Continuity of a series of functions at zero Let $\mathrm{sinc}(x) = \sin(x)/x$ if $x\neq 0$ and $\mathrm{sinc}(0) = 1$. This is a smooth function.
Let $(a_n)$ a real sequence such that $\sum_n a_n$ converge,
and let $$ f(x) = \sum_{n=0}^{+\infty} a_n \,\mathrm{sinc}(nx)^2. $$
I am trying to determine whether $f$ is continuous.
Here is what I have so far:


*

*$a_n \to 0$, so there exists $M$ such that $|a_n| \leq M$ for all $n$.
Let $\alpha>0$.
For all $|x| \geq \alpha$, we have $|a_n \mathrm{sinc}(nx)^2| \leq M/(\alpha n)^2$,  so that the series converges uniformly on $(-\infty,-a]\cup[a,\infty)$.
Hence $f$ is defined and continuous on $\mathbb{R} \setminus \{0\}$.

*If the series $\sum |a_n|$ converges, then $|a_n \mathrm{sinc}(nx)^2| \leq |a_n|$ so that the series converges uniformly on $\mathbb{R}$.
In this case, $f$ is continuous on $\mathbb{R}$.
However in the case where $\sum_n |a_n|$ diverges, I don't know how to proceed.
Maybe $f$ can be discontinuous at zero, but I have no idea on how to construct a conter-example.
 A: $f$ is continuous at $0$.
I will assume that $\sum a_n$ converges to $0$ (if it's not you can change its first term to make it converge to $0$ without changing the continuity at $0$ of anyone involved)
Let $b_n = \sum_{k=0}^n a_n$ and $u_n(x) = \sin_c(nx)^2 - \sin_c((n+1)x)^2$. 
Then, using an Abel summation, we have 
$f_n(x) = \sum_{k=0}^n a_k \sin_c(kx)^2 = \sum_{k=0}^{n-1} b_k u_k(x) + b_n\sin_c(nx)^2$.
Since this last term converges to $0$, taking the limit, we get
$f(x) = \sum_{k=0}^\infty b_ku_k(x)$.
The crucial point is that because $\sin_c(x)^2$ has bounded total variation, there is a $B > 0$ such that $\sum_{k=0}^\infty |u_k(x)| < B$ for all $x$.  
Now let us prove $f$ is continuous at $0$ : Let $\epsilon > 0$.
Let $n$ be an integer such that $|b_k| \le \epsilon/2B$ when $k \ge n$.
Then $|f(x)| \le |\sum_{k=0}^n b_ku_k(x)| + \epsilon/2$ for any $x$.
Since $u_k(0) = 0$ and the $u_k(x)$ are continuous, there is a $\delta > 0$ such that $|\sum_{k=0}^n b_ku_k(x)| \le \epsilon/2$ for $|x| < \delta$.
For those $x$, we get $|f(x)| \le \epsilon$, and this proves that $f$ is continuous at $0$.
