Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 34 / 3

For each sample point put $A_n={\max }_{0\le k<2^n}|X_{(k+1)2^{-n}}-X_{k{2}^{-n}}|$. First take $s<t$ in the dyadic rationals, set $h=t-s$, and choose $N\ge 0$ with $2^{-N}\le h<2^{1-N}$. The left grid approximations $s_N=2^{-N}\lfloor 2^{N}s\rfloor$ and $t_N=2^{-N}\lfloor 2^{N}t\rfloor$ are at most two grid steps apart, so $|X_{t_N}-X_{s_N}|\le 2A_N$.

For either endpoint $r$, successive left approximations $r_n,r_{n+1}$ are equal or one step apart on the level-$(n+1)$ grid. Since $r$ is dyadic these approximations eventually equal $r$ exactly. The triangle inequality therefore givesThe second inequality is valid also for $\alpha =0$. Thus the required bound holds simultaneously for every dyadic pair whenever $K_{\alpha }$ is finite; if it is infinite the inequality holds in the extended sense. For $s=t$ the left side is zero. This is dyadic increment chaining, with the stated constant two.

For a maximum of finitely many nonnegative quantities, its $p$th power is at most the sum of their $p$th powers. Hence the increment assumption givesBy the triangle inequality in the Lp norm, the partial sums defining $K_{\alpha }$ have norms bounded by a geometric series. Monotone convergence of their $p$th powers givesIn particular $K_{\alpha }$ is finite almost surely, so the preceding pathwise estimate has a common probability-one domain for all dyadic pairs.

First construct a centred Gaussian process on the countable dyadic set with covariance function $\mathbb{E}(B_s{B}_t)=\min (s,t)$. These consistent finite-dimensional normal distributions exist because, for coefficients $a_j$,the Kolmogorov extension theorem realizes them on one probability space. Every increment has the normal distribution $N(0,|t-s|)$, so for every finite $p\ge 1$,Use the preceding estimate with $\beta =1/2$, $p>2$ and $0<\alpha <1/2-1/p$. It yields an almost surely Hölder continuous extension from the dense dyadic set to $[0,1]$. Approximate arbitrary times by dyadic times; continuity of the paths and of the Gaussian covariance function preserves all finite-dimensional laws. Disjoint increments are jointly Gaussian with zero cross-covariances, hence independent, with the required variances. Together with $B_0=0$ and continuity, this constructs Brownian motion.

For any $\alpha <1/2$, choose $p$ large enough. Taking a countable increasing sequence of exponents tending to $1/2$ gives simultaneous Hölder regularity of every order $\alpha <1/2$ on a common probability-one set; the same argument on each compact time interval gives local Brownian Hölder regularity. This method does not give the endpoint exponent. In fact that endpoint fails: if a path had a finite $1/2$-Hölder seminorm $K$, all its $2^n$ adjacent normalized increments at level $n$ would have magnitude at most $K$. These are independent standard normal random variables, so for a fixed integer $m$ the probability that their maximum is at most $m$ is $\mathbb{P}(|Z|\le m{)}^{2^n}\to 0$. The event of a finite $K$ is contained in the union over $m$ of events with this bound at every level, each of probability zero. Thus Brownian paths are not $1/2$-Hölder continuous on $[0,1]$ almost surely.