Oh sorry, I probably meant to say C^\infty. On 6/19/07, Ki Song <kiwisquash@math.sunysb.edu> wrote:
If we are only assuming that the function if real-analytic, then nothing really, other than the fact that all of its derivatives vanish at the point. (eg. Take any partition of unity.)
On 6/19/07, David Wilson <davidwwilson@comcast.net> wrote:
Let [1] d(uv) = d(u) v + u d(v)
Then [2] d(u^2) = 2u d(u)
If d has a Maclaurin series, [2] implies that its coefficients are all 0, so d = 0 within its radius of convergence.
My real analysis is rusty, I'm not sure what this implies about the original question.
----- Original Message ----- From: "Dan Asimov" <dasimov@earthlink.net> To: "math-fun" < math-fun@mailman.xmission.com> Sent: Sunday, June 17, 2007 12:58 PM Subject: [math-fun] Continuous derivations on the reals
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?
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