# $\mathcal{C}^{\alpha}$ Besov spaces: Definition

I'm reading an article for my future thesis (I'm a third-year undergraduate) where the authors define the generalized Holder Spaces as a special class of Besov Spaces.

Define $\chi,\tilde{\chi}\in C_{c}^{\infty}(\mathbb{R}^d)$ such that Supp$(\chi)\subset B(0,8/3) \setminus B(0,3/4)$ and Supp$(\tilde{\chi}) \subset B(0,4/3),$ and such that $\tilde{\chi}(x)+\sum^{+ \infty}_{k=0} \chi(x/2^k)=1 \quad \forall x\in \mathbb{R}^d$.

Define $\chi_{-1}:=\tilde{\chi},$ and $\chi_k(\cdot):=\chi(\cdot/2^k).$ Now, $\forall f \in C^{\infty}(\mathbb{T}^d),$ set $\delta_k(f):=\mathscr{F}^{-1}(\hat{f} \cdot \chi_k$), where $\hat{f}(k):=\int_{\mathbb{T}^d}f(x)e^{-2\pi i k\cdot x} dx$ and $\mathscr{F}^{-1}(g)(x):=\sum_{k\in \mathbb{z}^d} g(k)e^{2 \pi i k\cdot x}.$ Heuristically, $\delta_k(f)$ is just a part of the frequencies of the smooth function $f.$

Now, for every $\alpha \in \mathbb{R},$ we can define the $\mathcal{C}^{\alpha}$ norm of a smooth function, which is $\|f\|_{\mathcal{C}^{\alpha}}:=sup_{k\geq1} 2^{\alpha k} \|\delta_k(f)\|_{L^{\infty}}.$

The space $\mathcal{C}^{\alpha}$ is defined as the completion of $C^{\infty}(\mathbb{T}^d)$ with respect to this norm.

The definition given is a bit different from the one which is mostly given in literature: the space of all tempered distributions such that the above-mentioned norm (which is well-defined because the Fourier antitransform of a compactly-supported distribution is a function) is finite.

Now, me questions are:

1. Why the $\mathcal{C}^{\alpha}$ norm is finite for every smooth function defined on the torus?

2. Why does it coincide (well, I don't think they precisely coincide, but they should be equivalent or at least generate the same completion) with the classical $\alpha-$Holder norm ($\|f\|_{\mathcal{C}^{\alpha}}=sup_{x\neq y} \frac{|f(x)-f(y)|}{|x-y|^{\alpha}})$?

3. Do the $\mathcal{C}^{\alpha}$ spaces respectively defined via completion of $C^{\infty}$ and via the tempered distributions coincide?

The article can be found here.

1. As for the first point I will just provide a hint: you can write each Littlewood-Paley block $\delta_k(f)$ as a convolution between the distribution $f$ and a smooth function $\varrho_k$ which I leave to you to think about. Crucial point is that $\varrho_k(x) = 2^{dk}\varrho(2^kx)$ for every $k$ and a fixed smooth function $\varrho.$ Now since $\varrho$ has zero Fourier coefficients near zero you can invert the laplacian, i.e. for any $\zeta \in \mathbb{N}$ find $\varphi$ such that $\Delta^\zeta \varphi = \varrho$ and now you can use integration by parts to find your bound (note that the $\mathcal{C}^{\alpha}$ norm is bounded by the supremum norm for $\alpha < 0$).
2. This is a theorem, for $\alpha \in (0,1)$. A reference could be the book by Chemin Bahouri and Danchin (Fourier Analysis and nonlinear PDEs) or some lecture notes on the net. BUT: you do need to take the space defined through tempered distributions,NOT the completion of $C^{\infty}.$ Why is the content of the next point.
3. $C^{\infty}$ is not dense in $\mathcal{C}^{\alpha}$ if you define the latter with tempered distributions. You should think of the difference between Lipschitz functions and $C^1$ functions. A striking example in the non-integer case is provided here. The result that makes everything work AS THOUGH $C^{\infty}$ would be dense is interpolation: if $f_n \to f$ in the sense of distributions and $$\sup_n\| f_n\|_{\mathcal{C}^{\alpha}} < {+} \infty$$ then $f_n \to f$ in $\mathcal{C}^{\beta}$ for any $\beta < \alpha.$
• If I understand the argument, can we proceed as follows? We fix an integer $\beta > d + \frac{\alpha}{2}$, then we can write \begin{align*} \delta_k f &= \delta_k (-\Delta)^{-\beta} g, \quad\text{with}\quad g=(-\Delta)^{\beta} f \in C^\infty(\mathbb{T}^d) \\ &=\sum_{|n|\sim 2^k} \chi(n/2^k) |n|^{-2\beta} \widehat{g}(n) e_n. \end{align*} Its $L^\infty$-norm is bounded by $$\leq C_{\chi} 2^{-2\beta k + dk} \sum_{n\in\mathbb{Z}^d} | \widehat{g}(n)| \lesssim 2^{-2\beta k + dk}$$ This would give us $2^{\alpha k} \| \delta_k f\|_\infty$ uniformly bounded in $k$. Sep 10, 2021 at 16:06