From: chernoff@math.berkeley.edu (Paul R. Chernoff) Subject: Re: Lie product formulae Date: 17 Dec 2000 22:15:02 GMT Newsgroups: sci.math Summary: [missing] In article <91iteg$kvh$1@wanadoo.fr>, moubinool.omarjee wrote: > > >A_1 , A_2 ,...,A_k , matrices > >show that > > >( exp(A_1/p)exp(A_2/p)...exp(A_k/p) )^p -------> exp(A_1 +A_2 +...+ A_k) > >when p------> +oo > > ------------------------------------------------------------ Much more general results are known. Here is a proof of a simple generalization. Th: Let X be a Banach space, B(X) the bounded linear operators on X equipped with the operator norm, and F : [0,oo) --> B(X) an operator-valued function such that F is norm-differentiable at t = 0, with derivative F'(0) = C. Also, assume F(0) = I, the identity operator. Then for any real number t >= 0, F(t/n)^n --> exp(tC) in operator norm as n --> oo. Proof: The assumption on F says that, for t >= 0, F(t) = I + tC + tE(t), where ||E(t)|| --> 0 as t --> 0. Hence F(t/n) = 1 + (t/n)C + (t/n)E(t/n), and by a straightforward estimate using the binomial theorem, F(t/n)^n = (I + (t/n)C)^n + Error(n), where Error(n) --> 0 as n --> oo. Finally, (I + (t/n)C)^n --> exp(tC) in operator norm as n --> oo. The proof is the same as the case when C is a scalar: Just expand (I + (t/n)C)^n using the binomial theorem and observe what happens as n --> oo: one gets convergence to the usual series for exp(tC). The case in the original question corresponds to X = finite dimensional space over R or C, F(t) = exp(tA_1)exp(tA_2)...exp(tA_k). Here it's easy to see that the hypotheses of the Theorem above are satisfied, with F'(0) = A_1 + ...+A_k . -------------------------------------------- -- # Paul R. Chernoff chernoff@math.berkeley.edu # # Department of Mathematics 3840 # # University of California "Against stupidity, the gods themselves # # Berkeley, CA 94720-3840 struggle in vain." -- Schiller #