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If I remember correctly, for the binary digits of a real number in $[0,1]$, I was told that satisfying the law of the iterated logarithm (LIL) is stronger than being normal. That is, supposedly, some normal numbers may satisfy a weaker law, but not LIL. Is that correct, and can you give an example? What would be that "weaker law"?

Context:

Think about the Liouville numbers $\lambda(k)$, $k=1,2$ and so on, oscillating somewhat randomly between $-1$ and $+1$. Create a number $\omega$ whose $k$-th binary digit is $d_k=(\lambda(k)+1)/2$. Let

$$S_n=\Big(\sum_{k=1}^n d_k\Big) -\frac{n}{2}.$$

LIL says $$\lim\sup_{n\rightarrow\infty} \frac{|S_n|}{\sqrt{n\log \log n}} = C$$ with $0<C<1$.

If $w$ satisfies this law, then RH is proved. Of course no one knows, but it is conjectured that it does not. Even if $w$ satisfies the weaker version with $\sqrt{n\log\log n}$ replaced by $\sqrt{n}\cdot n^\epsilon$ for any $\epsilon>0$, then RH would also be true (this is equivalent to RH, unless I am mistaken). Again, nobody knows is $w$ satisfies this weaker version. Could $w$ violate LIL, yet be normal? Can a number be normal, yet violate LIL?

More interesting stuff

This is not part of my question, you can skip it. I thought it is interesting enough to mention it though. Consider the following recursion:

$$ \begin{align} \text{If } z_k & <2y_k \text{ then } \\ & y_{k+1}=4y_k-2z_k, \\ & z_{k+1}=2z_k+3, \\ & d_{k}=1 \\ \text{Else } & \\ & y_{k+1}=4y_k,\\ & z_{k+1}=2z_k-1, \\ &d_{k}=0 \end{align} $$

The $d_k$'s are the binary digits of some number. If $y_0=2,z_0=5$, it seems LIL may be satisfied. If $y_0=90,z_0=91$, LIL may not be satisfied. The first number is $\sqrt{2}/2$, the latter is $(-45+3\sqrt{245})/2$, see here.

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  • $\begingroup$ what's ur definition of normal? $01010101\dots$ might be an example. if u want all finite strings to appear with the right density, then take a random string and repeat it a bunch of times, take a longer random string and repeat it a bunch of times, then take an even longer random string and repeat it a bunch of times, etc.. $\endgroup$
    – mathworker21
    Commented Jul 18, 2022 at 4:15
  • $\begingroup$ See definition at en.wikipedia.org/wiki/Normal_number. Your example is a simply normal number, not a normal number (in base 2). Also by Liouville number, I mean values of the Liouville function at integer arguments, see en.wikipedia.org/wiki/Liouville_function $\endgroup$
    – Vincent Granville
    Commented Jul 18, 2022 at 5:59
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    $\begingroup$ Take a simple symmetric random walk, reflected at 0. That is, at each step flip a fair coin to see whether the next step is +1 or -1, unless the current position is 0, in which case the next step must be +1. With probability 1 this gives you a normal sequence - you get a one-sided LIL instead of the standard one (the number of +1 is always at least as big as the number of -1). You could consider many other similar variations of SSRW - e.g. reflect it (or condition it) so that its position at time $n$ is never outside $[-\sqrt{n}, \sqrt{n}]$, or $[-n^{1/4}, n^{1/4}]$, or the like. $\endgroup$
    – James Martin
    Commented Jul 18, 2022 at 9:01
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    $\begingroup$ Thank you James. You may consider writing this is an answer, it does answer my question. $\endgroup$
    – Vincent Granville
    Commented Jul 18, 2022 at 9:11
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    $\begingroup$ Normality just requires that the relative number of zeros and ones (or blocks of such digits) converges to the expected value. This convergence could be as slow as you wish. An LIL in contrast requires convergence of the order of square-root cancellation, which is much more to ask for. I am sure you can construct normal numbers for which the convergence towards normality is arbitrarily slowly. By the way, if I remember correctly then it's known that the convergence towards normality for Champernowne's example of a normal number is only of logarithmic order, not square-root order, so no LIL. $\endgroup$
    – Kurisuto Asutora
    Commented Jul 27, 2022 at 8:53

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