Harmonic Besov and Triebel–Lizorkin Spaces on the Ball

Abstract

Harmonic Besov and Triebel–Lizorkin spaces on the unit ball in ${\mathbb R}^d$ with full range of parameters are introduced and studied. It is shown that these spaces can be identified with respective Besov and Triebel–Lizorkin spaces of distributions on the sphere. Frames consisting of harmonic functions are also developed and frame characterization of the harmonic Besov and Triebel–Lizorkin spaces is established.

Appendix

1.1 Proof of Lemma 2.3

Assume $g\in \Pi _N$. Let $\lambda \in C^\infty ({\mathbb R}_+)$ be an admissible function of type (a) in the sense of Definition 2.1. Set

$$\begin{aligned} \Lambda _N(x\cdot y):=\sum _{k\ge 0} \lambda (k/N)Z_k(x\cdot y). \end{aligned}$$

Clearly, $\Lambda _N*g=g$ and by Theorem 2.2 for any $M>0$ there exists a constant $c>0$ such that for $x, y, z\in {{\mathbb S}^{d-1}}$

$$\begin{aligned} |\Lambda _N(x\cdot z) - \Lambda _N(y\cdot z)| \le \frac{c \rho (x, y)N^{d}}{(1+N\rho (x, z))^M}, \quad \hbox {if}\;\; \rho (x, y)\le N^{-1}. \end{aligned}$$

(9.1)

Fix $0<{\varepsilon }<1$. For $y\in {{\mathbb S}^{d-1}}$ we have

$$\begin{aligned} |g(y)| \le \inf _{u\in G(y, {\varepsilon }N^{-1})}|g(u)| + \sup _{u\in G(y, {\varepsilon }N^{-1})}|g(y)-g(u)| \end{aligned}$$

and hence

$$\begin{aligned} H(x)&:= \sup _{y\in {{\mathbb S}^{d-1}}}\frac{|g(y)|}{(1+N\rho (x, y))^{(d-1)/t}} \le \sup _{y\in {{\mathbb S}^{d-1}}}\frac{\inf _{u\in G(y, {\varepsilon }N^{-1})}|g(u)|}{(1+N\rho (x, y))^{(d-1)/t}}\\&\quad + \sup _{y\in {{\mathbb S}^{d-1}}}\frac{\sup _{u\in G(y, {\varepsilon }N^{-1})}|g(y)-g(u)|}{(1+N\rho (x, y))^{(d-1)/t}} =: H_1(x)+H_2(x). \end{aligned}$$

To estimate $H_1(x)$ we first note that

$$\begin{aligned} \inf _{u\in G(y, {\varepsilon }N^{-1})}|g(u)| \le \left( \frac{1}{|G(y, {\varepsilon }N^{-1})|}\int _{G(y, {\varepsilon }N^{-1})}|g(u)|^t \mathrm{d}\sigma (u)\right) ^{1/t}, \end{aligned}$$

implying

$$\begin{aligned} H_1(x)&\le \left( \frac{|G(x, \rho (x, y)+{\varepsilon }N^{-1})|}{|G(y, {\varepsilon }N^{-1})|(1+N\rho (x, y))^{d-1}}\right) ^{1/t}\nonumber \\&\quad \times \left( \frac{1}{|G(x, \rho (x, y)+{\varepsilon }N^{-1})|} \int _{G(y, {\varepsilon }N^{-1})}|g(u)|^t \mathrm{d}\sigma (u) \right) ^{1/t}. \end{aligned}$$

(9.2)

Since $G\big (x, \rho (x, y)+{\varepsilon }N^{-1}\big ) \subset G\big (y, 2\rho (x, y)+{\varepsilon }N^{-1}\big )$, we have

$$\begin{aligned} \big |G\big (x, \rho (x, y)+{\varepsilon }N^{-1}\big )\big |&\le \big |G\big (y, 2\rho (x, y)+{\varepsilon }N^{-1}\big )\big |\\&\le (2/{\varepsilon })^{d-1}(1+ N\rho (x, y))^{d-1}|G(y, {\varepsilon }N^{-1})|. \end{aligned}$$

