Combinations of Continuous Functions

1. Mar 15, 2012

kingstrick

1. The problem statement, all variables and given/known data

Let g: ℝ→ℝ satisfy the relation g (x+y) = g(x)g(y) for all x, y in ℝ. if g is continuous at x =0 then g is continuous at every point of ℝ.

2. Relevant equations

3. The attempt at a solution

Let W be an ε-neighborhood of g(0). Since g is continuous at 0, there is a δ-neighborhood V of 0 = f(c)...not sure where to proceed from here.

2. Mar 15, 2012

sunjin09

you want to prove lim_{y->0}g(x+y)=g(x), now you know how to use your condition on g.

3. Mar 15, 2012

kingstrick

Okay, so continuing with my proof:

At x = 0, then g(0+y) = g(o)g(y) = g(0)+g(y)
→ g(0) = g(y) (g(0)-1) ... then what? I dont understand how to proceed...

4. Mar 15, 2012

sunjin09

First, check your derivation here for mistakes, and find what value g(0) may be. THEN use the hint I gave you for ARBITRARY x and y→0

5. Mar 15, 2012

kingstrick

so am i missing a concept, when a problem says x→0 for the lim g, does that mean to apply to g(x) and g(y) not just g(x)?

6. Mar 16, 2012

kingstrick

I just don't see how to determine the value of g(o)....I am stumped!

7. Mar 16, 2012

kingstrick

so is g(0) = 1?

8. Mar 16, 2012

sunjin09

You got it right. Since g(x)=g(x+0)=g(x)g(0), you have either g(0)=1 ( or g(x)=0 for all x's which is trivial.)
Now you can try to prove continuity by proving $\lim_{y\rightarrow0}g(x+y)=\lim_{y\rightarrow0}g(x)g(y)=g(x)g(0)=g(x)$, that's about it.

9. Mar 16, 2012

kingstrick

Thank you so much,i finally understand.