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Suppose that $f(x)$ is a continuous function on $\mathbb{R}$ satisfying the following.

  1. $f(0) = 0$.
  2. There exists some $L>0$ such that $\displaystyle \lim_{x \to 0} \dfrac{f(x)}{x} = L$.
  3. There exists some $M>0$ such that $|f(x)| \leq M$ for any $x \in \mathbb{R}$.

Then, the function $\displaystyle F(x) = \sum_{n=1}^{\infty}\frac{f(nx)}{n^2}$ is non-differentiable at $x=0$.

I know the function $F(x)$ uniformly converges and is continuous on $\mathbb{R}$. However, I do not know how to prove the statement above.

I can find the solutions for a specific case when $f(x) = \sin(x)$ (e.g. Differentiability of the sum of ∑∞n=1sin(nx)n2), but they use evaluation equation for $\sin x$. Although the solution might be similar to the case for $\sin x$, I cannot prove this statement. If you don't mind, I would appreciate your help.

Kotov
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1 Answers1

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Thank you for giving me an informative hint! Now, I was able to solve this problem. Here is my answer.

From hypothesis (2), for any natural number $n$, there exists some $\delta_n$ such that $$\begin{eqnarray} \frac{f(x)}{x} &>& L - \dfrac{1}{n} \qquad \left(0 < |x| < \delta_n\right)\\ \therefore \frac{f(nx)}{n^2x} &>& \dfrac{L}{n} - \dfrac{1}{n^2} \qquad \left(0 < |x| < \frac{\delta_n}{n}\right)\\ \therefore \sum_{n=1}^{m}\frac{f(nx)}{n^2x} &>& \sum_{n=1}^{m} \left( \frac{L}{n} - \frac{1}{n^2}\right) \left(0 < |x| < \min_{1 \leq n \leq m}\frac{\delta_n}{n}\right) \end{eqnarray} $$ Since $f(x)$ is continuous at $x=0$, $\frac{\delta_m}{m} \to 0$ when $m \to \infty$. Thus $$\lim_{x\to 0} \dfrac{F(x)}{x} = \lim_{m\to \infty}\sum_{n=1}^{m}\frac{f(nx)}{n^2x} > \lim_{m\to \infty} \left( \frac{L}{n} - \frac{1}{n^2}\right) = + \infty.$$

Is there anything wrong with it?

Kotov
  • 75
  • 6