Cosine series with non-negative coefficients, that is continuous at $0$ but not everywhere

Question

Is there an cosine series with non-negative terms, that is continuous at $x=0$, but not continuous everywhere?

More specifically, do there exist $a_n\geq0$ such that

$$f(x) = \sum_{n=1}^\infty a_n \cos(nx) $$

• converges for $x=0$
• converges almost everywhere
• is continuous at $x=0$
• is not a continuous function?

For motiviation, see this related question.

Answer 1

If the series converges at $x=0$, then $\sum_na_n$ converges. And since $a_n\ge 0$, it follows that $(a_n)\in\ell^1$. But then the series converges at every $x$ and defines a continuous function $f$. To see the latter, let $(x_k)$ be a sequence in $\Bbb R$ that converges to $x\in\Bbb R$. Then $$ f(x_k)-f(x) = \sum_na_n(\cos(nx_k)-\cos(nx)). $$ Now, the $n$-th summand converges to zero as $k\to\infty$ is bounded by $2|a_n|$. Hence, by Lebesgue's majorized convergence theorem, it follows that $f(x_k)\to f(x)$ as $k\to\infty$.

Answer 2

If $f(x)$ is continous at $x=0$ then we know

$$ \exists L \land \forall \epsilon \exists \delta \gt 0 \ \land \ if \ |x| \lt \delta \ then \ |f(x)-L| \lt \epsilon$$

This immediately implies that there exists infinitly many $x \neq 0$ such that $ |\sum_{n=0}^{\infty} a_n cos(nx) - L| \lt \epsilon $.

Answer 3

For each $n,$ $|a_n\cos (nx)|\le a_n,$ and we're given $\sum a_n <\infty.$ By the Weierstrass M test, $\sum a_n\cos (nx)$ converges uniformly on $\mathbb R.$ Since each summand is continuous on $\mathbb R,$ so is $f(x).$ Thus the answer to your question is no.