The theory of Besov and Triebel-Lizorkin spaces usually proceeds by taking a dyadic decomposition of unity, i.e. some non-negative functions $\psi_0,\psi \in C_c^\infty(\mathbb{R})$ such that \begin{equation*} \psi_0(x) + \sum_{n=1}^\infty \psi(2^{-n}x) = 1. \end{equation*}

My question is whether or not one may define equivalent Besov/Triebel-Lizrokin norms using a different ('logarithmic') decomposition of unity such as say \begin{equation*} \psi_0(x) + \sum_{n=1}^\infty \psi(3^{-n}x) = 1. \end{equation*}

Giving explicit definitions, one usually defines the Littlewood-Paley blocks of a function $f \in C_c^\infty(\mathbb{R})$ (taking it to be smooth/compactly supported only for illustration) via the fourier transform $\hat{f}$ \begin{gather*} \hat{f_0}(\xi) = \psi_0(\xi)\hat{f}(\xi), \\ \hat{f_n}(\xi) = \psi(2^{-n}\xi) \hat{f}(\xi), \end{gather*} or using the shorthand for Fourier multipliers, $f_0=\psi_0(D)f$, and $f_n = \psi(2^{-n}D)f$.

Then the Besov, and Triebel-Lizorkin norms ($s=0$ for illustration) are defined by \begin{gather*} \left\|f\right\|_{B^0_{p,q}} = \left(\sum_{n=0}^\infty \left\|f_n\right\|_{L^p}^q\right)^{\frac{1}{q}}, \\ \left\|f\right\|_{F^0_{p,q}} = \left\|\left(\sum_{n=0}^\infty |f_n(x)|^q\right)^{\frac{1}{q}}\right\|_{L^p}. \end{gather*}

I am aware of the Book by Triebel (Theory of Function Spaces II) which gives many equivalent characterisation of these norms. But missing in this book is the question of when instead of a dyadic decomposition of unity, I use some other ('logarithmic') decomposition of unity, such as say

\begin{equation*} \psi_0(x) + \sum_{n=1}^\infty \psi(3^{-n}x) = 1. \end{equation*}

If one then defines $f_n = \psi(3^{-n}D)f$, this likely still produces an equivalent norm, but does anyone know a reference for such a statement?


1 Answer 1


The norms should be equivalent with a constant depending only on the ratio between the two bases (in this case, 2 and 3). I'll just consider the Besov case.

I'll adopt the notation $\hat{f}_{n,b}(\xi)=\psi(b^{-n}\xi)\hat{f}(\xi)$ for $b>1$ and $n\geq 1$, with the obvious modification when $n=0$; note that $\hat{f}_{0,b}$ is independent of $b$. For any other $b'$ and any $n\geq 1$, there are $O_{b/b'}(1)$-many $n'\geq 1$ such that $\operatorname{supp}_\xi(\psi(b^{-n}\xi))\cap\operatorname{supp}_\xi(\psi(b'^{-n'}\xi))\neq\emptyset$; when this holds, I'll write $n\sim_{b,b'}n'$. As a consequence,


and, since each summand appears now $O_{b/b'}(1)$-many times,


By symmetry, we have an identical reverse inequality, which gives your equivalence.

EDIT: to add more explanation about the first display, write $\psi_{b,n}(\xi)=\psi(b^{-n}\xi)$. Then $\psi_{b,n}=\psi_{b,n}\sum_{n'\sim_{b,b'}n}\psi_{b',n'}$, so


using Young, since $\psi_{n,b}(D)g=\psi_{n,b}^\vee*g$ and $\|\psi_{n,b}^\vee\|_1=\|\psi^\vee\|_1$. Finally, by the triangle inequality and Cauchy-Schwarz,


  • $\begingroup$ why is it clear that if say $\mathrm{supp}(\psi) \subset \mathrm{supp}(\phi)$ then $\left\|\psi(D)f\right\|_{L^p} \lesssim \left\|\phi(D)f\right\|_{L^p}$? $\endgroup$
    – vmist
    Commented Jan 17 at 12:46
  • 1
    $\begingroup$ The estimate is instead something like "$\psi\phi=\psi$ implies $\|\psi(D)f\|_{L^p}\lesssim\|\phi(D)f\|_{L^p}$." To prove the latter, write $\|\psi(D)f\|_{L^p}=\|\psi(D)[\phi(D)f]\|_{L^p}\lesssim\|\phi(D)f\|_{L^p}$, where the inequality is Young's inequality (since $\psi(D)$ is given by convolving with a unit-$L^1$ function). $\endgroup$ Commented Jan 17 at 18:41

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