Why Brownian Paths Are Nowhere Differentiable

From a Derivative to a Local Oscillation Bound

Brownian increments over a time interval of length \(h\) have standard deviation \(\sqrt{h}\), rather than a size proportional to \(h\). This scaling suggests that a Brownian path should not have a finite derivative. The difficulty is that a statement about the difference quotient at one fixed time does not by itself rule out a derivative at some random time. We will prove a simultaneous result: on one event of probability one, every sufficiently short dyadic interval has oscillation larger than any prescribed integer multiple of its length. A deterministic argument then excludes a finite derivative at every interior time.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), and let \((B_t)_{t\geq 0}\) be standard Brownian motion: \(B_0=0\), its paths are continuous almost surely, and its increments over disjoint time intervals are independent, with \(B_v-B_u\) normally distributed with mean \(0\) and variance \(v-u\) when \(0\leq u<v\). For a continuous path \(f\) on an interval \(J\), define its oscillation there by \(\operatorname{osc}_J(f)=\sup_{s,u\in J}|f(s)-f(u)|\).

The final implication follows directly from the definition of the limit: choose \(\delta\) so that the difference quotient has absolute value at most \(|L|+1\) for \(0<|h|<\delta\), and then choose an integer \(M\geq |L|+1\). We need only this necessary consequence of differentiability. If it holds, then all path values in a short interval to the right of \(t\) lie within distance \(Mh\) of \(f(t)\), and hence the oscillation on that interval is at most \(2Mh\).

A Small-Oscillation Estimate

We first bound the chance that Brownian motion stays within a narrow range on one fixed interval. The estimate uses many independent increments inside that interval; bounding just the increment between the two endpoints would not give a sufficiently strong probability bound.

Proof. Set \(m=\lceil 1/\ell\rceil\) and divide \(J\) into \(m\) equal subintervals. Each has length \(\ell/m\). If the oscillation on \(J\) is at most \(K\ell\), then each increment across one of these subintervals has absolute value at most \(K\ell\). Each increment is normal with variance \(\ell/m\), so after dividing by its standard deviation, this constraint is \(|Z|\leq K\sqrt{\ell m}\). Since \(m\leq 1/\ell+1\), we have \(\ell m\leq 1+\ell\leq 2\). Thus the probability of the constraint for any one increment is at most \(p_K\). The \(m\) increments are independent, so the probability that all satisfy the constraint is at most \(p_K^m\), giving the asserted bound. Finally, \(p_K<1\) because a standard normal variable has positive probability of lying outside every finite interval. \(\square\)

Uniform Oscillation on Countably Many Intervals

A dyadic interval of level \(n\) is an interval \([k2^{-n},(k+1)2^{-n}]\), where \(k\) is an integer. Fix a positive integer \(T\) and a positive integer \(K\). There are exactly \(T2^n\) such intervals in \([0,T]\) at level \(n\). The lemma bounds the probability of small oscillation on each one, and the union bound gives \[ \mathbb{P}\left(\text{some level-\(n\) dyadic interval in \([0,T]\) has oscillation at most \(K2^{-n}\)}\right) \leq T2^n p_K^{\,2^n}. \]

The sum over \(n\geq 0\) of the right-hand side is finite. Indeed, if \(c=-\log p_K>0\), then \(p_K^{2^n}=e^{-c2^n}\). The exponential decay in \(2^n\) dominates the factor \(2^n\), so \(\sum_{n\geq0}T2^n e^{-c2^n}<\infty\). Consequently, with probability one, only finitely many levels have any such small-oscillation interval. To see this conclusion directly, the probability that such an interval occurs at some level \(n\geq N\) is at most the tail sum \(\sum_{n\geq N}T2^n p_K^{2^n}\), which tends to zero as \(N\) tends to infinity.

Proof. For any fixed pair \((T,K)\), the preceding summable bound shows that, with probability one, all sufficiently fine levels have no dyadic interval with oscillation at most \(K2^{-n}\). There are only countably many pairs of positive integers \((T,K)\). The intersection of their probability-one events still has probability one: the complement is a countable union of probability-zero events, and therefore has probability zero. Intersect also with the probability-one event on which the Brownian path is continuous. On the resulting event, the stated oscillation bound holds simultaneously for every such \(T\) and \(K\). \(\square\)

The Countable Reduction and Nowhere Differentiability

The oscillation theorem concerns only countably many dyadic intervals and integer bounds. This is the countable reduction: rather than take an uncountable union over possible derivative times, we establish one simultaneous event using countably many interval tests. On that event, a deterministic argument rules out every candidate time, including times that are not rational.

Proof of nowhere differentiability. Fix a path in the probability-one event of the theorem, and suppose, for contradiction, that it has a finite derivative at some interior time \(t>0\). Choose an integer \(T>t\). By the consequence of differentiability above, there are an integer \(M\geq1\) and a \(\delta>0\) such that \(|B_s-B_t|\leq M|s-t|\) whenever \(|s-t|<\delta\). Choose \(h>0\) small enough that \(h<\delta\), \(t+h\leq T\), and the dyadic intervals used below are at levels at least \(n_0\) for \(K=16M\).

The interval \([t+h/4,t+3h/4]\) has length \(h/2\). It contains a dyadic interval \(J\) whose length \(\ell\) satisfies \(h/8<\ell\leq h/4\): choose the largest dyadic length not exceeding \(h/4\), and take a grid interval of that length inside the interval of length \(h/2\). Such an interval fits because its length is at most half the containing interval's length. Every \(s\in J\) satisfies \(|s-t|\leq h<\delta\), so \(|B_s-B_t|\leq Mh\). For any \(s,u\in J\), the triangle inequality now gives \(|B_s-B_u|\leq2Mh\). Taking the supremum yields \(\operatorname{osc}_J(B)\leq2Mh\). But \(\ell>h/8\), so \(16M\ell>2Mh\). The simultaneous oscillation theorem, applied with integer \(K=16M\), gives \(\operatorname{osc}_J(B)>16M\ell>2Mh\), a contradiction. Thus no such \(t\) exists. Since \(T\) can be chosen to contain any finite interior time and the oscillation event holds for every integer \(T\), almost every Brownian path has no finite derivative at any interior time. \(\square\)

Why the Fixed-Time Argument Is Not Enough

A calculation at a fixed deterministic time can show that the difference quotient fails to converge there with probability one. It does not, on its own, show that there is no random time at which a derivative exists: the exceptional null set could depend on the time, and an uncountable union of null sets need not be null. The argument here avoids that gap. It first proves a probability-one statement from a countable collection of dyadic interval events, then uses a pathwise contradiction that applies to every time.

Common mistake. It is not valid to say, “For each fixed \(t\), differentiability has probability zero; therefore differentiability at some \(t\) has probability zero.” The first assertion involves a separate null set for each \(t\), and uncountably many such sets may have a union of positive probability. The correction is to prove a simultaneous path property using countably many tests, as above. Another tempting shortcut is to use only the increment between the endpoints of each short interval. Its probability of being at most \(K\ell\) is of order \(\sqrt{\ell}\), which is not strong enough to sum over all intervals at a level. Subdividing each interval supplies many independent increments and the summable bound \(p_K^{\lceil1/\ell\rceil}\).