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Is it possible to have a continuous periodic function whose Fourier series $(a_n)_{n\in \mathbb{Z}}$ is such that $$ n . a_n \rightarrow 1?$$

I know that there exist continuous periodic functions with non-summable Fourier series, but I suspect that the sequence cannot have the proposed asymptotic behavior.

Edit. My initial question was for the condition $ |n . a_n |\rightarrow 1$, for which Conrad provides clear examples of continuous functions below, by playing on the phase of $a_n$.

Goulifet
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Much more is true, namely, there are continuous periodic functions for which $\Sigma{|a_n|^{(2-\epsilon)}}$ diverges for any positive $\epsilon$

Construction starts by analyzing the Hardy-Littlewood series: $\Sigma e^{(icn\log n)}\frac {e^{inx}}{n^{\frac{1}{2}+\alpha}}$, where $n \geq 1$, $\alpha$ real, $c>0$ and showing for example that it converges uniformly (so to a continuos periodic function on $[0, 2\pi]$) for $0 < \alpha <1$, function which is otherwise quite complicated; in particular $\alpha = \frac{1}{2}$, gives your required example.

Conrad
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    Also replacing $n^{\frac{1}{2}+\alpha}$ with $n^{\frac{1}{2}}(\log {n})^{\beta}$, where $\beta > 1$, gives a continuous periodic function and now the Fourier coefficients satisfy the strong property above for any $\epsilon > 0$ which is best we can do as continuous functions are in $L^2$ so their Fourier coefficients are square summable. – Conrad Feb 22 '19 at 00:43
  • This is very interesting, this definitely answers my initial question. But your examples are constructed by playing with the phase and I realize that I am more interested by the case where $n.a_n$ (not $\lvert n a_n\rvert$) as a limit. I update the question accordingly. What do you think about this new question? – Goulifet Feb 22 '19 at 16:51
  • for any continuous function (Riemann integrable enough with statement true at points of continuity) for which the Fourier coefficients are $O(\frac{1}{n})$, the Fourier series converges uniformly to it (at points of continuity respectively, uniformly on any compact segment of continuity), so you need strong oscillation in $a_n$ if the sum of $a_n$ absolutely diverges – Conrad Feb 22 '19 at 17:23
  • Note also that if you are willing to accept a slightly weaker condition (namely $na_n$ converging to zero and $a_n$ decreasing positive - so for example stuff like $a_n = \frac{1}{n\log {n}}$ or even $a_n = \frac{1}{n\log {\log {n}}}$ for $n$ large enough to make $a_n$ positive, the sin series, $\Sigma {a_n\sin{nx}}$ converges uniformly to a continuos function on $[0, 2\pi]$ (the cos series or the exponential Fourier series never converge at $0$ if $\Sigma {a_n}$ doesn't, obviously) and $\Sigma {a_n}$ is divergent and formed of positive terms – Conrad Feb 22 '19 at 21:56
  • I see. But I am more interested by simple conditions ensuring that the function is not continuous. I guess I understand globally why oscillations are necessary in your examples, but do you think there exists no continuous function such that $n,a_n$ has a nonzero limit? – Goulifet Feb 23 '19 at 18:11
  • The only possibility would be a sin series and then I think the proof that shows that to get uniform limit (which is necessary by the above) you need the stronger condition mentioned that works when the coefficients decrease can be adapted here, so I think that there is no such continuous function indeed – Conrad Feb 23 '19 at 18:23
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    So you get a condition as desired: for any coefficients $a_n$ positive, decreasing, $a_n = O(\frac{1}{n})$ but not $o(\frac{1}{n})$ all the three Fourier series (exponential, sin, cos) give you functions that are unbounded at the origin for the cos and exponential, but bounded for the sin and continuous everywhere else (up to the period $2\pi$ of course); as noted I am fairly confident, the decreasing condition can be removed for the sin series (as for the others there is nothing else needed) assuming as we do that $a_n = O(\frac{1}{n})$ and will look into that to make sure – Conrad Feb 24 '19 at 03:47
  • Could you tell how do you prove the discontinuity (or even unboundedness) around 0 for this case? Do you simply show that there are different right and left limits at the origin or something similar? – Goulifet Feb 27 '19 at 15:25
  • You need some lower bounds obviously, so $c \leq na_n \leq C$ for positive constants $c,C$ and $n$ large enough, condition which follows from either $a_n$ positive, decreasing, $a_n = O(\frac{1}{n})$ but not $a_n = o(\frac{1}{n})$ or from your limit condition $na$ converges to 1 (or some strictly positive number) and then you just notice that $\sum_{n}^{2n} a_k\sin{k\pi/(4n)} \geq \frac {c\sqrt{2}}{4}$ for all $n$ large enough and that contradicts uniform continuity on $[0, 2\pi]$ (since the series is $0$ at $x = 0$) which would happen if the sin series is continuos at zero hence everywhere – Conrad Feb 27 '19 at 15:49
  • remembering that continuity everywhere implies uniform continuity everywhere for Fourier series with coefficients that satisfy $a_n = O(\frac{1}{n})$ - for cos or exponential series, the sum at $0$ is obviously $\Sigma a_n$ so clearly you get infinity there in this case, but for the sin series, the result is bounded but discontinuous – Conrad Feb 27 '19 at 15:55
  • Nice, thanks a lot! – Goulifet Feb 27 '19 at 16:03
  • Conrad: Would you have a reference where the Hardy-Littlewood series you mentioned are studied? – Goulifet Jul 17 '19 at 20:15
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    One reference (and while dense and not recommended as a starter I find it invaluable in anything dealing with trigonometric series) is the classic book by Zygmund - Trigonometric series - (I have the 1988 edition which combines both volumes); Chapter 5, volume 1 has the stuff above – Conrad Jul 17 '19 at 20:36
  • Got it.I started reading this book very recently, and that's true, it is extremely riche and dense! Thanks for your help. – Goulifet Jul 18 '19 at 17:25
  • no problem and happy to be of help – Conrad Jul 18 '19 at 17:37