= Solution
For $t>e$, put $\Phi(t)=\sqrt{2t\log\log t}$. Fix $a>1$ and $\varepsilon>0$, and set
$$
E_n=\left\{\sup_{0\leq s\leq a^n}B_s
\geq(1+\varepsilon)\Phi(a^n)\right\}
$$
for $n$ large enough that $a^n>e$. The <Brownian reflection principle> and the Gaussian tail estimate give
$$
\begin{aligned}
\mathbb P(E_n)
&=2\mathbb P\left(B_{a^n}\geq(1+\varepsilon)\Phi(a^n)\right)\\
&\leq2\exp\bigl(-(1+\varepsilon)^2\log\log(a^n)\bigr)\\
&=2(n\log a)^{-(1+\varepsilon)^2}.
\end{aligned}
$$
Since $(1+\varepsilon)^2>1$, the sum of these probabilities is finite. The <Borel-Cantelli first lemma> implies that, <almost surely>, for all sufficiently large $n$,
$$
\sup_{s\leq a^n}B_s<(1+\varepsilon)\Phi(a^n).
$$
Now take $t\in[a^{n-1},a^n]$. The function $\Phi$ is increasing for $t>e$, so
$$
\frac{B_t}{\Phi(t)}
\leq(1+\varepsilon)\frac{\Phi(a^n)}{\Phi(a^{n-1})}.
$$
Here 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,
$$
\frac{\Phi(a^n)}{\Phi(a^{n-1})}
=\sqrt{a\,\frac{\log(n\log a)}{\log((n-1)\log a)}}
\longrightarrow\sqrt a.
$$
Thus, for each fixed pair $(a,\varepsilon)$,
$$
\limsup_{t\to\infty}\frac{B_t}{\Phi(t)}
\leq(1+\varepsilon)\sqrt a
\quad\text{almost surely}.
$$
Use the countable choices $a_m=1+1/m$ and $\varepsilon_m=1/m$, intersect their probability-one events, and let $m\to\infty$. This proves the <Brownian upper law of the iterated logarithm>:
$$
\boxed{\limsup_{t\to\infty}\frac{B_t}{\sqrt{2t\log\log t}}
\leq1\quad\text{almost surely}.}
$$
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 $t\geq e^e$. No monotonicity at the small-time edge of the logarithmic expression is required.
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