On Sunday 17 June 2007 17:58, Dan Asimov wrote:
But what are the continuous derivations d: R -> R ?
I.e., for all real u,v, we must have
d(uv) = d(u)*v + u*d(v)
The 0 function works. What is the general solution?
Pedestrian answer follows after spoiler-space: ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... d(ab) = a d(b) + b d(a) => [triv] d(ab)/ab = d(a)/a + d(b)/b [a,b nonzero] => [e(x) := d(x)/x] e(ab) = e(a) + e(b) [a,b nonzero] => [f(x) := e(exp x)] f(a+b) = f(a) + f(b) [a,b real] => [famous result of Cauchy, given continuity of f] f(a) = ka [some constant k] => [unwind definitions] d(a) = k a log a [a positive] and it's easy to check that we must also have d(-a) = d(a), whence d(a) = k a log a [all a, with the obvious convention at 0] and we can also easily check that this is a solution. We're done. (There's an even more pedestrian proof, via proving that d(a^r) = r a^(r-1) d(a) successively for r a positive integer, then any integer, then any rational, then any real, at which point again we're done modulo checking what sign changes do.) -- g