 Solution

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.