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Differentiability on R

  1. Jan 8, 2008 #1
    1. The problem statement, all variables and given/known data
    Suppose f:(0,[tex]\infty[/tex])->R and f(x)-f(y)=f(x/y) for all x,y in (0,[tex]\infty[/tex]) and f(1)=0. Prove f is continuous on (0,[tex]\infty[/tex]) iff f is continuous at 1.
    2. Relevant equations
    I think I ought to use these defn's of continuity: f continuous at a iff f(x)->f(a) as x->a
    or f is cont at a iff for Xn->a, f(Xn)-f(a) as Xn->[tex]\infty[/tex]

    3. The attempt at a solution
    The forward direction is immediate. For the backwards direction, we want to show that f(x)->f(a) as x->a for a in (0,[tex]\infty[/tex]). So since f cont at 1, f(x)->f(1)=0 as x->1. I tried to manipulate this but couldn't find a way to make x->a instead of x->1.

    Then I used the other definition and let Xn=1+1/n and Yn=a(Xn)=a+(a/n). Now Yn->a so just want to show that f(Yn)->f(a) as n->[tex]\infty[/tex]. But f(Yn)->a*0=0 as n->[tex]\infty[/tex]...

    I know I have to use f(x)-f(y)=f(x/y) somehow. So I went backwards: So I want to show that f(x)-f(a)->0 as x->a. So that means I want f(x/a)->0 as x->a. But now I don't see how to incorporate the fact that f is continuous at a.


    I know this is related to the log function but don't think this problem requires me to appeal that fact.... Note Xn and Yn are sequences indexed by n (I'm noob at this latex).


    Thanks for helping.
     
    Last edited: Jan 8, 2008
  2. jcsd
  3. Jan 8, 2008 #2
    take any a in (0, inf), then a !=0 so x/a makes sense and we have f(x) - f(a) = f(x/a), so f(x) = f(x/a) + f(a), now take the limit as x->a. Note as x->a, x/a -> 1
     
  4. Jan 9, 2008 #3
    Thank you, Dan. It makes so much sense now.
     
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