Are there non-zero real numbers equal to their "average decimal digit"?

Question

For some values of $x \in \mathbb{R}$, we can define what I'll call the "average digit" of $x$ and denote as $\theta(x)$ via

$$\theta(x) = \lim_{n \to \infty} \frac{a_1 + \dots + a_n}{n}$$

where the $a_i$ are the uniquely defined as digits in the decimal expansion of $x$, that is $x = N.a_1a_2\dots$, for some integer $N$, where we do not allow infinitely trailing $9$s (for example, for $x=1$, we have $a_i=0 \forall i$, rather than $a_i = 9 \forall i$. In other words, we are using that $1=1.000...$, not that $1 = 0.999...$)

This limit is well-defined since $\theta_n := \frac{a_1 + \dots + a_n}{n}$ is a Cauchy sequence. (it’s not, my mistake! Let’s just say:)

$\theta(x)$ is defined for whichever $x$ this limit exists.

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Problem: I was wondering whether there are any fixed points of $\theta$. That is, is there some number which describes what its average digit is? Well, we of course have $\theta(0) = 0$, but is there anything else?

I've been playing around with this for a bit, and I think I might have found an irrational number which is a fixed point of $\theta$. Intuitively it seems like it has to be, but I need to do a little more to prove it rigorously - hence the question. (N.B. its possible to show that all non-zero fixed points of $\theta$ are irrational, see this question.)

Here's the formal definition of the number:

Define $a_1 = 1$, $x_1 = \frac{1}{10}$ and, recursively for all $i > 1$, $a_i = \begin{cases}1 & \frac{a_1 + \dots + a_{i-1} + 0}{i} < x_{i-1} \\ 0 & \text{otherwise}\end{cases}$, $x_i = x_{i-1} + \frac{a_i}{10^i}$.

Then $x_i$ is increasing and bounded above (by $1$, say) so $x_i \to x$. $x$ is my number which I think obeys $\theta(x) = x$.

Intuitively, what we're actually doing here in this number's definition is the following: if we have some sequence of decimal digits $(a_i)$ of a real number $x$, we can define approximations $x_n = \sum_{i=1}^n \frac{a_i}{10^i}$, and $\theta_n = \sum_{i=1}^n \frac{a_i}{n}$ which will converge to $x$ and $\theta(x)$ respectively. We observe that if $a_n = 0$, then $x_n = x_{n-1}$ but $\theta_n < \theta_{n-1}$. So, if our integers $a_n$ are mostly zeroes, the $x_n$s will often stay constant for several terms in the sequence while the $\theta_n$s decrease slightly. Whenever $\theta_n$ drops below $x_n$, we'll insert a $1$ into the sequence to "correct course" and keep the $(\theta_n)$ sequence above the $(x_n)$ sequence.

The integers $i$ for which $a_i = 1$ are $1, 11, 20, 30, 40, 50, 60, 70, 80, 90, 100, \dots$. You can find these integers as a sequence $p_k$ by recursively defining $p_1 = 1$, $p_k = \min(n \in \mathbb{N} : \frac{k-1}{n} < \sum_{i=1}^{k-1} \frac{1}{10^{p_i}})$. As you can see the terms of $a_i$ quickly turn into successive multiples of $10$, and so we can approximate

$$x \approx \frac{1}{10} + \frac{1}{10^{11}} + \sum_{k=2}^{\infty} \frac{1}{10^{10k}} = 0.10000000001\overline{0000000010}.$$

This rational on the right-hand side has $\theta(x) = \frac{1}{10} \approx x$.

The sequence $p_k$ isn't going to be multiples of $10$ forever though - eventually this will break and instead of getting a rational with period $10$ with an "average digit" of $\frac{1}{10}$, we'll get a rational with period $10^{11}$ with an "average digit" of $\frac{1}{10} + \frac{1}{10^{11}}$, and so on.

What I've thus got from this recursive definition is that these $a_i$s define a sequence $x_n \to x$, with $\theta(x_n) \to x$. The only bit missing is that I'd like to say $\lim_{n \to \infty} \theta(x_n) = \theta(x)$.

