The critical Hölder boundary at zero

On the probability-one event of [F1], for every $T>0$ and every $0<\gamma <1\mathrm{/}2$ there is a finite constant $K$ with $\mathrm{\mid }{B}_t-B_s\mathrm{\mid }\le K\mathrm{\mid }t-s{\mathrm{\mid }}^{\gamma }$ on $[0,T]$; since $[0,T]$ contains $0$ and the exponents are ordered, this is precisely assertion 1.

On the probability-one event of [F2], put $r(t)=B_t\mathrm{/}\sqrt{2t\log \log (1\mathrm{/}t)}$ for $0<t<e^{-1}$. For every $\epsilon >0$ the limsup and liminf bounds imply $-1-\epsilon <r(t)<1+\epsilon$ for all sufficiently small $t$, and $r(t)>1-\epsilon$ occurs at arbitrarily small positive times. Hence ${\mathrm{lim\hspace{0.17em}sup}}_{t\downarrow 0}\mathrm{\mid }r(t)\mathrm{\mid }=1$; in particular $\mathrm{\mid }r(t)\mathrm{\mid }>1\mathrm{/}2$ at arbitrarily small positive times.

Fix any finite $C\ge 0$ and $\delta >0$. Since $\sqrt{2\log \log (1\mathrm{/}t)}\to \mathrm{\infty }$, choose $\eta >0$ smaller than $\delta$ and $e^{-1}$ such that this factor exceeds $2C$ whenever $0<t<\eta$. Step 1.2 supplies such a $t$ with $\mathrm{\mid }r(t)\mathrm{\mid }>1\mathrm{/}2$. Then $\mathrm{\mid }{B}_t\mathrm{\mid }\mathrm{/}\sqrt{t}=\sqrt{2\log \log (1\mathrm{/}t)}\mathrm{\mid }r(t)\mathrm{\mid }>C$. As this works for every $C$ and $\delta$, the ratio is unbounded in every right neighborhood of zero and assertion 2 follows. A negative $C$ cannot bound the nonnegative ratio either.

Intersect the events in steps 1.1 and 1.2 with $\{B_0=0\}$, also of probability one by the Brownian definition. Both assertions then hold simultaneously; since $B_0=0$, the critical bound written with $\mathrm{\mid }{B}_t\mathrm{\mid }$ is precisely the pointwise Hölder bound at zero. The correct exponent comparison is downward: for any $0<\alpha <1\mathrm{/}2$, choose a rational $q$ with $\alpha <q<1\mathrm{/}2$ using [F3] and an integer $N\ge \max (1,T)$. A bound with exponent $q$ on $[0,N]$ implies $\mathrm{\mid }{B}_t-B_s\mathrm{\mid }\le K{N}^{q-\alpha }\mathrm{\mid }t-s{\mathrm{\mid }}^{\alpha }$ on $[0,T]$, since $\mathrm{\mid }t-s{\mathrm{\mid }}^{q-\alpha }\le N^{q-\alpha }$; the diagonal is immediate. Thus rational exponents above each desired exponent suffice, not exponents below it. In this proof [F1] already supplies the single event for every exponent and horizon, so no further uncountable intersection is made. The normalizer in step 1.2 is used only at positive $t<e^{-1}$. AC is inherited through [F4] and the two Brownian suppliers; the arbitrarily-small-time argument requires no selected sequence of times.