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Is there an example of a periodic continuous function on $\mathbb{R}$ such that its Fourier series is uniformly convergent (to $f$), but it is not absolutely convergent ?

Phil-W
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1 Answers1

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$\sum \frac {\sin (nx)} {n\log \, n}$ is such a series. The proof is not obvious but it is a well known result. See Edwards, Fourier Series. p. 166.

Kavi Rama Murthy
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  • If you assume the well-known fact that $\sum \frac {\sin (nx)} {n}$ has partial sums that are uniformly bounded, the proof is immediate with a summation by parts. With oscillating coefficients we can construct examples for which $a_n$ ~ $n^{-a}, a > \frac{1}{2}$ and even $a_n$ ~ $n^{-\frac{1}{2}}{(\log n})^{-1-b}, b>0$ which is best we can do since by Plancherel we know that $\Sigma{|a_n|^2} < \infty$ when the Fourier series converges uniformly, so in particular we get examples of continuos functions with Fourier series for which $\Sigma{|a_n|^{(2-b)}} = \infty$ for any $b>0$ – Conrad Mar 28 '19 at 12:31
  • Use $\Sigma{\frac {1} {n\log , n}} = \infty$ and $x=\frac{\pi}{2}$ and the absolute sum is the odd part of $\Sigma{\frac {1} {n\log , n}}$ – Conrad Mar 28 '19 at 12:47
  • @Conrad You are right. – Kavi Rama Murthy Mar 28 '19 at 13:00
  • The tricky part is uniform convergence near $0$ and for that some subtle result or method is needed since the result is not true for cosine series where full period interval pointwise, uniform and absolute convergence are obviously equivalent when the coefficients are positive – Conrad Mar 28 '19 at 14:12
  • @Conrad In the book I have referred to it is shown that if $(a_n)$ decreases to $0$ then $\sum a_n \sin(nx)$ is uniformly convergent iff $na_n \to 0$. – Kavi Rama Murthy Mar 28 '19 at 23:21
  • Yes definitely, but the point is that $na_n \to 0$ implies uniform convergence of the sin series is fairly subtle and requires either a careful split of the partial sums or when $na_n$ also decreases the use of the uniform boundnes mentioned above which is one of those well known facts cited everywhere; the other implication is fairly easy as you can truncate the sin series on a length ~n interval starting at index ~n at $x=\frac{\pi}{n}$ and estimate from below by ~$na_n$ so uniform convergence implying uniform Cauchy, requires that to go to $0$. – Conrad Mar 28 '19 at 23:58