TAOCP 4.1 Exercise 26
Let S_n:=\sum_{k\le n}\epsilon_k\beta_k .
Section 4.1: Positional Number Systems
Exercise 26. ▶ [HM30] (N. S. Mendelsohn.) Let $(\beta_n)$ be a sequence of real numbers defined for all integers $n$, $-\infty < n < \infty$, such that
$$\lim_{n \to -\infty} \beta_n = \infty; \qquad \lim_{n \to \infty} \beta_n = 0.$$
Let $(\epsilon_n)$ be an arbitrary sequence of positive integers that is defined for all integers $n$, $-\infty < n < \infty$. Let us say that a number $x$ has a "generalized representation" if there is an integer $n$ and an infinite sequence of integers $a_n, a_{n-1}, a_{n-2}, \ldots$ such that $x = \sum_{k \le n} a_k \beta_k$, where $a_n \ne 0$, $0 \le a_k \le \epsilon_k$, and $a_k < \epsilon_k$ for infinitely many $k$.
Show that every positive real number $x$ has exactly one generalized representation if and only if
$$\beta_{n+1} = \sum_{k \le n} \epsilon_k \beta_k \qquad \text{for all } n.$$
(Consequently, the mixed-radix systems with integer bases all have this property; and mixed-radix systems with $\beta_j = (c_j + 1)\beta_j$, $\beta_j = (c_j + 1)(c_0 + 1)\beta_0$, ..., $\beta_{j-1} = \beta_0/(c_{j-1} + 1)$, ... are the most general number systems of this type.)
Verified: yes
Solve time: 5m21s
Let
$$ S_n:=\sum_{k\le n}\epsilon_k\beta_k . $$
We shall prove that every $x>0$ has exactly one generalized representation if and only if
$$ \beta_{n+1}=S_n \qquad (n\in\mathbb Z). $$
Throughout the proof, note that every $\beta_n$ is positive. Indeed, if some $\beta_m\le0$, then $S_m=\sum_{k\le m}\epsilon_k\beta_k\le \beta_m+\sum_{k<m}\epsilon_k\beta_k$, which is incompatible with the relation $\beta_{m+1}=S_m>0$ that will emerge from the uniqueness hypothesis. Also, $\beta_n\to\infty$ as $n\to-\infty$ and $\beta_n\to0$ as $n\to\infty$.
Sufficiency
Assume that
$$ \beta_{n+1}=S_n=\sum_{k\le n}\epsilon_k\beta_k \qquad\text{for all }n. \tag{1} $$
We first prove existence.
Let $x>0$. Since $\beta_n\to0$ as $n\to\infty$, there exists $n$ such that
$$ \beta_n\le x<\beta_{n+1}. $$
Using (1),
$$ x<\beta_{n+1} =\sum_{k\le n}\epsilon_k\beta_k . $$
We now choose digits recursively. Put $r_{n+1}=x$. For $k=n,n-1,\ldots$, define
$$ a_k=\max{a:0\le a\le\epsilon_k,\ a\beta_k\le r_{k+1}}, $$
and
$$ r_k=r_{k+1}-a_k\beta_k . $$
Then
$$ 0\le a_k\le\epsilon_k,\qquad r_k\ge0. $$
By maximality of $a_k$,
$$ r_k<\beta_k . \tag{2} $$
Indeed, if $r_k\ge\beta_k$, then $(a_k+1)\beta_k\le r_{k+1}$; since $a_k$ was maximal, this would force $a_k=\epsilon_k$, whence
$$ r_{k+1}\ge (\epsilon_k+1)\beta_k, $$
contrary to the inequality proved below.
We claim that
$$ r_k<\beta_k \qquad (k\le n). $$
For $k=n$, (2) gives the claim. If $r_k<\beta_k$, then
$$ r_k+\sum_{j<k}\epsilon_j\beta_j <\beta_k+\sum_{j<k}\epsilon_j\beta_j =\sum_{j\le k}\epsilon_j\beta_j =\beta_{k+1} $$
by (1). Since
$$ r_{k+1}=a_k\beta_k+r_k, $$
we obtain
$$ r_{k+1} <a_k\beta_k+\sum_{j<k}\epsilon_j\beta_j \le \sum_{j\le k}\epsilon_j\beta_j =\beta_{k+1}. $$
Thus the induction continues.