We use the above in (9.2) and enlarge the region of integration in (9.2) from $B(y, {\varepsilon }N^{-1})$ to $G(x, \rho (x, y)+{\varepsilon }N^{-1})$ to bound $H_1(x)$ by

$$\begin{aligned}&c {\varepsilon }^{(-d+1)/t}\sup _{y\in {{\mathbb S}^{d-1}}} \left( \frac{1}{|G(x, \rho (x, y)+{\varepsilon }N^{-1})|} \int _{G(x, \rho (x, y)+{\varepsilon }N^{-1})} |g(u)|^t \mathrm{d}\sigma (u)\right) ^{1/t}\\&\qquad \le c {\varepsilon }^{(-d+1)/t} \mathcal {M}_tg(x). \end{aligned}$$

Therefore,

$$\begin{aligned} H_1(x) \le c {\varepsilon }^{(-d+1)/t} \mathcal {M}_tg(x). \end{aligned}$$

(9.3)

We now estimate $H_2(x)$. Using (9.1) we obtain

$$\begin{aligned} \sup _{u\in G(y, {\varepsilon }N^{-1})}|g(y)-g(u)|&\le \sup _{u\in G(y, {\varepsilon }N^{-1})} \int _{{\mathbb S}^{d-1}}|\Lambda _N(y\cdot z)- \Lambda _N(u\cdot z)||g(z)| \mathrm{d}\sigma (z)\\&\le c\sup _{u\in G(y, {\varepsilon }N^{-1})} \int _{{\mathbb S}^{d-1}}\frac{N^{d}\rho (y, u) |g(z)|}{(1+N\rho (y, z))^M} \mathrm{d}\sigma (z)\\&\le c{\varepsilon }\int _{{\mathbb S}^{d-1}}\frac{N^{d-1}|g(z)|}{(1+N\rho (y, z))^M} \mathrm{d}\sigma (z) \end{aligned}$$

and choosing $M := (d-1)/t+d$ we get

$$\begin{aligned} H_2(x) \le c {\varepsilon }\sup _{y\in {{\mathbb S}^{d-1}}} \int _{{\mathbb S}^{d-1}}\frac{N^{d-1}|g(z)|}{(1+N\rho (y, x))^{\frac{d-1}{t}}(1+N\rho (y, z))^{\frac{d-1}{t}+d}} \mathrm{d}\sigma (z). \end{aligned}$$

Clearly, $1+N\rho (x, z) \le (1+N\rho (y, x))(1+N\rho (y, z))$ and hence

$$\begin{aligned} H_2(x)&\le c {\varepsilon }\sup _{y\in {{\mathbb S}^{d-1}}} \int _{{\mathbb S}^{d-1}}\frac{N^{d-1}|g(z)|}{(1+N\rho (x, z))^{\frac{d-1}{t}}(1+N\rho (y, z))^{d}} \mathrm{d}\sigma (z)\\&\le c {\varepsilon }\sup _{z\in {{\mathbb S}^{d-1}}}\frac{|g(z)|}{(1+N\rho (x, z))^{\frac{d-1}{t}}} \sup _{y\in {{\mathbb S}^{d-1}}}\int _{{\mathbb S}^{d-1}}\frac{N^{d-1}}{(1+N\rho (y, z))^{d}} \mathrm{d}\sigma (z)\\&\le c' {\varepsilon }H(x), \end{aligned}$$

where for the last inequality we used (2.17). From this and (9.3) we infer

$$\begin{aligned} H(x) \le c {\varepsilon }^{(-d+1)/t} \mathcal {M}_t g(x) + c' {\varepsilon }H(x). \end{aligned}$$

Here the constants c and $c'$ are independent of ${\varepsilon }$. Consequently, choosing ${\varepsilon }$ so that $c' {\varepsilon }= 1/2$ and taking into account that $H(x)<\infty$ we obtain (2.15). $\square$