We define $f_n(x) = \frac{n x^2 \sin(nx)}{n^4 + x^4}$. We have to prove that $$ \sum_{n=1}^\infty f_n $$ converges uniformly to a derivative function on $\mathbb{R}$. To prove that, it suffices to show that there exists $x_0 \in \mathbb{R}$ so that $\sum_{n=1}^\infty f_n(x_0)$ converges (just take $x_0 = 0$) and that $$ \sum_{n=1}^\infty f_n'$$ converges uniformly to a certain function on $\mathbb{R}$. If we do the derivative, we obtain $$ f_n'(x) = \frac{(2nx \sin(nx) + n^2x^2 \cos(nx))(n^4+x^4) – 4nx^5 \sin(nx)}{(n^4+x^4)^2}. $$ Now, to show the uniform convergence of the series, I think we must use the Weierstrass criterion, but I don't know how to obtain an upper bound valid on $\mathbb{R}$ for $f_n'$.
Can anyone help me? Thank you.
Best Answer
1) Uniform convergence of $\sum_{n=1}^\infty f_n$ in $\mathbb{R}$: note that $$\sum_{n=1}^\infty f_n(x)=x^2\sum_{n=1}^\infty \frac{n \sin(nx)}{n^4 + x^4}$$ and $\left|\frac{n\sin(nx)}{n^4 + x^4}\right|\leq \frac{1}{n^3}$, then by the $M$-test the series converges uniformly in $\mathbb{R}$ to some continuous function $F$.
2) As regards the differentiability of $F$, all you need is the convergence of $\sum_{n=1}^\infty f_n'$ in the compact set $[-R,R]$ for all $R>0$: for $x\in [-R,R]$, and for $n\geq1$, $$|f'_n(x)|\leq \frac{(2nR + n^2R^2)(n^4+R^4) + 4nR^5}{n^8}\leq \frac{C_R}{n^2}$$ for some constant $C_R$ (which depends on $R$). Then $\sum_{n=1}^\infty f_n$ converges uniformly to the derivative of $F$ on $[-R,R]$ for all $R>|x_0|$ (recall that differentiability is a local property).