Find kk so that |an|≤Cn−k|an|≤Cn−k, |bn|≤Cn−k|bn|≤Cn−k => f∈C1(T)f∈C1(T) (anan,bnbn are the Fourier coefficients of f)

Question

I'm trying to find k that when |an|≤Cn−k$|a_n|\le C{n}^{-k}$,|bn|≤Cn−k$|b_n|\le C{n}^{-k}$, (|a0|≤C$|a_0|\le C$), f satisfies f∈C1(T)$f\in C^1(T)$ which can proof Fourier series is converged. I feel k may be 1$1$ but I don't know how to proof it.

Answer 1

(Apologies if this has already been said - MSE is telling me an answer already exists but I can't see it...)

No, f=∑∞1cos(nx)n$f=\sum _1^{\mathrm{\infty }}\frac{\cos (nx)}{n}$ shows that k=1$k=1$ does not even imply f$f$ is continuous. (Because it's clear that σn(f,0)→∞${\sigma }_n(f,0)\to \mathrm{\infty }$.) Similarly, if f=∑sin(nx)n2$f=\sum \frac{\sin (nx)}{n^2}$ then f$f$ is certainly continuous, but not continuously differentiable; thus k=2$k=2$ is not enough.

Otoh if k>2$k>2$ then f∈C1$f\in C^1$, roughly because ∑|n|−k+1<∞$\sum |n{|}^{-k+1}<\mathrm{\infty }$.

Answer 2

If f$f$ is in L2$L^2$ then ∑a2k+b2k<∞$\sum a_k^2+b_k^2<\mathrm{\infty }$ Since derivative of f$f$ has Fourier coefficients k∗ak$k\ast a_k$ if the derivative belongs to L2$L^2$ then ∑(k∗ak)2+(k∗bk)2<∞$\sum (k\ast a_k{)}^2+(k\ast b_k{)}^2<\mathrm{\infty }$ . Now, using Cauchy-Schwartz one gets this estimate

∑|ak|+|bk|=∑k∗|ak|/k+k∗|bk|/k<=(∑(k∗ak)2+(k∗bk)2)∗(∑1/k2)$$\sum |a_k|+|b_k|=\sum k\ast |a_k|/k+k\ast |b_k|/k<=(\sum (k\ast a_k{)}^2+(k\ast b_k{)}^2)\ast (\sum 1/k^2)$$

So, if ∑(k∗ak)2+(k∗bk)2<∞$\sum (k\ast a_k{)}^2+(k\ast b_k{)}^2<\mathrm{\infty }$ then ∑|ak|+|bk|<∞$\sum |a_k|+|b_k|<\mathrm{\infty }$ which implies that f$f$ has absolutely convergent Fourier series and is continuous. In short, if |k∗ak|<1/k1+s$|k\ast a_k|<1/k^{1+s}$ and |k∗bk|<1/k1+s$|k\ast b_k|<1/k^{1+s}$ where s>1/2$s>1/2$ then f$f$ is in C0(T)$C^0(T)$ Using this trick multi times, one gets - if |k∗ak|<1/km+s$|k\ast a_k|<1/k^{m+s}$ and |k∗bk|<1/km+s$|k\ast b_k|<1/k^{m+s}$ then f$f$ is in Cm−1(T)$C^{m-1}(T)$.