned this theorem yet. $\endgroup$
    – user2467
    Nov 22, 2010 at 0:11
  • $\begingroup$ @Daniel: The lemma you could prove and apply here is that if $(g_n)$ is a sequence of continuous functions converging uniformly to $g$ on the interval $[a,b]$, then $\int_a^b g(t)dt=\lim_{n\to\infty}\int_a^b g_n(t)dt$. It is pretty straightforward. If $|g(t)-g_n(t)|\lt\varepsilon$ for all $t$, then what can you say about $|\int_a^b g(t)-g_n(t)dt|$? In the application to your problem, $(g_n)$ will be a sequence of partial sums. $\endgroup$
    – Jonas Meyer
    Nov 22, 2010 at 0:17
  • $\begingroup$ @George: [clarification of uneditable comment I'm deleting.] Yes, continuity allows the integrability of the series of derivatives to follow, so that the FTC can be applied, and without knowing this a priori it takes considerably more work. (And in general, the derivative need not even be Riemann integrable. E.g., one could consider a constant sequence consisting of an everywhere differentiable function without integrable derivative, as seen e.g. in mathoverflow.net/questions/6711/integrability-of-derivatives/…) Thanks for the comment. $\endgroup$
    – Jonas Meyer
    Nov 22, 2010 at 1:08
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You had to show that both $(f_n)$ and $(f_n')$ converge uniformly because you want to apply a theorem like this:

Theorem 10.7 (from "Real Analysis" by N.L. Carothers):

Suppose that $(f_n)$ is a sequence of real-valuled functions, each having a continuous derivative on $[a,b]$, and suppose that the sequence of derivatives $(f_n')$ converges uniformly to a function $g$ on $[a,b]$. If $(f_n(x_0))$ converges at any point $x_0$ in $[a,b]$, then in fact, $(f_n)$ converges uniformly to a differentiable function $f$ on $[a,b]$. Moreover $f' = g$. That is, $(f_n')$ converges uniformly to $f'$ on $[a,b]$.

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Well, to see that $f(x)$ *converges uniformly* please apply the Weierstrass M-Test.

Now, i am aware of the fact that term by term differentiation and integration is possible if the series is uniformly convergent. But i don't know much about this, i don't want to answer that, but i am sure someone here will help you out regarding this.

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  • $\begingroup$ To differentiate term by term you need the series of derivatives to also converge uniformly, and this is probably what you meant. A uniform limit of differentiable functions need not be differentiable (e.g., see the Weierstrass approximation theorem and the existence of continuous nowhere differentiable functions). $\endgroup$
    – Jonas Meyer
    Nov 22, 2010 at 0:27
  • $\begingroup$ (For a lower tech example, consider the sequence of functions $g_n(x)=\frac{\sin(nx)}{n}$, which converges uniformly to $0$ but whose sequence of derivatives does not converge.) $\endgroup$
    – Jonas Meyer
    Nov 22, 2010 at 0:45
  • $\begingroup$ @jonas: yes u are right! $\endgroup$
    – anonymous
    Nov 22, 2010 at 0:51

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