三维西尔平斯基垫圈上的正交指数函数

Pub Date : 2024-04-30 DOI:10.1007/s11785-024-01536-y
Zhi-Min Wang
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引用次数: 0

摘要

让 \(\xi \in \mathbb {R}\), and\(\rho _i\in \mathbb {R}\) with \(0<|\rho _i|<1\) for \(1\le i\le 3\).对于扩展实矩阵 $$\begin{aligned}M= (开始)\rho _1^{-1}&{}0&{}\xi\0&{}\rho _2^{-1}&{}-\xi\0&{}0&;{}\rho _3^{-1} \end{bmatrix}\in M_3(\mathbb {R}) \end{aligned}$$ and an integer digit set \(D=\{(0,0,0)^t, (1,0,0)^t, (0,1,0)^t, (0,0、1)^t \}子集 \mathbb {Z}^3\), let \(\mu _{M,D}\) be the self-affine measure defined by \(\mu _{M,D}(\cdot )=\frac{1}{|D||}\sum _{d\in D}\mu _{M,D}(M(\cdot )-d)\).在本文中,我们证明如果 \(\rho _1=\rho _2\),那么 \(L^2(\mu _{M,D})\) 允许一个无限正交的指数函数集,当且仅当\(|/rho _i|=(p_i/q_i)^{\frac{1}{r_i}}\) for some \(p_i、q_i,r_i\in \mathbb {N}^+\) with \(\gcd (p_i,q_i)=1\) and \(2|q_i\), \(i=1,2\).特别是,如果 \(\rho _1,\rho _2,\rho _3\in \{frac{p}{q}:p,q\in 2\mathbb {Z}+1\}\) 并且 \(\rho _1=\rho _2/),那么在 \(L^2(\mu _{M,D})\) 中最多存在 4 个相互正交的指数函数,而数字 4 是最好的。
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Orthogonal Exponential Functions on the Three-Dimensional Sierpinski Gasket

Let \(\xi \in \mathbb {R}\), and \(\rho _i\in \mathbb {R}\) with \(0<|\rho _i|<1\) for \(1\le i\le 3\). For an expanding real matrix

$$\begin{aligned} M=\begin{bmatrix} \rho _1^{-1}&{}0&{}\xi \\ 0&{}\rho _2^{-1}&{}-\xi \\ 0&{}0&{}\rho _3^{-1} \end{bmatrix}\in M_3(\mathbb {R}) \end{aligned}$$

and an integer digit set \(D=\{(0,0,0)^t, (1,0,0)^t, (0,1,0)^t, (0,0,1)^t \}\subset \mathbb {Z}^3\), let \(\mu _{M,D}\) be the self-affine measure defined by \(\mu _{M,D}(\cdot )=\frac{1}{|D|}\sum _{d\in D}\mu _{M,D}(M(\cdot )-d)\). In this paper, we prove that if \(\rho _1=\rho _2\), then \(L^2(\mu _{M,D})\) admits an infinite orthogonal set of exponential functions if and only if \(|\rho _i|=(p_i/q_i)^{\frac{1}{r_i}}\) for some \(p_i,q_i,r_i\in \mathbb {N}^+\) with \(\gcd (p_i,q_i)=1\) and \(2|q_i\), \(i=1,2\). In particular, if \(\rho _1,\rho _2,\rho _3\in \{\frac{p}{q}:p,q\in 2\mathbb {Z}+1\}\) and \(\rho _1=\rho _2\), then there exist at most 4 mutually orthogonal exponential functions in \(L^2(\mu _{M,D})\), and the number 4 is the best.

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