What I'm not sure how to formally codify the fact that we are inserting $1$s near-periodically (we've already seen that $x \approx \frac{1}{10} + \frac{1}{10^{11}} + \sum_{k=2}^{\infty} \frac{1}{10^{10k}} = 0.10000000001\overline{0000000010}.$, and that this approximation is very good) and we are inserting these $1$s at a frequency that happens to agree with the earlier parts of $x$ that we've already written down. If we can express the intuitive idea here formally, I think that's what will allow us to prove that $\theta(x)=x$.

I might have just been staring at this for too long and gotten tired, but I can't make progress on this last bit. Can anyone see what we need to do here?

Of course, if I'm wrong and this number isn't a fixed point of $\theta$, explaining what's gone wrong here would also be a valid answer. Perhaps you can show $0$ is the only fixed point of $\theta$, that would certainly be a valid answer also.)

Answer 1

The fixed points of $\theta$ are actually an uncountable, dense, measure $0$ subset of $(0,9)$. This shouldn't be a surprise because $\theta$ sends every open interval surjectively to $[0,9]$.

Let $S = \{x\in(0,9) \mid \theta(x)=x\}$. That $S$ has measure $0$ is obvious, since the set of points such that $\theta(x) = 4.5$ has full measure. The method of proof of the other properties is inspired by Riemann's paradox.

Theorem: For each $x\in(0,9)$ and $\epsilon>0$, the set $S\cap (x-\epsilon,x+\epsilon)$ is uncountable.

Proof: Fix $x_0\in(0,9)$ and $\epsilon>0$. We will show that for each $s\in(0,1)$, there is a distinct point in $x_s \in S\cap(x_0-\epsilon,x_0+\epsilon)$. We will construct a sequence $x_n$ such that $\lim\limits_{n\rightarrow \infty} x_n = x_s$ and $\theta(x_s) = x_s$.

Let $d_n$ be the $n$th digit of $x_0$, and let $s_n$ be the $n$th digit of $s$, and let $N=\lceil -\log_{10}\epsilon\rceil$. For $n\ge 1$, we'll define $x_n$ as follows:$$ x_n = a_{0}.a_{1}a_{2}a_{3}\dots a_n = \sum_{k=0}^n \frac{a_{k}}{10^k} $$ where $a_{k}$ are defined as: $$ a_n = \begin{cases} d_n & n \le N\\ s_k & n > N \text{ and }n = 2^k+N\text{ for some }k\ge 0\\ 9 & \theta_{n-1}(x_{n-1}) < x_{n-1}\\ 0 & \theta_{n-1}(x_{n-1}) > x_{n-1} \end{cases} $$ Obviously, $x_n$ converges in the limit, and we'll call the limit $x_s$. We also have that for any $s,s'\in(0,1)$, $x_s\ne x_{s'}$, and that $|x_s - x_0| <\epsilon$. Then, all that remains to show is that $\theta(x_s) = x_s$. Note that $\theta_n(x_n) = \theta_n(x_s)$. Thus, it suffices to show that $\theta_n(x_n)$ converges to $x_s$. We will show this by showing that $$ \Delta_n = \theta_n(x_n) - x_n $$ decreases to $0$. Note that $$ \theta_n(x_n) = \frac{a_1+\cdots + a_n}{n} = \frac{a_1+\cdots + a_{n-1}}{n} + \frac{a_n}{n} = \frac{n-1}n \theta_{n-1}(x_{n-1}) + \frac{a_n}{n} $$ and $x_n = x_{n-1} + \frac{a_n}{10^n}$. Then \begin{eqnarray} \Delta_n &=& \frac{n-1}n \theta_{n-1}(x_{n-1}) + \frac{a_n}{n} - x_{n-1} - \frac{a_n}{10^n}\\ &=& \theta_{n-1}(x_{n-1}) - x_{n-1} - \frac1n\theta_{n-1}(x_{n-1}) + a_n\left(\frac1n - \frac{1}{10^n}\right) \\ &=& \Delta_{n-1} -\frac1n\theta_{n-1}(x_{n-1}) + a_n\left(\frac1n - \frac{1}{10^n}\right) \end{eqnarray}

