This is my second question in a short time on Besov spaces. I apologize. I am having a rough time with them and I really need to understand this spaces quickly.

The Besov spaces $B^s_{\infty,\infty}(\mathbb{R^n})$ for $s \in \mathbb{R}$ are defined by the following:

A tempered distribution $u \in \mathcal{S}'(\mathbb{R^n})$ is said to be in the Besov space $B^s_{\infty,\infty}(\mathbb{R^n})$ if we have that $$ \|u\|_{B^s_{\infty,\infty}(\mathbb{R^n})}:=\sup_{j \in \mathbb{N}} 2^{js} \|\varphi_j(D)u\|_{L^{\infty}} < \infty $$

Where $\varphi_j$, $j \ge 0$ is a Littlewood–Paley partition of unity, and

$\varphi_j(D)u:=\mathcal{F}^{-1} \varphi_j \mathcal{F}u$


My question is: given $u \in B^s_{\infty,\infty}(\mathbb{R^n})$, is its distributional derivative $\partial_{x_i}u$ in the space $B^{s-1}_{\infty,\infty}$?

How I tried

My calculations show that $\varphi_j(D)u_{x_i} = (\varphi_j(D)u)_{x_i}$, so for a schwartz function $\phi \in \mathcal{S}$ we have


Taking the supremum on $\phi$ with $\|\phi\|_{L^1}=1$ in the expression above we should obtain $\|\varphi_j(D)u_{x_i}\|_{L^{\infty}}$ but from this I do not get any good decay for $\|\varphi_j(D)u_{x_i}\|_{L^{\infty}}$, since the differential operators are unbounded in $L^1$


This result is true for $s>1$ non integer since it can be proved that the Besov spaces $B^s_{\infty,\infty}$ coincide with the Hölder spaces $C^s$ for $s>0$ non integer. For integer $s\ge 1$ it is also true since they coincide with the Zygmund spaces defined in my other question here

Besov–Zygmund spaces and the Inverse Function Theorem, is the Inverse Zygmund?

In the classical references I do not find this result, so I begin to doubt whether it is true, since it should be easy to prove and should be among the basic properties on Besov spaces that are in any book, but it seems it is not the case (maybe it is just too easy for the books and I don't see it).

Than you for any help.


1 Answer 1


Theorem 9 of Chapter 3 of Peetre's ``New Thoughts on Besov Spaces'':

For any multi-index $\alpha$, we have the continuous map

\begin{equation*} D^{\alpha}: B_p^{s,q}(\mathbb{R}^n) \rightarrow B_p^{s-|\alpha|,q}(\mathbb{R}^n) \end{equation*}

Proof Sketch:

The $B_{p,q}^s(\mathbb{R}^n)$ norm of a function $u$ consists of terms of the form $\| \varphi_j(D)u \|_{L^p}$, so showing that $\partial_{x_i}:B_{p,q}^s(\mathbb{R}^n) \rightarrow B_{p,q}^{s-1}(\mathbb{R}^n)$ continuously, reduces to bounding $\|\partial_{x_i} \varphi_j(D)u \|_{L^p}$ by $\|\varphi_j(D)u \|_{L^p}$. The required bound is

\begin{equation*} \|\partial_{x_i} \varphi_j(D)u \|_{L^p} \leq C 2^j \|\varphi_j(D)u \|_{L^p} \end{equation*} for some uniform constant $C>0$. Such inequalities that bound a (semi-) norm by a weaker norm are commonly called Bernstein inequalities. In this case, $\varphi_j$ is compactly supported, so $\varphi_j(D)u$ is a band-limited function with bandwidth $\approx 2^k$. The inequality can be proved for a fixed $j_0$ and extended using the homogeneity of $\partial_{x_i}$.


If you search the literature for Bernstein inequalities, you may need to search for entire functions of exponential type in place of band-limited functions.

  • 1
    $\begingroup$ Could you perhaps add some detail, or else consider making this a comment? Thanks $\endgroup$
    – user452
    Oct 14, 2014 at 13:14
  • $\begingroup$ I understand what you are saying in your note, but for someone who is not familiar with these methods, it would be useful to add more detail to the note. $\endgroup$
    – robjohn
    Oct 15, 2014 at 0:18

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