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

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.

• If you take a bounded even function which is $2 \pi$-periodic, continuous at zero and disctontinuous somewhere else, then its Fourier series will converge to it almost everywhere, and will be made only by cosines. Aug 27, 2018 at 9:52
• But will we know its Fourier coefficients are $\geq0$? For the example of a symmetric step function, this is not the case. Aug 27, 2018 at 9:57
• By "is not a continuous function" you mean that there is no continuous function $g$ such that $f=g$ a.e.? Aug 27, 2018 at 10:09
• OP, I suggest editing your post title and maybe also your post to point out the non-negativity assumption more clearly. Aug 27, 2018 at 10:30
• This is simple. First, if $(a_n)\in\ell^1$, then the series obviously converges for every $x$. If $x_k\to x$, then $f(x_k)-f(x) = \sum_na_n(\cos(nx_k)-\cos(nx))$. Each summand converges to zero and is bounded by $2|a_n|$. Hence, Lebesgue's majorized convergence theorem implies that the whole thing tends to zero as $k\to\infty$. Aug 27, 2018 at 10:49

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 $k\to\infty$.

• Marked as answered, but I still don't get how you used Lebesgues domainated convergence theorem here. (Assuming that's what you mean.) That theorem is about convergence of integrals, we are talking about convergence of limits. Aug 29, 2018 at 7:18
• @Neuromath The theorem is stated for integrals with respect to arbitrary measures. Here, it is the counting measure $\mu(A) = \#(A\cap\Bbb Z)$. Aug 29, 2018 at 15:57

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.

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 $|\sum_{n=0}^{\infty} a_n cos(nx) - L| \lt \epsilon$.

• I don't know how to find non continuous cases for $a_n$ - part of your question. Just thought this might be helpful for other aspects of the question. Aug 27, 2018 at 10:49
• Also as some comented ask, not at no point do we consider when $x=0$ because our $\delta =0$ which is not found in definition of continuity. Aug 27, 2018 at 10:52
• Woops yes I redact last comment because it explicitly $f(x)=L$ to be cont at $x$. Aug 27, 2018 at 15:11