Combinations of Continuous Functions

[kingstrick]

Homework Statement

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.

Homework Equations

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.

 

[sunjin09]

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

 

[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 don't understand how to proceed...

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

Homework Statement

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.

Homework Equations

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.

 

[sunjin09]

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 don't understand how to proceed...

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

 

[kingstrick]

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

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)?

 

[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)?

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

 

[kingstrick]

so is g(0) = 1?

 

[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.

 

[kingstrick]

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.

Thank you so much,i finally understand.