The Sharp Hölder Exponent of Brownian Paths

Why the Exponent One Half Is a Boundary

Brownian increments over a time interval of length \(h\) have standard deviation \(\sqrt{h}\). This scaling suggests that the natural path regularity boundary should be Hölder order \(1/2\): a Hölder bound of order \(\alpha\) would require increments of size at most a constant times \(h^\alpha\). When \(\alpha>1/2\), that permitted size is smaller than the typical scale \(\sqrt h\), and the discrepancy becomes more pronounced as intervals shrink. We will turn this observation into a probability-zero result by examining many independent increments on the same fine grid.

Work on a probability space \((\Omega,\mathcal F,\mathbb P)\) carrying a standard Brownian motion \((B_t)_{t\ge0}\). We use that \(B_t-B_s\) has distribution \(N(0,t-s)\) for \(0\le s<t\), that increments over disjoint time intervals are independent, and that the paths are continuous. No conditioning or filtration is needed in the argument.

The key scaling calculation is simple, but the independence across the grid is essential. Fix \(a<b\), put \(L=b-a>0\), and divide \([a,b]\) into \(m\) equal intervals of length \(h=L/m\). Define the standardized increments

For each fixed \(m\), these \(m\) variables are independent and each has the standard normal distribution. If the path has a Hölder bound with constant \(C\) and order \(\alpha\), then every grid increment satisfies \(|B_{a+kh}-B_{a+(k-1)h}|\le Ch^\alpha\). Dividing by \(\sqrt h\) gives the necessary condition \(|Z_{m,k}|\le Ch^{\alpha-1/2}\) for every \(k\).

The Fixed-Order Result

Proof. Fix a finite constant \(C\ge0\), and let \(H_{\alpha,C}\) be the event that the Hölder inequality of order \(\alpha\), with this particular constant \(C\), holds for every pair of times in \([a,b]\). We show that \(\mathbb P(H_{\alpha,C})=0\). On the grid with \(m\) intervals, membership in this event requires \(|Z_{m,k}|\le x_m\) for all \(k\), where \(x_m=C(L/m)^{\alpha-1/2}\). Since \(\alpha-1/2>0\), \(x_m\) tends to zero.

Let \(Z\) have the standard normal density \(\varphi(x)=(2\pi)^{-1/2}e^{-x^2/2}\). For any \(x\ge0\), \(\varphi(u)\le(2\pi)^{-1/2}\) for all \(u\), so integrating over \([-x,x]\) gives \(\mathbb P(|Z|\le x)\le 2x/\sqrt{2\pi}\). Independence of the \(m\) grid increments therefore gives

Write \(\delta=\alpha-1/2>0\) and \(A=2C L^\delta/\sqrt{2\pi}\), which is a fixed finite nonnegative number. The last bound is \((A m^{-\delta})^m\). Because \(m^{-\delta}\to0\), there is an integer \(m_0\) such that \(A m^{-\delta}\le1/2\) for all \(m\ge m_0\); if \(A=0\), the bound is already zero. Thus the probability is at most \(2^{-m}\) for all sufficiently large \(m\), and hence is zero. This conclusion holds for each fixed \(C\). If a path has some finite Hölder constant, it has a Hölder constant that is a nonnegative integer at least as large. The event of any finite constant is therefore contained in the countable union of the events \(H_{\alpha,C}\) over nonnegative integer \(C\). Each has probability zero, so their union has probability zero. \(\square\)

The theorem concerns a fixed \(\alpha\) and a fixed interval. To get a statement about every order above \(1/2\), rather than an uncountable collection of separate probability-one statements, first use rational orders.

Proof. For every rational \(\alpha>1/2\), the theorem says the event of Hölder continuity of order \(\alpha\) has probability zero. There are countably many such rationals, so with probability one none of those events occurs. Now suppose, on a path in this probability-one event, that the path were Hölder of some order \(\beta>1/2\) with finite constant \(C\). Choose a rational \(\alpha\) with \(1/2<\alpha<\beta\). For \(d=|t-s|\le L=b-a\), we have \(d^\beta\le \max(1,L^{\beta-\alpha})d^\alpha\). Consequently the order-\(\beta\) bound would imply an order-\(\alpha\) bound with a finite constant, a contradiction. Thus no such \(\beta\) exists. \(\square\)

Worked Increment Calculations

What the Sharp Exponent Means

The earlier dyadic bounds for Brownian Hölder continuity establish that, on a fixed compact interval, Brownian paths are almost surely Hölder of every order \(\alpha<1/2\). Combined with the result proved here, this means the supremum of the pathwise Hölder orders is \(1/2\): orders below \(1/2\) are available, and orders above \(1/2\) are not. A supremum does not by itself say whether the endpoint order is attained. In fact, the same grid idea also rules out order \(1/2\): for a fixed finite \(C\), the necessary constraints are \(|Z_{m,k}|\le C\), so their joint probability is \(\mathbb P(|Z|\le C)^m\), which tends to zero because a standard normal variable has positive probability of exceeding every finite \(C\) in absolute value. Taking the countable union over integer \(C\) excludes any finite endpoint constant.

This argument is deliberately about whether a finite Hölder constant exists. It does not estimate the precise size of the largest increments on small intervals or give a refined modulus of continuity; those are stronger questions. Its central ingredients are the Brownian scaling law, independent increments on disjoint intervals, and a Gaussian probability bound at the standardized scale.

A Common Scaling Pitfall

A frequent error is to standardize a Brownian increment by multiplying its Hölder threshold by \(\sqrt h\), rather than dividing by the standard deviation \(\sqrt h\). For example, with \(\alpha=3/4\), \(C=1\), and \(h=1/16\), the Hölder condition gives \(|\Delta B|\le h^{3/4}=1/8\). Since \(\Delta B/\sqrt h\) is standard normal and \(\sqrt h=1/4\), the correct standardized threshold is \((1/8)/(1/4)=1/2\). The incorrect multiplication gives \((1/8)(1/4)=1/32\), which is not the threshold for the standard normal variable. In general the standardized threshold is \(Ch^\alpha/\sqrt h=Ch^{\alpha-1/2}\). The exponent \(\alpha-1/2\), rather than \(\alpha\), is what makes the probability argument work.