Consider the simple symmetric random walk on $\mathbb{Z}$. That is, let $X\_1, X\_2, \dots$ be i.i.d. random variables with
$$
P(X\_i=1)=P(X\_i=-1)=1/2,
$$ 
and define $S\_n=X\_1+\dots+X\_n$ with $S\_0=0$. As is well known, the sum $S\_n$ is (null) recurrent and satisfies the law of the iterated logarithm
$$
P\left(\limsup\_{n\to \infty} \frac{S\_n}{\sqrt{2n \log \log n}}=1\right)=1,
$$


Assume now that we restrict ourselves to the subset of realisations of $S\_n$ where each realisation satisfies 


1. $\frac{1}{n} S\_n \to 0$.
2. For any integer $m$, there are infinitely many values of $n$ such that $S\_n=m$ or $S\_n=-m$.


Does the law of the iterated logarithm take a stronger form in this case, in the sense:



> 
> 
> > 
> > Q: If $S\_n$ is satisfies the criteria above, does this imply
> >  $$
> > \limsup\_{n\to \infty} \frac{|S\_n|}{\sqrt{2n \log \log n}}\leq 1?
> > $$
> > 
> > 
> > 
> 
> 
> 


Or are there even in this case subsets of realisations for which this is not true?



Surely recurrent random walks and the law of the iterated logarithm
Consider a path satisfying $S\_{k^3} = (-1)^k k^2$ for all sufficiently large $k$. Since $$(k+1)^3 -k^3 \approx 3k^2 \gg 4k \approx |S\_{(k+1)^3} - S\_{k^3}|$$
this is clearly possible. Moreover, we can ensure that $|S\_n| \le |S\_{k^3}|$ for all $1 \le n \le k^3$ simply by replacing all excursions outside that value by a zigzag path of alternating increments $\pm 1$. To see that (1) is satisfied, note that for $k^3 \le n \le (k+1)^3$ we have by assumption $|S\_n| \le |S\_{(k+1)^3}| = (k+1)^2$, so that $$\frac{|S\_n|}{n} \le \frac{(k+1)^2}{k^3} \to 0.$$
(2) is also clearly satisfied. And since $|S\_n| = n^{2/3}$ when $n = k^3$ the limsup in the LIL is infinite.


