Dyadic Bounds for Brownian Hölder Continuity

From Gaussian Increments to Pathwise Bounds

Brownian increments are small on average at short time scales, but a pathwise continuity bound must control many increments at once. At dyadic level \(n\), there are exponentially many adjacent grid intervals. A Gaussian tail estimate makes the chance of a large increment decrease much faster than this number grows. Borel–Cantelli then gives a bound that holds at every sufficiently fine dyadic scale, almost surely. A chaining argument transfers the grid bounds to arbitrary times.

Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, and let \((B_t)_{t\geq 0}\) be standard one-dimensional Brownian motion with state space \(\mathbb{R}\) and continuous sample paths. Thus \(B_0=0\), and for \(0\leq s<t\), \(B_t-B_s\) has the normal distribution with mean \(0\) and variance \(t-s\). We use the continuous version of Brownian motion; its existence is established by the earlier material on the Kolmogorov continuity criterion.

The Grid Estimate

We first record the Gaussian estimate used at each scale. If \(Z\) is standard normal, then for \(x>0\),

Proof. The moment generating function of \(Z\) is \(\mathbb{E}[e^{\theta Z}]=e^{\theta^2/2}\). For \(\theta=x>0\), Markov's inequality gives \(\mathbb{P}(Z>x)=\mathbb{P}(e^{xZ}>e^{x^2})\leq e^{-x^2}\mathbb{E}[e^{xZ}]=e^{-x^2/2}\). Applying the same argument to \(-Z\), which is also standard normal, gives \(\mathbb{P}(Z<-x)\leq e^{-x^2/2}\). Adding the two tail probabilities proves the estimate. \(\square\)

Fix a positive integer \(M\). At level \(n\), divide \([0,M]\) into intervals of length \(2^{-n}\), with grid points \(k2^{-n}\) for \(0\leq k\leq M2^n\). There are \(M2^n\) adjacent increments, and each has variance \(2^{-n}\). For a fixed \(\gamma\in(0,\tfrac12)\), let \(A_n\) be the event that at least one adjacent increment has absolute value greater than \(2^{-n\gamma}\). Standardizing one increment gives the threshold \(2^{n(1/2-\gamma)}\). The Gaussian bound and the union bound therefore give

This sequence of probabilities is summable. Indeed, if \(a=1-2\gamma>0\) and \(b_n=2M2^n e^{-\frac12 2^{an}}\), then \(b_{n+1}/b_n=2\exp(-\tfrac12(2^a-1)2^{an})\), which tends to zero. Thus from some level onward \(b_{n+1}\leq \tfrac12b_n\), so the tail is bounded by a convergent geometric series.

Proof. Fix \(M\) and \(\gamma\in(0,\tfrac12)\). Since \(\sum_n\mathbb{P}(A_n)<\infty\), the first Borel–Cantelli lemma implies that only finitely many \(A_n\) occur almost surely. In this use, the lemma follows directly from the union bound: the probability that some \(A_n\) occurs for \(n\geq r\) is at most \(\sum_{n\geq r}\mathbb{P}(A_n)\), which tends to zero as \(r\to\infty\). Therefore, almost surely, there is a finite integer \(N\) such that every adjacent level-\(n\) increment is at most \(2^{-n\gamma}\) for every \(n\geq N\).

We show how this grid bound controls arbitrary times. Take \(s,t\in[0,M]\) with \(s<t\), and first suppose \(0<t-s\leq 2^{-N}\). Choose the integer \(m\geq N\) such that \(2^{-(m+1)}<t-s\leq 2^{-m}\). For \(u\in[0,M]\), set \(u_n=2^{-n}\lfloor 2^n u\rfloor\), where \(\lfloor x\rfloor\) is the greatest integer at most \(x\). The points \(u_n\) and \(u_{n+1}\) are either identical or adjacent on the level-\((n+1)\) grid. Continuity gives \(B_{u_n}\to B_u\). Hence, using the grid bounds and summing a geometric series,

The level-\(m\) grid indices of \(s_m\) and \(t_m\) differ by at most two: their difference is at most \(2^m(t-s)+1\leq2\). Thus \(|B_{t_m}-B_{s_m}|\leq2\cdot2^{-m\gamma}\). Applying the preceding approximation bound once at \(s\) and once at \(t\), and using the triangle inequality, gives

The last inequality follows from \(2^{-m}<2(t-s)\). For larger gaps, \(t-s>2^{-N}\), continuity makes \(K=\sup_{0\leq u\leq M}|B_u|\) finite, and \(|B_t-B_s|\leq2K\leq 2K2^{N\gamma}|t-s|^\gamma\). Combining the two bounds proves a finite \(\gamma\)-Hölder constant on \([0,M]\).

To obtain the asserted simultaneous statement, take the intersection of these probability-one events over all positive integers \(M\) and all rational \(\gamma\in(0,\tfrac12)\). This is a countable intersection, so it still has probability one. For any real \(\alpha\in(0,\tfrac12)\), choose a rational \(\gamma\) with \(\alpha<\gamma<\tfrac12\). On \([0,M]\), \(|t-s|\leq M\), and therefore \(|t-s|^\gamma\leq M^{\gamma-\alpha}|t-s|^\alpha\). The \(\gamma\)-Hölder bound implies the \(\alpha\)-Hölder bound. Every compact interval \([0,T]\) is contained in \([0,M]\) for some integer \(M\), completing the proof. \(\square\)

Worked Examples of the Argument

Why the Scale and the Union Bound Matter

The standard deviation of a Brownian increment over a level-\(n\) interval is \(2^{-n/2}\), not \(2^{-n}\). Comparing the threshold \(2^{-n\gamma}\) with that standard deviation gives the standardized threshold \(2^{n(1/2-\gamma)}\). When \(\gamma<\tfrac12\), this grows with \(n\), and its squared value in the Gaussian tail is \(2^{n(1-2\gamma)}\). That rapid tail decay is what offsets the \(M2^n\) intervals being checked at level \(n\).

Common mistake. A frequent error is to standardize the increment as though its standard deviation were \(2^{-n}\), leading to the incorrect threshold \(2^{n(1-\gamma)}\). The correct calculation is \(2^{-n\gamma}/2^{-n/2}=2^{n(1/2-\gamma)}\). Another error is to apply a tail bound to one increment and then conclude that the entire grid is controlled. At level \(n\) there are \(M2^n\) increments, so the union bound must include that factor. In this argument the corrected bound is \(2M2^n\exp(-\tfrac12 2^{n(1-2\gamma)})\), and it is the summability of these level-wide probabilities that permits Borel–Cantelli.

The proof establishes every exponent strictly below one half, but does not establish a Hölder bound at exponent one half. The next question is whether that endpoint is possible, and how to formulate the sharp regularity statement. That requires a different level of analysis than the summable dyadic bounds used here.