A continuous nowhere differentiable singular one simplex

Each term is nonnegative and at most $2^{-n-1}$ by [F3]. Thus the sum is well-defined and $0\le T\le 1$. By [F2] it is continuous, so it is a singular one-simplex. Direct substitution gives $T(0)=T(1)=0$ and $T(1\mathrm{/}2)=1\mathrm{/}2$, because every term with $n\ge 1$ vanishes there. The path is therefore nonconstant despite its equal endpoints.

To spell out the differentiability obstruction supplied by [F2], take the nested adjacent dyadic interval $[u_N,v_N]$ of length $2^{-N}$ containing the parameter, using the interval to the right at a dyadic point and the interval to the left at $1$. All summands of index at least $N$ vanish at its endpoints. Each earlier summand is affine there with slope ${\epsilon }_n\in \{-1,1\}$. The secant slope of $T$ is ${\sum }_{n<N}{\epsilon }_n$. Nested intervals preserve the earlier slopes, so consecutive secant slopes differ by one in absolute value and cannot converge to a finite value. If a finite derivative existed, the two endpoint quotients would tend to it, and their convex combination, this secant slope, would also tend to it. At a dyadic point or endpoint the appropriate one-sided quotient gives the same contradiction. This verifies exactly the finite-derivative assertion needed here.

A smooth extension in [F1] would give a finite derivative at every interior parameter, contradicting step 2.1. Thus this example meets the stronger nowhere-differentiable requirement, not only failure at one cusp. There is no empty-domain case for a singular one-simplex. A point target would yield a constant smooth map and is not this witness. No infinite selections are used: the series and dyadic intervals are specified arithmetically.