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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.

Neuromath
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    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. – Crostul Aug 27 '18 at 09:52
  • But will we know its Fourier coefficients are $\geq0$? For the example of a symmetric step function, this is not the case. – Neuromath Aug 27 '18 at 09:57
  • By "is not a continuous function" you mean that there is no continuous function $g$ such that $f=g$ a.e.? – amsmath Aug 27 '18 at 10:09
  • You cannot have the coefficients $a_n$ be summable(necessary for continuity at 0), positive and decreasing, at least. This would imply $n a_n \to 0$ and thus $f$ is continuous. – Calvin Khor Aug 27 '18 at 10:11
  • @CalvinKhor What about $a_n = 1/n$ ? – amsmath Aug 27 '18 at 10:13
  • @amsmath that's not convergent at 0 (the harmonic series appears), contrary to assumptions in OP. I have edited my comment just in time to clarify this – Calvin Khor Aug 27 '18 at 10:14
  • I've recently learned that desmos implements summation, so if anyone else would like to explore with a graphing tool, https://www.desmos.com/calculator/uuew6rewh8 – Calvin Khor Aug 27 '18 at 10:19
  • Take the Fourier series of a periodic function with discontinuities. –  Aug 27 '18 at 10:25
  • Please take a look at Gibbs-Phenomenon https://en.wikipedia.org/wiki/Gibbs_phenomenon – denklo Aug 27 '18 at 10:28
  • At least $(a_n)$ cannot be $\ell^1$ because in this case $f$ is a continuous function and the series converges everywhere. Hence, $f(0)$ is not defined (by the series). – amsmath Aug 27 '18 at 10:28
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    OP, I suggest editing your post title and maybe also your post to point out the non-negativity assumption more clearly. – Calvin Khor Aug 27 '18 at 10:30
  • @amsmath thats what I said, pretty much. The non-negativity is the interesting part because there are examples like the squarewave in the above graph. – Calvin Khor Aug 27 '18 at 10:32
  • @CalvinKhor Which graph do you mean? And for my claim non-negativity is not needed. This is a simple consequence of Lebesgue's majorized convergence theorem. – amsmath Aug 27 '18 at 10:34
  • @amsmath the desmos link above. I know the proof. – Calvin Khor Aug 27 '18 at 10:35
  • @CalvinKhor, thanks I will edit the title – Neuromath Aug 27 '18 at 10:37
  • @CalvinKhor In order to avoid misunderstandings, please don't change the meaning of your comments by editing. Thanks. – amsmath Aug 27 '18 at 10:38
  • It still is not clarified what "continuous at zero" means if the function is not defined there. – amsmath Aug 27 '18 at 10:40
  • @amsmath my apologies, you can think of the edit mark as an admission of error. I'm only allowed to edit the comments for 5 minutes, will you give me the same leeway as this website? – Calvin Khor Aug 27 '18 at 10:42
  • Yes, the original question required the function to be defined at zero. @amsmath I think you may have answered the question. Can you provide a link to the theorem that if $a_n$ are in $l^1$ then $f$ must be continuous? – Neuromath Aug 27 '18 at 10:45
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    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 $k\to\infty$. – amsmath Aug 27 '18 at 10:49
  • @amsmath Thank you, I think that is the answer (in fact a stronger one): if the series converges at $x=0$, then $(a_n)\in l^1$, so $f(x)$ will be continuous everywhere. Could you post this as an answer, so I can mark it as answered? – Neuromath Aug 27 '18 at 12:31
  • @amsmath, Actually though I don't see how Lebesgues dominated convergence theorem is relevant - that is about limits of integrals, rather than showing continuity? But I can see an argument where you bound $|f(x')-f(x)| < \epsilon$ by finding $N$ where $2\sum_{n>N} a_n \leq \epsilon/2$, and $\delta$ where $|\sum_{n \leq N} a_n(\cos(nx')-\cos(nx))| \leq \epsilon/2$ if $|x'-x|<\delta$ by continuiity of a finite sum – Neuromath Aug 27 '18 at 16:06

3 Answers3

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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 $k\to\infty$.

amsmath
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  • 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. – Neuromath Aug 29 '18 at 07:18
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    @Neuromath The theorem is stated for integrals with respect to arbitrary measures. Here, it is the counting measure $\mu(A) = #(A\cap\Bbb Z)$. – amsmath Aug 29 '18 at 15:57
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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.

zhw.
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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 $.

marshal craft
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  • 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. – marshal craft Aug 27 '18 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. – marshal craft Aug 27 '18 at 10:52
  • Woops yes I redact last comment because it explicitly $f(x)=L$ to be cont at $x$. – marshal craft Aug 27 '18 at 15:11