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Consider series $$\sum_{n=1}^\infty \frac{\sin nx^2}{1 + n^3}$$ Is its limit continuously differentiable?

We have for all $x \in \mathbb{R}$, $$ \left|\frac{\sin nx^2}{1 + n^3}\right| \leq \frac{1}{1 + n^3} < \frac{1}{n^3}$$ By the Weierstrass M-test, as $\sum \frac{1}{n^3}$ is convergent, our series is uniformly convergent. Thus, its limit function is continuous.

To show that its limit is continuously differentiable, if $f_n(x) = \frac{\sin nx^2}{1 + n^3}$, it suffices to show that $\sum f_n'$ converges uniformly on $\mathbb{R}$, where $$f_n'(x) = \frac{2nx\cos nx^2 }{1 + n^3}$$ (As $f_n'$ is continuous for each $n \in \mathbb{N}$, uniform convergence would also imply that the derivative is continuous).

However, I don't know where this series converges uniformly or not. We do have $$ \left|\frac{2nx\cos nx^2 }{1 + n^3}\right| \leq \frac{2n|x|}{1 + n^3}$$ By the comparison test, we can conclude this series converges pointwise. Does it converge uniformly? If not, how do we prove whether or not the original series is continuously differentiable?

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You can consider a compact set $K=\left[a,b\right]$ with $\left(a,b\right) \in \mathbb{R}^{2}$ and $a<b$. The function $\displaystyle x \mapsto \frac{2nx\cos\left(nx^2\right)}{1+n^3}$ is odd and then we can study it only on $\mathbb{R}^{+}$. For $x \in K$ you now have $$ \left|\frac{2nx\cos\left(nx^2\right)}{1+n^3}\right| \leq \frac{2nb}{1+n^3} $$ Hence

$$ \left\|\frac{2nx\cos\left(nx^2\right)}{1+n^3}\right\|_{\infty,K} \leq \frac{2nb}{1+n^3}\underset{(+\infty)}{\sim}\frac{2b}{n^2} $$

The series $\displaystyle \sum_{n \geq 1}^{ }\frac{1}{n^2}$ converges hence the series $\displaystyle \sum_{n \geq 0}^{ }f'_n$ converges normally so also uniformly on every compact set of $\mathbb{R}$ ( remember it was odd ). Then it is differentiable on $\mathbb{R}$ because it is a local property.

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  • $\begingroup$ Great, I didn't realize that it is enough to show uniform convergence on any bounded interval, but I see now that it is. Out of curiosity, is $\sum f_n'$ uniformly convergent on $\mathbb{R}$? Or indeed, is there an easy way to answer that? $\endgroup$ – Anu Jan 31 '18 at 12:30
  • $\begingroup$ I cannot tell you like that but my feeling is that it is not because you will have difficulties majoring $x$ on $\mathbb{R}^{+}$. But I might be wrong because I have not proved it. $\endgroup$ – Atmos Jan 31 '18 at 12:33

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