Find F: R->R satisfying F(x+y)=F(x)+F(y) and F(xy)=F(x)F(y)

  • Thread starter Thread starter iN10SE
  • Start date Start date
Click For Summary
SUMMARY

The function F: R->R satisfying the conditions F(x+y)=F(x)+F(y) and F(xy)=F(x)F(y) can only be either F(x)=0 or F(x)=x. The proof involves showing that if F(0)=0, then F must be of the form F(x)=xg(x), where g(x) is non-zero for all x. The discussion highlights that F must be odd and that F(1) can only be 0 or 1, leading to the conclusion that F is linear and can only take the specified forms. The hint regarding positive real numbers being squares is crucial in establishing the necessary properties of F.

PREREQUISITES
  • Understanding of functional equations, specifically Cauchy's functional equation.
  • Knowledge of properties of odd functions.
  • Familiarity with basic algebraic structures and mappings.
  • Concept of continuity and its implications in functional equations.
NEXT STEPS
  • Study Cauchy's functional equation and its solutions in depth.
  • Explore the implications of continuity on functional equations.
  • Investigate the properties of odd functions and their applications.
  • Learn about the axiom of choice and its role in constructing discontinuous functions.
USEFUL FOR

Mathematicians, students of advanced mathematics, and anyone interested in functional equations and their properties.

  • #31
I screwed up most of all by throwing a monkey wrench into peoples thinking by suggesting it might not be true. Mea culpa again.
 
Physics news on Phys.org
  • #32


intuitively, the solution is f(kx)=kf(x). i am able to prove it for all integers k.but for a rational number k i am facing problem.so i assumed k to be p/q and later that i can't proced.would you give me a small hint
 
  • #33
Given that f(x+ y)= f(x)+ f(y), you can prove, by induction on the positive integer n, that f(nx)= nf(x), x any real number. From that, for n a non-zero integer, f(n/n)= nf(1/n)= f(1)= 1. Thus, f(1/n)= 1/f(n). You can get the case for f(m/n) from that.

This is an entirely different problem from before, isn't it? Requiring that f be continuous at x= 0 rather than that f(xy)= f(x)f(y).
 

Similar threads

Find f(x,y) given partial derivative and initial condition
  • · Replies 6 ·
Replies
6
Views
2K
Find f s. t. ||f||=1 and f(x) < 1 with ||x||=1
  • · Replies 20 ·
Replies
20
Views
3K
Find g(x)/h(y) for a given F(x,y)
  • · Replies 3 ·
Replies
3
Views
1K
Proof given ##x < y < z## and a twice differentiable function
  • · Replies 8 ·
Replies
8
Views
2K
Prove that ##\lim\limits_{x \to \infty} f(x) = 0##
  • · Replies 3 ·
Replies
3
Views
977
Find the constrained maxima and minima of ##f(x,y,z)=x+y^2+2z##
  • · Replies 2 ·
Replies
2
Views
2K
Find a function satisfying these conditions: f(x)f(f(x)) = 1 and f(2020) = 2019
  • · Replies 14 ·
Replies
14
Views
2K
Prove that f is a homeomorphism iff g is continuous, fg=1 and gf=1
  • · Replies 5 ·
Replies
5
Views
2K
Is f(z) =(1+z)/(1-z) a real function?
  • · Replies 17 ·
Replies
17
Views
3K
Condition for f(x,y,z) = f(x,y,z(x,y)) being extremized
  • · Replies 15 ·
Replies
15
Views
3K