0

There is a proposition in Fourier analysis that states that, for some $\epsilon > 0$, $M > 0$, if $c_k \leq M/|k|^{n+1+\epsilon}$, where $c_k$ is the k-th Fourier coefficient of $f$, then $f \in C^n$ a.e. I am trying to find a function $f \not\in C^n$ that satisfies $c_k \leq M/|k|^{n+1}$.

The reason why we need the $\epsilon$ in the bound for $c_k$ is that to show that $s_nf \to f$, we need that $|s_af - s_bf| \leq |\sum_{a<|k|<b} c_k e^{ikx}| \leq \sum_{a<|k|<b} |c_k| < \infty$ for $n=0$ case, but this is true for $c_k = M/|k|^{1+\epsilon} < \infty$ iff $\epsilon > 0$ and not true for $\epsilon = 0$ since the harmonic series diverges.

It is therefore clear why the epsilon is necessary, but finding an example of it breaking down for $\epsilon = 0$ has eluded me.

Tomo
  • 2,141

1 Answers1

1

Hint: Try $f(x) = |x|$ and $n = 1$.

Julien Clancy
  • 4,868
  • But though $f$ is not $C^1$, it is $C^1$ a.e., no? – Tomo Apr 05 '13 at 06:50
  • There is no such thing as $C^1$ almost everywhere. In any case, it is not equal to any continuously differentiable function almost everywhere. Observe that $f'$ has a jump discontinuity at zero. Assume that $f' \equiv g'$ a.e. and derive a contradiction for some $\delta$ and $\varepsilon = 1/2$. – Julien Clancy Apr 05 '13 at 06:55
  • I threw out that answer earlier because of this misunderstanding. thanks Julien. =) – Tomo Apr 05 '13 at 06:57
  • No problem, any time :). Is there any chance you could help me with this question? If you are who I think you are. http://math.stackexchange.com/questions/351866/decay-of-fourier-coefficients-and-smoothness – Julien Clancy Apr 05 '13 at 06:58
  • If you are not that person then this is rather embarrassing. But it's worth a shot. – Julien Clancy Apr 05 '13 at 07:01
  • 1
    Lol Julien what is the chance that the same person who answered your Galois question last night also asked this one tonight? As for that problem, when I did it I differentiated termwise, so I will need to go back to it and either find a justification for termwise differentiation or find another way. – Tomo Apr 05 '13 at 07:11
  • I actually ended up doing the Galois question another way; I don't know how to prove casus irreducibilis. I figured the chances were quite low, especially given your username. – Julien Clancy Apr 05 '13 at 07:15
  • What a skeptic. – Tomo Apr 05 '13 at 07:18