2

For $\alpha\in(0,1]$, define the $\alpha$-Hölder space on the 1-torus $\mathbb{T}:=\mathbb{R}/2\pi\mathbb{Z}$ as the space: $$C^{0,\alpha}(\mathbb{T}):=\{f\in C(\mathbb{T})\ |\ \sup_{s,t\in\mathbb{T}\\s\neq t}\frac{|f(s)-f(t)|}{|s-t|^\alpha}<+\infty\}.$$ Define: $$\|\cdot\|_{C^{0,\alpha}(\mathbb{T})}:C^{0,\alpha}(\mathbb{T})\to[0,+\infty), f\mapsto \|f\|_\infty+\sup_{s,t\in\mathbb{T}\\s\neq t}\frac{|f(s)-f(t)|}{|s-t|^\alpha}.$$ Then $(C^{0,\alpha}(\mathbb{T}),\|\cdot\|_{C^{0,\alpha}(\mathbb{T})})$ is a Banach space.

Define $$\forall n\in\mathbb{Z}, \forall t\in\mathbb{T},e_n(t):=e^{int}$$ and if $f\in L^1(\mathbb{T})$, the Fourier tranform of $f$ will be denoted by $\hat{f}$, i.e. $$\forall n\in\mathbb{Z},\hat{f}(n)=\int_{-\pi}^\pi f(t)e_{-n}(t)\frac{\operatorname{d}t}{2\pi}.$$

I know from the answers to this question that: $$\forall f\in C^{0,\alpha}(\mathbb{T}), \|\sum_{n=-N}^N \hat{f}(n)e_n-f\|_\infty \to0, N\to +\infty$$

While, since $C^{0,1}(\mathbb{T})$ is essentially the Sobolev space $W^{1,\infty}(\mathbb{T})$, I know that: $$\exists f\in C^{0,1}(\mathbb{T}), \|\sum_{n=-N}^N \hat{f}(n)e_n-f\|_{C^{0,1}(\mathbb{T})}\not\to0, N\to+\infty.$$

For which $\alpha\in(0,1)$, is it true that $$\forall f\in C^{0,\alpha}(\mathbb{T}), \|\sum_{n=-N}^N \hat{f}(n)e_n-f\|_{C^{0,\alpha}(\mathbb{T})}\to0, N\to +\infty?$$

Bob
  • 5,705
  • What is the explicit bound in the Dirichlet kernel + Holder continuity proof of the convergence of Fourier series ? If $f$ is $\alpha$-Holder continuous $1$-periodic then $|f(x)-f \ast D_N(x)| \le h_{\alpha,N}( |f|\infty,\sup{x,y} \frac{|f(x)-f(y)|}{(x-y)^\alpha})$ where $D_N(x) =\sum_{|n| \le N} e^{2i \pi nx} $, $f \ast D_N(x)= \int_0^1 f(y) D_n(x-y)dy=\sum_{|n| \le N} \hat{f}(n)e^{2i \pi nx}$. I'm asking the exact form of that function $h_{\alpha,N}$. – reuns Jan 29 '19 at 00:22
  • I think that an explicit form for $h_{\alpha,N}$ can be deduced from my answer in the linked question. It is hidden inside the constant $C$ in the last estimate: it is enough to unpack the other constants which $C$ is built from – Bob Jan 29 '19 at 00:31

0 Answers0