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We use dyadic quadratic variation of Brownian motion. For fixed rational $0\le u<v$, set $\ell =v-u$ and let $Q_n$ be the sum of the squares of the $2^n$ Brownian increments on its equal-length partition. Independence and the normal fourth moment giveFor every $\eta >0$, Chebyshev's inequality bounds $\mathbb{P}(|Q_n-\ell |>\eta )$ by $2{\ell }^2{\eta }^{-2}{2}^{-n}$, a summable sequence. The Borel-Cantelli lemma implies $Q_n\to \ell$ almost surely. Intersect over positive rational $\eta$ and all rational intervals to obtain one probability-one event on which every such interval has this positive quadratic-variation limit.

If a path were Hölder of exponent $\alpha >1/2$ on a nonempty open interval, choose a compact rational subinterval $[u,v]$ within it and a finite Hölder constant $C$ there. Every partition increment would have absolute value at most $C(\ell 2^{-n}{)}^{\alpha }$, soThis contradicts $Q_n\to \ell >0$. The same pathwise event excludes every exponent greater than $1/2$, without needing an uncountable intersection of events. ThereforeThis is the quadratic variation obstruction to Hölder regularity. It rules out a continuously differentiable trajectory on any interval, since a continuous derivative is bounded on smaller compact intervals and gives a Lipschitz bound. By itself it does not rule out differentiability at an isolated point. The stronger nowhere differentiability of Brownian motion can also be seen directly as follows.

On a mesh of width $h=2^{-n}$ in $[0,M]$, the probability that any specified three consecutive increments all have absolute value at most $Kh$ is at most $C{K}^3{h}^{3/2}$: each increment is $N(0,h)$ and the three are independent. There are at most $M/h$ possible blocks, so the probability of any such block is at most $CM{K}^3{h}^{1/2}$, summable in $n$. By Borel-Cantelli lemma, for every pair of positive integers $M,K$ there are eventually no such blocks. But a finite derivative at some time $t<M$ would imply $|B_s-B_t|\le C_t|s-t|$ near $t$. The three mesh increments beginning at the mesh point immediately to the left of $t$ would then all be at most $6C_{t}h$ for all sufficiently fine meshes. Choosing an integer $K\ge 6C_t$ contradicts the preceding event. Thus finite derivatives do not exist anywhere almost surely, including a finite right derivative at $0$. The Hölder estimates below $1/2$ express continuity with a roughness scale; they do not imply differentiability.

For every $w\in C^{\alpha }[0,1]$, its finite Hölder seminorm gives the deterministic boundWhen $\alpha >1/2$, the right-hand side tends to zero. Thus under any probability measure on the stated coordinate sigma-algebra, these measurable finite-coordinate sums converge to zero on every path, and hence almost surely.

If the coordinate process were a Brownian motion, part (b) would make the same sums converge to one in $L^2$, and hence in probability. Almost sure convergence to zero also implies convergence in probability to zero. Uniqueness of a limit in probability is a contradiction. Therefore no such Brownian measure exists:This is the quadratic variation obstruction to Hölder regularity. It does not address the critical exponent $\alpha =1/2$, where the displayed deterministic bound does not tend to zero.