Suppose there is a function sequence defined on $[0,1]$ and $f_1(t)=t$. For each $f_n(t)$, there is a set of points $T_n=\{0,2^{-n},2\times2^{-n},\cdots,1\}$, such that between each of these points, $f_n(t)$ is linear. For example, since $f_1(t)=t$, it is linear between $[0,1/2]$ and $[1/2,1]$. The construction from $f_n(t)$ to $f_{n+1}(t)$ is: use refinement of $T_n$, i.e., $T_{n+1}=\{0,2^{-(n+1)},\cdots,1\}$, for all points belonging to $T_n\cap T_{n+1}$, $f_{n+1}(t)=f_n(t)$. While for $t=(2k-1)2^{-(n+1)}$, which are unique in $T_{n+1}$, $f_{n+1}(t) = f_n(t)+(-1)^{k+n+1}2^{-n/2-1}$. Then all points within intervals determined by $T_{n+1}$ are built by linear function using end points.
Now we want to prove that the function sequence will not converge to a differentiable function. One inequality is that for $t=k2^{-n}$, $|f_m(t+2^{-m})-f_m(t)|\ge 2^{-m/2}/8$, $m$ maybe $n+1$ or $n+2$. Can one actually prove that this inequality holds for $m=n+1$, or $n+2$?
Or, could anyone prove that this function sequence cannot converge to a function that satisfies Holder inequality?