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Combinations of Continuous Functions

  1. Mar 15, 2012 #1
    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. jcsd
  3. Mar 15, 2012 #2
    you want to prove lim_{y->0}g(x+y)=g(x), now you know how to use your condition on g.
     
  4. Mar 15, 2012 #3
    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...

     
  5. Mar 15, 2012 #4
    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
     
  6. Mar 15, 2012 #5
    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)?
     
  7. Mar 16, 2012 #6
    I just don't see how to determine the value of g(o)....I am stumped!
     
  8. Mar 16, 2012 #7
    so is g(0) = 1?
     
  9. Mar 16, 2012 #8
    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.
     
  10. Mar 16, 2012 #9
    Thank you so much,i finally understand.
     
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