Lévy’s Modulus of Continuity

Small-Scale Oscillation of Brownian Paths

Continuity says that the increments of a Brownian path become small when the time gap becomes small. It does not specify how quickly they become small, or whether one can use a single bound for every pair of times in an interval. Lévy’s modulus of continuity answers both questions: it gives the exact almost-sure scale of the largest increment over all time gaps of length at most \(h\), including its sharp constant.

Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space carrying a standard Brownian motion \((B_t)_{t\geq0}\), with continuous paths, \(B_0=0\), and independent increments \(B_t-B_s\sim N(0,t-s)\) for \(0\leq s<t\). For a fixed \(T>0\) and \(0<h<T\), define the modulus on \([0,T]\) by

The supremum is over all pairs in the interval, not just over a prescribed grid. Continuity ensures that the path is bounded and uniformly continuous on \([0,T]\), so this supremum is finite and tends to zero as \(h\downarrow0\). The theorem below gives its more precise rate.

The logarithm is natural. The factor \(2\) inside the square root and the limiting constant \(1\) are both part of the sharp statement. The theorem is about an almost-sure limit as \(h\) decreases, not an identity for any particular positive \(h\).

Estimates Used in the Proof

The proof compares increments on a fine grid with increments at arbitrary times. We first establish a bound for the largest Brownian displacement from the start of a time interval. For \(l>0\) and \(x>0\),

Proof. Fix \(\theta>0\). At any finite increasing set of times \(0=t_0<t_1<\cdots<t_m\leq l\), the random variables \(Z_k=\exp(\theta B_{t_k}-\theta^2t_k/2)\) form a nonnegative martingale with \(Z_0=1\). Indeed, the Gaussian increment \(B_{t_k}-B_{t_{k-1}}\) is independent of the earlier values and has exponential moment \(\exp(\theta^2(t_k-t_{k-1})/2)\); conditioning therefore gives \(\mathbb{E}[Z_k\mid B_{t_0},\ldots,B_{t_{k-1}}]=Z_{k-1}\). For \(a>0\), the probability that \(\max_k Z_k\geq a\) is at most \(1/a\): partition that event by the first index \(k\) at which \(Z_k\geq a\); on each such event, conditional expectation gives \(\mathbb{E}[Z_m\mathbf{1}_{\{\text{first index }k\}}]=\mathbb{E}[Z_k\mathbf{1}_{\{\text{first index }k\}}]\geq a\mathbb{P}(\text{first index }k)\). Summing and using \(\mathbb{E}Z_m=1\) proves the bound. If \(B_{t_k}\geq x\), then \(Z_k\geq\exp(\theta x-\theta^2l/2)\). Thus the probability that the maximum over the finite set is at least \(x\) is at most \(\exp(-\theta x+\theta^2l/2)\). Taking \(\theta=x/l\) gives \(\exp(-x^2/(2l))\). Apply the same argument to \(-B\), add the two probability bounds, and take an increasing sequence of finite grids whose union is dense in \([0,l]\). Path continuity identifies the supremum on that union with the supremum on the interval. This proves the stated estimate. \(\square\)

We will also use a lower bound for a standard normal tail. If \(Z\sim N(0,1)\), then for \(x\geq1\), integrating its density over \([x,x+1/x]\) gives

In particular, there is a constant \(c>0\), independent of \(x\), such that \(\mathbb{P}(|Z|>x)\geq c x^{-1}e^{-x^2/2}\) for \(x\geq1\). To see this, expand \((x+1/x)^2/2=x^2/2+1+1/(2x^2)\), and use \(1/(2x^2)\leq1/2\). The resulting constant can be chosen as \(c=e^{-3/2}/\sqrt{2\pi}\), since the two-sided tail is at least the one-sided tail.

Proof of the Exact Modulus

Write \(L(h)=\sqrt{2h\log(T/h)}\), and first consider the geometric sequence \(h_n=T r^{-n}\), where \(r>1\) is fixed. The upper bound is obtained by approximating arbitrary times by grid points at each scale \(h_n\). Fix \(\varepsilon>0\). Choose \(\delta>0\) small enough that

Such a choice is possible: as \(\delta\downarrow0\), \(a\) tends to \((1+\varepsilon/2)^2>1\), while \(b\) tends to infinity. For a scale \(h=h_n\), take grid points spaced \(\delta h\) apart, including the endpoint \(T\) if needed. There are at most a constant depending on \(\delta\) times \(T/h\) grid points. Any two points \(s,t\) with \(|t-s|\leq h\) can be rounded down to grid points \(s',t'\); the rounded points are at distance at most \((1+2\delta)h\), and each rounding error is at most \(\delta h\).