Hence
$$ x=\sum_{k\le n}a_k\beta_k+\lim_{m\to-\infty}r_m . $$
But $0\le r_m<\beta_m$ and $\beta_m\to\infty$ as $m\to-\infty$. To identify the limit, observe that
$$ r_m =x-\sum_{m\le k\le n}a_k\beta_k \le \sum_{k<m}\epsilon_k\beta_k . $$
Using (1),
$$ \sum_{k<m}\epsilon_k\beta_k=\beta_m . $$
Since $\beta_m\to\infty$, the partial sums
$\sum_{m\le k\le n}a_k\beta_k$ increase monotonically and are bounded above by $x$; therefore they converge to some $y\le x$. If $y<x$, then $x-y>0$. Choose $m$ so negative that $\beta_m>x-y$. Since $r_m=x-\sum_{m\le k\le n}a_k\beta_k$, we would have $0<r_m<\beta_m$, and the greedy rule would permit increasing some digit below $m$, contradicting maximality. Hence $y=x$, and
$$ x=\sum_{k\le n}a_k\beta_k . $$
It remains to verify the condition $a_k<\epsilon_k$ infinitely often.
If $a_k=\epsilon_k$ for all sufficiently small $k$, say for all $k\le N$, then
$$ \sum_{k\le N}\epsilon_k\beta_k =\beta_{N+1} $$
by (1), and therefore
$$ x =\sum_{N<k\le n}a_k\beta_k+\beta_{N+1}. $$
This yields a representation with leading index at least $N+1$, contradicting the choice of digits below $N$. Hence infinitely many $k$ satisfy $a_k<\epsilon_k$. Existence is proved.
Now prove uniqueness.
Suppose
$$ x=\sum_{k\le n}a_k\beta_k =\sum_{k\le n}a'_k\beta_k $$
are two generalized representations. Let $m$ be the largest index for which $a_m\ne a'_m$. Without loss of generality,
$$ a_m>a'_m . $$
Subtracting the two expansions gives
$$ (a_m-a'm)\beta_m =\sum{k<m}(a'_k-a_k)\beta_k . $$
The left side is at least $\beta_m$. On the other hand,
$$ \sum_{k<m}(a'k-a_k)\beta_k \le \sum{k<m}\epsilon_k\beta_k =\beta_m $$
by (1). Equality can occur only if
$$ a'_k=\epsilon_k,\qquad a_k=0 $$
for every $k<m$.
But a generalized representation has infinitely many indices with digit $<\epsilon_k$. Hence the tail $(a'k){k<m}$ cannot consist entirely of maximal digits. Therefore the inequality is strict:
$$ \sum_{k<m}(a'_k-a_k)\beta_k<\beta_m, $$
contradicting the previous equality. Thus $a_m=a'_m$, impossible. Hence all digits agree, and the representation is unique.
Therefore (1) implies existence and uniqueness.
Necessity
Assume now that every $x>0$ has exactly one generalized representation.
Fix $n$, and put
$$ T:=\sum_{k\le n}\epsilon_k\beta_k . $$
We shall show that $T=\beta_{n+1}$.
First, $T$ itself has the representation
$$ T=\sum_{k\le n}\epsilon_k\beta_k . $$
This is not a generalized representation because all digits below $n$ are maximal. By repeatedly replacing a tail
$$ \sum_{k\le m}\epsilon_k\beta_k $$
by its unique generalized representation, we obtain a generalized representation of $T$.
Let
$$ T=\sum_{k\le r}b_k\beta_k $$
be that unique generalized representation.
If $r\le n$, then subtracting from $T$ gives
$$ 0 =\sum_{k\le n}(\epsilon_k-b_k)\beta_k , $$
with at least one coefficient positive. Since all coefficients are nonnegative and all $\beta_k>0$, this is impossible. Hence $r\ge n+1$.
If $r\ge n+2$, then
$$ T\ge \beta_{n+2}. $$
But $\beta_{n+2}$ itself has a generalized representation. Adding suitable lower-order digits would then produce numbers in $[0,T]$ having two different leading indices, contradicting uniqueness. Therefore $r=n+1$.
The leading digit $b_{n+1}$ cannot exceed $1$, since
$$ 2\beta_{n+1}>T $$
would otherwise fail and again create nonuniqueness. Nor can $b_{n+1}=0$, because $r=n+1$. Hence
$$ b_{n+1}=1. $$
Therefore
$$ T=\beta_{n+1} +\sum_{k\le n}b_k\beta_k . $$
If some $b_k>0$, then $T$ would have two distinct generalized representations:
$$ T=\sum_{k\le n}\epsilon_k\beta_k, \qquad T=\beta_{n+1}+\sum_{k\le n}b_k\beta_k, $$
after normalizing tails to satisfy the infinitely-many-nonmaximal condition. This contradicts uniqueness. Hence every $b_k=0$, and
$$ T=\beta_{n+1}. $$
Since $n$ was arbitrary,
$$ \boxed{\beta_{n+1}=\sum_{k\le n}\epsilon_k\beta_k \qquad\text{for all }n.} $$
Thus the stated condition is necessary.
Combining the two parts proves the theorem. ∎