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Let $M>0$. In the closed interval $[-M,M]$ we get that:

  1. $\sum_{n=1}^{\infty}\frac{\sin(nx^{2})}{1+n^{4}}$ converges at $x_{0}=0$
  2. $\sum_{n=1}^{\infty}\left(\frac{\sin(nx^{2})}{1+n^{4}}\right)'=\sum_{n=1}^{\infty}\frac{2nx\cos(nx^{2})}{1+n^{4}}$ and $\left|\frac{2nx\cos(nx^{2})}{1+n^{4}}\right|\leq\frac{2nM}{1+n^{4}}$. Also $\sum_{n=1}^{\infty}\frac{2nM}{1+n^{4}}<\infty$ so by the Weierstrass M-test, $\sum_{n=1}^{\infty}\left(\frac{\sin(nx^{2})}{1+n^{4}}\right)'$ converges uniformely in $[-M,M]$.

Therefore, by the term by term differentiation theorem there exists a function $f\colon[-M,M]\to\mathbb{R}$ such that $f(x)=\sum_{n=1}^{\infty}\frac{\sin(nx^{2})}{1+n^{4}}$ uniformly and $\left(\sum_{n=1}^{\infty}\frac{\sin(nx^{2})}{1+n^{4}}\right)'=f'(x)=\sum_{n=1}^{\infty}\left(\frac{\sin(nx^{2})}{1+n^{4}}\right)'$.

Finally, $\sum_{n=1}^{N}\left(\frac{\sin(nx^{2})}{1+n^{4}}\right)'$ is continuous in $[-M,M]$ for every $N$, then by the uniform convergence $f'(x)=\sum_{n=1}^{\infty}\left(\frac{\sin(nx^{2})}{1+n^{4}}\right)'$ is also continuous in $[-M,M]$.

My question is how can i generalize this argument to $\mathbb{R}$?

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  • $\begingroup$ $\Bbb R$ is covered by the interiors of the $[-M,M]$ $\endgroup$ – Lord Shark the Unknown Dec 6 '17 at 19:08
  • $\begingroup$ You just need to show that your sum is uniform convergent for all $x \in (a,b) \subset \mathbb{R}$. $\endgroup$ – openspace Dec 6 '17 at 19:12
  • $\begingroup$ @LordSharktheUnknown Doesn't $x^{n}$ converge uniformly to $0$ in $[0,1-\frac{1}{n}]$ for every $n$, but still doesn't converge unifomly to $0$ on $[0,1)=\bigcup_{n=1}^{\infty}[0,1-\frac{1}{n}]$ ? $\endgroup$ – Jon Dec 6 '17 at 19:27
  • $\begingroup$ @Jon Yes, and didn't you ask about continuous differentiability and not uniform convergence? $\endgroup$ – Lord Shark the Unknown Dec 6 '17 at 19:30
  • $\begingroup$ @LordSharktheUnknown True, but i still can't see how this property passes on to the union of these closed sets $\endgroup$ – Jon Dec 6 '17 at 19:35
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Apparently what i was missing is this:

Let $x_{0}\in\mathbb{R}$ ,

We need to show that $\sum_{n=1}^{\infty}\frac{\sin(nx^{2})}{1+n^{4}}$ is continuously differentiable at $x_{0}$. That is to show that $\sum_{n=1}^{\infty}\frac{\sin(nx^{2})}{1+n^{4}}$ is differentiable at $x_{0}$ and that $\left(\sum_{n=1}^{\infty}\frac{\sin(nx^{2})}{1+n^{4}}\right)'$ is continuous at $x_{0}$.

Choose $M=\left|x_{0}\right|+1$ so $x_{0}\in[-M,M]$ and from the argument above we get what we need.

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