There are at most \(C_\delta T/h\) ordered pairs of grid points at distance at most \((1+2\delta)h\), for some finite constant \(C_\delta\). Each corresponding Brownian increment is centered Gaussian with variance at most \((1+2\delta)h\). The elementary Gaussian bound \(\mathbb{P}(|Z|>x)\leq2e^{-x^2/2}\), obtained by the exponential moment bound for \(Z\) and \(-Z\), therefore gives

To control the rounding errors, partition \([0,T]\) into intervals of length at most \(\delta h\). On each interval, the displacement from its left endpoint has the preceding supremum bound, applied to the Brownian increments from that endpoint. There are at most \(T/(\delta h)+1\) intervals. The probability that any such displacement exceeds \((\varepsilon/4)L(h)\) is at most

When the grid-increment bound and the cell-displacement bound both hold, the increment between arbitrary \(s,t\) with \(|t-s|\leq h\) is at most \((1+\varepsilon/2)L(h)+2(\varepsilon/4)L(h)=(1+\varepsilon)L(h)\). Since \(a>1\) and \(b>1\), the failure probabilities just obtained are summable along \(h_n=T r^{-n}\): each is bounded by a constant times a geometric sequence with ratio less than one. The Borel–Cantelli lemma implies that, almost surely, for all sufficiently large \(n\), \(\omega_B(h_n;T)\leq(1+\varepsilon)L(h_n)\).

For the lower bound, divide \([0,T]\) into \(N=\lfloor T/h\rfloor\) disjoint intervals of length \(h\). Their Brownian increments are independent, and each has the distribution \(\sqrt{h}Z\), where \(Z\sim N(0,1)\). Fix \(0<\varepsilon<1\), and set \(x=(1-\varepsilon)\sqrt{2\log(T/h)}\). For sufficiently small \(h\), \(x\geq1\), and the normal-tail estimate gives, for a constant \(c_\varepsilon>0\),

The same probability applies to each of the \(N\) intervals. The chance that none exceeds the threshold is \((1-p_h)^N\leq e^{-Np_h}\). As \(h\downarrow0\), \(N\) is at least \(T/(2h)\) once \(T/h\geq2\); hence \(Np_h\) is bounded below by a positive constant times \((T/h)^{1-(1-\varepsilon)^2}/\sqrt{\log(T/h)}\). Along \(h_n=T r^{-n}\), this grows exponentially in \(n\), up to the factor \(1/\sqrt{n\log r}\). Consequently the probabilities \(e^{-Np_{h_n}}\) are summable. Borel–Cantelli shows that almost surely, for all sufficiently large \(n\), at least one of these increments exceeds \((1-\varepsilon)L(h_n)\). It is among the increments included in the modulus, so \(\omega_B(h_n;T)\geq(1-\varepsilon)L(h_n)\).

It remains to pass from the geometric sequence to every small \(h\). If \(h_{n+1}\leq h\leq h_n\), monotonicity of the supremum gives \(\omega_B(h;T)\leq\omega_B(h_n;T)\) for the upper bound and \(\omega_B(h;T)\geq\omega_B(h_{n+1};T)\) for the lower bound. Also, uniformly for \(h\) in this range, the ratio \(L(h_n)/L(h)\) has upper limit at most \(\sqrt r\), and \(L(h_{n+1})/L(h)\) has lower limit at least \(1/\sqrt r\); the logarithmic factors in these ratios tend to one because their arguments differ by at most the fixed factor \(r\). Thus the almost-sure upper limit is at most \((1+\varepsilon)\sqrt r\), and the lower limit is at least \((1-\varepsilon)/\sqrt r\). Take a countable sequence of \(\varepsilon\) decreasing to zero and a countable sequence of \(r>1\) decreasing to one. The resulting probability-one events still have probability one, and these bounds yield the claimed limit \(1\). \(\square\)

Worked Interpretations

Equivalent Normalizations and a Common Pitfall

For a fixed \(T>0\), one may replace \(\log(T/h)\) by \(\log(1/h)\), provided \(h\) is expressed in a fixed time unit and is small enough that both logarithms are positive. Indeed,

Consequently the theorem also gives \(\omega_B(h;T)/\sqrt{2h\log(1/h)}\to1\) almost surely. The direction of the conversion matters: if \(R(h)=\sqrt{2h\log(T/h)}/\sqrt{2h\log(1/h)}\), then

Common mistake. Multiplying the theorem’s ratio by \(1/R(h)\) does not give the ratio with denominator \(\sqrt{2h\log(1/h)}\); it gives a different expression. The correct factor is \(R(h)\), as the displayed equality shows. Both \(R(h)\) and \(1/R(h)\) tend to one, so the mistaken multiplication may appear to preserve the final limit, but it does not justify that limit for the stated ratio.

The result is stronger than a Hölder bound: a Hölder estimate describes a power of \(h\), whereas this theorem resolves the additional logarithmic factor and its sharp constant. The logarithm reflects the growing number of opportunities for a large increment as the interval is examined at finer scales. The proof’s lower bound makes that role concrete through many independent increments; the upper bound ensures that allowing all times between grid points does not change the leading constant.