CF 104303E - 读中国数字

CF 104303E - \u8bfb\u4e2d\u56fd\u6570\u5b57

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Solution

Solution

Let $x \in [0,1)$ have ternary expansion

$x = 0.x_1 x_2 x_3 \cdots \quad (x_j \in {0,1,2}),$

where nonterminating representations are used.

Define $\omega = e^{2\pi i/3}$, so $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.

For each $j \ge 1$, define the ternary Rademacher functions

$r_j(x) = \omega^{x_j}.$

Each $r_j(x)$ depends only on the $j$th ternary digit and takes values in ${1,\omega,\omega^2}$.

Let $k$ be a nonnegative integer with ternary representation

$k = k_0 + 3k_1 + 3^2 k_2 + \cdots,$

where only finitely many $k_j$ are nonzero and each $k_j \in {0,1,2}$.

Define the ternary Walsh function $w_k(x)$ by

$w_k(x) = \prod_{j \ge 1} r_j(x)^{k_{j-1}}.$

Equivalently,

$w_k(x) = \omega^{\sum_{j \ge 1} k_{j-1} x_j} = \omega^{\sum_{j \ge 0} k_j x_{j+1}}.$

The exponent is taken modulo $3$, since $\omega^m$ depends only on $m \bmod 3$.

For each fixed $k$, the product is finite because $k_j = 0$ for all sufficiently large $j$.

Let $m \ge 1$ and restrict to the $\sigma$-algebra generated by the first $m$ ternary digits. Then $w_k$ depends only on $x_1,\dots,x_m$ whenever $k_j = 0$ for $j \ge m$.

For integers $k,\ell$ with ternary digits $(k_j)$ and $(\ell_j)$, the product satisfies

$w_k(x),\overline{w_\ell(x)} = \omega^{\sum_{j \ge 1} (k_{j-1}-\ell_{j-1})x_j}.$

Orthogonality is computed by integrating digit by digit. For each fixed $j$,

$\int_0^1 \omega^{a x_j},dx = \frac{1}{3}(1 + \omega^a + \omega^{2a})$

where $a \in {0,1,2}$. This equals $1$ when $a \equiv 0 \pmod 3$ and equals $0$ otherwise.

Since the ternary digits $x_j$ are independent and uniformly distributed under Lebesgue measure on $[0,1)$, the integral factorizes:

= \prod_{j \ge 1} \frac{1}{3}\left(1 + \omega^{k_{j-1}-\ell_{j-1}} + \omega^{2(k_{j-1}-\ell_{j-1})}\right).$$ Each factor equals $1$ if $k_{j-1} = \ell_{j-1}$ and equals $0$ otherwise. The product is therefore $1$ when $k = \ell$ and $0$ when $k \ne \ell$, yielding $$\int_0^1 w_k(x),\overline{w_\ell(x)},dx = \delta_{k\ell}.$$ The system ${w_k}$ is complete in $L^2[0,1)$ because it coincides with the character system of the compact abelian group $\prod_{j \ge 1} \mathbb{Z}/3\mathbb{Z}$ under the identification given by ternary expansion, and characters form an orthonormal basis for the corresponding $L^2$ space. Under the digit map between $[0,1)$ and ternary sequences, this group is measure-preservingly isomorphic to the unit interval with Lebesgue measure. Thus the ternary Walsh system is obtained by replacing binary digits and the sign group ${\pm 1}$ with ternary digits and the multiplicative group of third roots of unity, preserving orthogonality and completeness through digitwise independence. This completes the construction of a ternary generalization of Walsh functions. ∎