Week 10 Tutorial Solutions

  1. R(x)=0010011010
  2. R(x)=1010100101
  3. The rule does not make sense at the constant sequence x=11111111111111 because no x(n) is equal to 0. Define R at this point to take the value 000000000000 and think of the odometer resetting.
  4. R1([1])=[0] because if x(1)=0 then (R(x))(1)=1 and if (R(x))(1)=1 it must have been the case that γ(x)=1 and therefore x(1)=0.
  5. Let μ be the (1p,p) coin measure. Thus μ([1])=p and μ([0])=1p. If μ is R invariant then p=μ([1])=μ(R1([1]))=μ([0])=1p which is impossible unless p=0.5.
  6. Since R1([0])=[1] and R1([1])=[0] we have fR=1R1[0]1R1[1]=1[1]1[0] and from 1[1]1[0]=(1[0]1[1]) we see it is an eigenfunction of eigenvalue 1.
  7. Take f1=1[00]+11[10]+11[01]+11[11]f2=1[00]+i1[10]11[01]i1[11]f3=1[00]11[10]+11[01]11[11]f4=1[00]i1[10]11[01]+i1[11] which are eigenfunctions of R with eigenvalues 1, i, 1 and i respectively. As their eigenvalues are distinct they are linearly independent.
  8. From 1[0]1[1]=1[00]+1[01]1[10]1[11] we see that 1[0]1[1] and f3 span the same subspace.
  9. We can write 1[00]=14(f1+f2+f3+f4)1[10]=14(f1if2f3+if4)1[01]=14(f1f2+f3f4)1[11]=14(f1+if2f3if4) by some linear algebra.
  10. The collection D of cylinders (including the empty set) is a π-system so it suffices to check that μ(R1(C))=μ(C) for every cylinder C. But for every cylinder [ϵ(1)ϵ(r)] its inverse image under R is another cylinder of the same length, defined by the symbols δ(1)δ(r) that odometrically precede ϵ(1)ϵ(r). With respect to the fair coin measure, all cylinders of length r have the same measure. Thus μ is an invariant measure for R.
  11. Mimicking the answer to 7 with longer cylinders, we can find for every ηC with |η|=1 and η2n=1 for some nN an eigenfunction of R with eigenvalue η. As, for every cylinder set C, we can write 1C as a linear combination of such eigenfunctions, we can conclude that the eigenfunctions span L2(X,B,μ). There are therefore no other eigenvalues of R and R has discrete spectrum.