From this, you can see clearly that $|\Delta_n - \Delta_{n-1}|$ goes to $0$. Furthermore, if $\Delta_{n-1} < 0$, by the definition of $a_n$, if $n$ is not $N+2^k$ for some $k$, we have $$ \Delta_n = \Delta_{n-1} + \frac1n\left(9-\theta_{n-1}(x_{n-1})-\frac{9n}{10^n}\right) $$ Hence for $n$ sufficiently large, $\Delta_{n-1}<0$ implies $\Delta_n>\Delta_{n-1}$. Similarly, if $n$ is sufficiently large and not $N+2^k$ for any $k$, then $\Delta_{n-1}>0$ implies $\Delta_n<\Delta_{n-1}$. Furthermore, note that \begin{eqnarray} \left|\Delta_n - \Delta_{n-1}\right| &=& \left|-\frac1n\theta_{n-1}(x_{n-1}) + a_n\left(\frac1n - \frac{1}{10^n}\right)\right| \\&=& \frac1n \left|-\theta_{n-1}(x_{n-1}) + a_n(1-\frac{n}{10^n})\right|\asymp \frac1n \end{eqnarray} because $\limsup\limits_{n\rightarrow\infty}\theta_{n-1}(x_{n-1}) < 9$ and $\liminf\limits_{n\rightarrow\infty}\theta_{n-1}(x_{n-1})>0$.

Then, let $n_j$ be the subsequence of numbers such that $\{n_j\} = \mathbb{N}\setminus\{N+2^k\mid k\in\mathbb{N}\}$. The set of $n = N+2^k$ is sparse enough that it will not affect the convergence of $\Delta_n$. The sequence $\Delta_{n_j}$ satisfies:

1. $|\Delta_{n_{j-1}} - \Delta_{n_{j}}|$ goes to $0$.
2. If $\Delta_{n_{j-1}} > 0$ then $\Delta_{n_{j}} < \Delta_{n_{j-1}}$ except when $n_j = N+2^k$ for some $k$.
3. If $\Delta_{n_{j-1}} < 0$ then $\Delta_{n_{j}} > \Delta_{n_{j-1}}$ except when $n_j = N+2^k$ for some $k$.
4. $\sum_j |\Delta_{n_{j-1}} - \Delta_{n_{j}}|$ diverges (which follows from the difference being asymptotically like $\frac1j$).

This implies that $\Delta_{n_j}$ converges to $0$, whence we get that $\Delta_n$ converges to $0$, and therefore $$ x_s = \lim\limits_{n\rightarrow\infty} x_n = \lim\limits_{n\rightarrow\infty} \theta_n(x_n) = \lim\limits_{n\rightarrow\infty} \theta_n(x_s) = \theta(x_s)$$ as desired. $\square$

I think this procedure shows how one could generate other fixed points as well.

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Is $\theta$ actually well defined?

Actually, it is not always well-defined. There are numbers for which $\theta(x)$ does not converge. For instance, suppose for $n\ge 0$, we have $$ a_n = \begin{cases} 9 & n \in \{2^{2^{2k}}...2^{2^{2k+1}}-1\}\text{ for some }k\\ 0 & n \in \{2^{2^{2k+1}}...2^{2^{2k+2}}-1\}\text{ for some }k \end{cases} $$ If we evaluate $\theta_{2^{2^k}}(x)$ for even $k$, the result will approach $0$, while if we evaluate it at odd $k$, it approaches $9$. One could use similar trickery to, for any sequence of rational numbers $r_n$, generate a number $x$ such that $r_n$ is a subsequence of $\theta_n(x)$.

Answer 2

Fixed points of $\theta$ (other than zero) are not that hard to find. We construct one here using just the digits $1$ and $2$.

Start with $1.2$. The average of these digits exceeds the value of the number, so to bring the former down we insert $1$'s to the right until there are enough to pull the average digit below the value of the number (which turns out to be $1.2111$). Now append $2$'s until the average grows above the number value, then $1$'s until the average drops below, and continue. The difference between the average digit and the value of the number is bounded by $(2-1)/n$ where $n$ is the number of digits, so your result is convergent.

Using a MS Excel worksheet we render the first sixteen digits:

$1.211121111211112...$

Try it with other pairs of digits, for fun.