The problem involves a function g: ℝ→ℝ that satisfies the functional equation g(x+y) = g(x)g(y) for all real numbers x and y. The continuity of g at x = 0 is given, and the goal is to show that g is continuous at every point in ℝ.
Some participants have provided hints and guidance on how to approach proving continuity, particularly by examining the behavior of g at 0 and considering limits. There is an ongoing exploration of the implications of the derived values and conditions without reaching a consensus on the next steps.
Participants express uncertainty about the derivation of g(0) and its implications for the continuity proof. There is a mention of the trivial case where g(x) could be zero for all x, which is noted but not resolved.
sunjin09 said:you want to prove lim_{y->0}g(x+y)=g(x), now you know how to use your condition on g.
kingstrick said: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.
kingstrick said: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...
sunjin09 said: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 said: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)?
sunjin09 said: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 [itex]\lim_{y\rightarrow0}g(x+y)=\lim_{y\rightarrow0}g(x)g(y)=g(x)g(0)=g(x)[/itex], that's about it.