G(x)=Sum sign(m top n=0 bottom)(1/2^n)h(2^n x).for more inf. please check #30026,#30028,#30029.
show that (g(x_m)-g(0))/(x_m - 0)=m+1, and use this to prove that g'(0) does not exist.
any temptation to say something like g'(0)=oo should be resisted. setting x_m=-(1/2^m) in the previous argument produces difference heading towards -oo. the geomteric manifestation is the "cusp" that appears at x=0 in the graph of g.
Let x_m=-(1/2)^m. Then for any n<=m, we have |(2^n)x_m|<=1. Thus we have ...