 Solution

For $t>e$, put $\Phi (t)=\sqrt{2t\log \log t}$. Fix $a>1$ and $\epsilon >0$, and setfor $n$ large enough that $a^n>e$. The Brownian reflection principle and the Gaussian tail estimate giveSince $(1+\epsilon {)}^2>1$, the sum of these probabilities is finite. The Borel-Cantelli first lemma implies that, almost surely, for all sufficiently large $n$,Now take $t\in [a^{n-1},a^n]$. The function $\Phi$ is increasing for $t>e$, soHere the positive upper bound for $B_t$ can first be divided by $\Phi (t)$, and the denominator can then be bounded below by $\Phi (a^{n-1})$; this remains valid even when $B_t<0$. Moreover,Thus, for each fixed pair $(a,\epsilon )$,Use the countable choices $a_m=1+1/m$ and ${\epsilon }_m=1/m$, intersect their probability-one events, and let $m\to \infty$. This proves the Brownian upper law of the iterated logarithm:Only large times are involved. The hint's related monotonicity assertion for $\Phi (t)/t$ is also valid eventually: the derivative of $2\log \log t/t$ is $2(1/\log t-\log \log t)/t^2$, which is negative for $t\ge e^e$. No monotonicity at the small-time edge of the logarithmic expression is required.