# continuously differentiable series

I want to show that the following function is continuously differentiable:

$$g(x)=\sum_{n=1}^{\infty}\frac{1}{n^{2}}e^{\int_{0}^{x}t\sin(n/t)\,dt}.$$

I tried using the idea that series where the first n terms of the infinite series converge uniformly if it converges pointwise at a time and derivative series converges uniformly. So I tried to show derivative series converges by the Cauchy criterion but I can't seem to bound the tail.

Let $f_n(x)=\int_0^xt\sin(n/t)\,dt$. Since $t\sin(n/t)$ is even and continuous on $\mathbb{R}$ (assuming it takes the value $0$ at $t=0$), $f_n$ is odd and differentiable and $f_n'(x)=x\sin(n/x)$. Let $R>0$. Then $$|f_n(x)|\le\int_0^{|x|}t\,dt\le\frac{R^2}{2}\quad\forall x\in[-R,R].$$ By the Weierstrass convergence theorem (also known as the $M$-test) the series converges uniformly on $[-R,R]$ to a continuous function. This shows that $g$ is continuous on $\mathbb{R}$.
Moreover $$|\Bigl(e^{f_n(x)}\Bigr)'|=|f_n'(x)|e^{f_n(x)}|\le Re^{R^2/2}\quad\forall x\in[-R,R].$$ It follows that the series obtained by term by term differentiation is uniformly convergent on $[-R,R]$ to a continuous function, that $g$ is continuously differentiable and that $$g'(x)=\sum_{n=1}^\infty\frac{x\sin(n/x)}{n^2}e^{f_n(x)}.$$
• I have not shown uniform convergence on $\mathbb{R}$. What I have shown is that the series converges uniformly on each interval $[-R,R]$. This implies in particular that the series converges point wise in all of $\mathbb{R}$. And since the sum is differentiable in each interval $[-R,R]$, it is differentiable 0n $\mathbb{R}$. Nov 22 '12 at 17:01
• so uniform convergence on each [-r,r] does not imply uniform convergence on all of R but pointwise on all of R? Can you explain why? Also if you differentiate term by term shouldn't you have $g(x)=\sum_{n=1}^{\infty}\frac{1}{n^{2}}e^{\int_{0}^{x}tsin(n/t)dt}sin(n/x)$ don't you have to use the exponential rule since it's of the form e^u. Also can you refer me to some theorems regarding these results? Thanks. Nov 22 '12 at 18:14
• I'll try to explain it with a simpler example. The sequence $x^n$ converges uniformly to $0$ on $[0,a]$ for $0<a<1$, but converges point wise on $[0,1]$ to the discontinuous function that equals $0$ on $[0,1)$ and to $1$ for $x=1$. As for the derivatives, I have edited the answer. Nov 22 '12 at 18:52