= Solution
The <Borel-Cantelli first lemma> applied to that finite sum says that almost surely, for all sufficiently large $n$,
$$
S_{2^{-n-1}}\geq2^{-n/2}f(2^{-n}).
$$
For $2^{-n-1}\leq t\leq2^{-n}$, both the <Brownian running maximum> and the positive function $\sqrt t\,f(t)$ are nondecreasing. Thus eventually
$$
S_t\geq S_{2^{-n-1}}\geq\sqrt{2^{-n}}\,f(2^{-n})\geq\sqrt t\,f(t).
$$
This proves the lower limit bound one, with interpolation across all small real times rather than only the dyadic sequence.
For every positive integer $m$, the function $mf$ satisfies the same assumptions and integrability condition. Repeating the argument gives an eventual lower bound $S_t\geq m\sqrt t f(t)$ on a probability-one event. Take the countable intersection of these events. On it, the ratio is eventually at least any prescribed positive integer, proving the <integral lower envelope for the Brownian maximum>:
$$
\boxed{\liminf_{t\downarrow0}\frac{S_t}{\sqrt t\,f(t)}=\infty\quad\text{almost surely}.}
$$
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