4
$\begingroup$

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:

  1. $a_n \to 0$, 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 $\mathbb{R} \setminus \{0\}$.

  2. 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.

$\endgroup$
0

1 Answer 1

3
+100
$\begingroup$

$f$ 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 $0$.

$\endgroup$
1
  • $\begingroup$ Nice observation regarding the BV property! $\endgroup$
    – H. H. Rugh
    Commented Jun 18, 2017 at 15:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .