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I have read several times that there exist many continuous functions whose Fourier series diverge at some points (sometimes even on a dense subset of the domain!). I tried to find an explicit example of them and the closest thing I found is This example by Fejer, a link from MSE. According to a comment responding to the question, it seems that Fejer's example cannot be represented by a graph. Here are my questions:

1.)What is the reason that the function cannot be represented by graph?

2.)Is there any function with the properties I mentioned that can be represented by graph?

3.)What is the intuitive reason that those functions have divergent Fourier series at some points?

I am aware that this question is a duplication of some already existing topics on MSE but none of the links I read give me a satisfactory answer. Also, could anyone give me a rough description of Fejer's function's behavior that explains why it can't be graphed? Thank you in advance.

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BigbearZzz
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  • No, that comment in that other thread does not say it's impossible to represent the function by a graph. It says that drawing the graph on a computer is very slow. And it says that it is impossible to draw the graph and see from looking at the graph why the series diverges. – David C. Ullrich Oct 01 '15 at 13:52
  • Alright, then could you please explain why is it hard to draw the graph? Is it because of high variation or it has many fine details? – BigbearZzz Oct 01 '15 at 14:11
  • Yes. The function is continuous, but it can't be too continuous or the Fourier series would converge. So it's sort of continuous but not very, meaning that we need to plot a large number of points to get a good picture. And since the Fourier series diverges we can't calculate the values by adding the Fourier series... – David C. Ullrich Oct 01 '15 at 14:16
  • If $f$ is continuous at $x$ and its Fourier series does not converge at $x$ or converges to something other than $f(x)$, then $f$ cannot be of bounded variation on any interval $(x-\epsilon,x+\epsilon)$, no matter how small you take $\epsilon$ to be. (This is because if it were of bounded variation on some such interval, the Fourier series would converge at $x$ to $f(x)$.) So you'd have to move your pen infinitely far to plot the graph over any such small interval! – Disintegrating By Parts Oct 01 '15 at 20:10

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