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Prove that f(x) differentiable if f(a+b)=f(a)f(b)

Anonymous Thu 16 Jan 20:14:23 2020 No.11306727 View Reply Original Report

Quoted By: >>11306752 >>11306800 >>11306883 >>11306929

Given:

f:RR, f(0) exists, f(x)?0 and for all a, b ? R, f(a+b)=f(a)f(b)

How to prove that f(x) is differentiable in R?

Anonymous

Anonymous Thu 16 Jan 2020 20:28:42 No.11306752 Report

Quoted By:

>>11306727
take the log, and look up Cauchy functional equation, pretty sure you need f to be continuous

Anonymous

Anonymous Thu 16 Jan 2020 20:57:40 No.11306800 Report

Quoted By:

>>11306727
Find the general solution for f. f(0)=1 and f(2b)=f(b)^2. looks like a*e^-x

Anonymous

Anonymous Thu 16 Jan 2020 21:52:27 No.11306883 Report

Quoted By: >>11306889

>>11306727
( f(x+h) - f(x) ) / h
= ( f(x)f(h) - f(x) ) / h
f(x) ( (f(h) - f(0)) /h )
the limit h->0 exists and equals f(x)f'(0)

Anonymous

Anonymous Thu 16 Jan 2020 21:55:19 No.11306889 Report

Quoted By: >>11306903 >>11306919 >>11308303

>>11306883
f'(0) doesn't necessarily exist.

Anonymous

Anonymous Thu 16 Jan 2020 22:03:09 No.11306903 Report

Quoted By:

>>11306889
>Reading is hard.

Anonymous

Anonymous Thu 16 Jan 2020 22:10:15 No.11306919 Report

Quoted By: >>11308182

>>11306889
yea.. I see. ok how about this:
f is positive since f(a)= f(a/2)^2 and f(a)=/=0 for all a. so we can consider log•f. now log(f(a+b))= log(f(a)) + log(f(b)) . thus log•f is a linear map and therefore of the form log(f(a)) = ca for some constant c. now taking the exponential gives f(a) = exp(ca) (which is differentiable)

Anonymous

Anonymous Thu 16 Jan 2020 22:17:34 No.11306929 Report

Quoted By: >>11308188

>>11306727
>f:RR
what does this assumption mean?

Anonymous

Anonymous Fri 17 Jan 2020 11:03:21 No.11308177 Report

Quoted By:

I don't think this is true.

Anonymous

Anonymous Fri 17 Jan 2020 11:06:02 No.11308182 Report

Quoted By:

>>11306919
not every additive function is of the form cx, that's the problem.
If we take a discontinous solution to the Cauchy functional equation f, e^f will solve the problem's eqn, and will be discontinous everywhere, thus non-differentiable.

Anonymous

Anonymous Fri 17 Jan 2020 11:11:40 No.11308188 Report

Quoted By:

>>11306929
OP is trying to say that f maps from R to R.

Anonymous

Anonymous Fri 17 Jan 2020 12:34:08 No.11308303 Report

Quoted By:

>>11306889
Probably a typo
>f(0) exists
is given since it's a function over R and o.c. f(a+0)=f(a)f(0) & f(x)?0 -> f(0)=1

Anonymous

Anonymous ## Mod Fri 17 Jan 2020 13:20:02 No.11308368 Report

Quoted By:

This thread was moved to >>>/wsr/764762

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