[math-fun] Continuous derivations on the reals
Marc wrote: << Is there term for when f(a b) = a f(b) + b f(a)? And the answer, supplied by Neil, is "derivation". Okay, sure, if say C^oo(R) is the set of infinitely differentiable functions from R to R, then D: C^oo(R) -> C^oo(R) defined by D(f)(x) = f'(x) is perhaps the epitome of a derivation. -------------------------------------------------------------------- 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? --Dan
I thought 'derivation' was restricted to linear operators. Perhaps these operators should be called something like quasi-derivation, or pseudo-derivation? On 6/17/07, Dan Asimov <dasimov@earthlink.net> wrote:
Marc wrote:
<< Is there term for when f(a b) = a f(b) + b f(a)?
And the answer, supplied by Neil, is "derivation".
Okay, sure, if say C^oo(R) is the set of infinitely differentiable functions from R to R, then D: C^oo(R) -> C^oo(R) defined by
D(f)(x) = f'(x)
is perhaps the epitome of a derivation. --------------------------------------------------------------------
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?
--Dan
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- If I'm not working, then I must be depressed.
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
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?
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?
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- If I'm not working, then I must be depressed.
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?
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- If I'm not working, then I must be depressed.
-- If I'm not working, then I must be depressed.
Quoting David Wilson <davidwwilson@comcast.net>:
Let [1] d(uv) = d(u) v + u d(v)
Then [2] d(u^2) = 2u d(u)
not when the quantities are noncommutative, as is typically the case when the object of discussion is a Lie Algebra. As for this implying a vanishing Maclaurin series, what does this say about ordinary functions, say polynomials, which do indeed have (non-zero) derivatives? - hvm ------------------------------------------------- www.correo.unam.mx UNAMonos Comunicándonos
participants (5)
-
Dan Asimov -
David Wilson -
Gareth McCaughan -
Ki Song -
mcintosh＠servidor.unam.mx