Can all null-homotopy be made differentiable on arbitrary metric space? Let $M$ be a metric, and assume that it is simply connected. For a closed curve $f$, we define it to be differentiable iff for any $x$ then $\lim\limits_{h\rightarrow 0}\frac{d(f(x),f(x+h))}{h}$ exist, where we take $f$ to be extended into a periodic function with period $1$ for the sake of having both side when taking the derivative at the boundary.
The question is:
Given a closed curve $f$ that is differentiable. Does there always exist a null-homotopy of $f$ such that all curves along the homotopy process is differentiable.
I believe the answer to be no, but I am still unsure either way. Any helps would be welcome. Thank you.
 A: Consider the "snowflake" metric  $d(a,b)=|a-b|^\alpha$ on the closed unit  disk $D\subset \mathbb R^2$, where $\alpha \in(0,1)$ can be any fixed number.  Let this metric space be denoted by $X_1$, and its subset $\partial D\subset X_1$ by $Y_1$.  It is easy to see that $X_1$ contains no differentiable curves except constant ones, since differentiability with respect to its metric implies having zero derivative in the Euclidean sense. 
Also consider the Euclidean (restriction) metric on the complement of open unit disk, $\overline{\mathbb R^2\setminus D}$. Let this metric space be denoted by $X_2$, and its subset $\partial D\subset X_2$ by $Y_2$. 
The idea is to sew $X_1$ and $X_2$ along $\partial D$, thus obtaining a counterexample: $X_2$ contains differentiable closed curves which would have to pass through $X_1$ to be nullhomotoped. 
Sewing is messy, but luckily it was already done in detail: see Theorem 1 in A sewing problem in metric spaces by Haissinsky. 
It applies here with the sewing map $f:Y_1\to Y_2$ being the identity map (which is quasisymmetric, as required). The theorem produces a metric on  the quotient  $\widehat{X}=(X_1\sqcup X_2)/f$ which is locally bi-Lipschitz equivalent to the original metrics away from $\partial D$. This is the desired counterexample.
