Monday, 22 December 2014

real analysis - Solutions to matrix-valued multiplicative Cauchy equation under local boundedness

Let f:(0,)Rn×n, nN satisfy the functional equation f(x+y)=f(x)f(y). In general, f need not be measurable (by the usual constructions of non-measurable solutions based on a Hamel basis of R over Q).



Is it true that if f is uniformly bounded on (0,) (or only on some subinterval of (0,)) then f is measurable. (It then can also be shown that f is continuous and of the form f(x)=exp(Ax) for some ARn×n.)



Remarks:





  1. For n=1 this follows from here. For n2 the above functional equation translates into a system of one-dimensional functional equations fij(x+y)=nk=1fik(x)fkj(y). Are similar arguments applicable?

  2. For "n=" the above claim is not true, see e.g. Doob "Topics in the theory of Markoff chains" (1942), p. 2.



Edit: For the case of stochastic solutions (f(x) is a stochastic matrix for each x>0) the above question was solved in the affirmative by W. Doeblin in the 30s.)

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