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Topology-Continuos Functions

  1. Dec 10, 2009 #1
    1. The problem statement, all variables and given/known data
    Let p:C[0,1]->C[0,1] the function that doing this "match":
    For each f in C[0,1] , p(f)=f(x^2)

    We need to prove that p is a continuous function.


    2. Relevant equations
    C[0,1] is the metric space of all the functions that are continuos in [0,1].
    The distance between two functions g,f in C[0,1] is:
    max{|f(t)-g(t)|} where t is in [0,1] ...

    3. The attempt at a solution
    I'm pretty sure we need to use the fact that if x is in [0,1] then x^2 is also in [0,1] ...
    Maybe we should try using uniform continuity or Lifchitz Condition...

    TNX to all the helpers!
     
  2. jcsd
  3. Dec 10, 2009 #2
    I realy think Lifchitz condition is the way...My intuition tells me that the maximum distance between f(x^2) and g(x^2) must be less the g(x) and f(x) but I realy don't know how to write it a formal way...
    Hope someone will be able to help me

    TNX
     
  4. Dec 10, 2009 #3
    The Lipschitz condition is a red herring.

    As you say, given [tex]f, g \in C[0,1][/tex], the distance between them is [tex]\|g - f\|_{C[0,1]} = \sup \{ |g(t) - f(t)| : t \in [0,1] \}[/tex]. This is the supremum of a set.

    What is the distance between [tex]p(g)[/tex] and [tex]p(f)[/tex]? It is also the supremum of a set. What is the relation between these two sets?
     
  5. Dec 10, 2009 #4
    That one set contains all the squares of the other one?
    This is excatly what is missing in the way to soloution...

    :(

    TNX
     
  6. Dec 10, 2009 #5
    No, that's not right.

    Write it out and look at it carefully. What is squared is not what you think is squared.
     
  7. Dec 10, 2009 #6
    Their values are squared! It's the definition of f(x^2) no?
    Can you help me out here? I'm kind of hopeless...
    TNX
     
  8. Dec 10, 2009 #7
    Well there is my try: Let f , g be: f(x)=x , g(x)=2x and let's check their values for 0,0.25,0.5 and 1:
    We'll get: g(t)-f(t)={0, 0.25, 0.5, 1} ... Now: f(x^2)=x^2 , g(x^2)=2x^2
    So : g(t^2)-f(t^2)={0, 0.0625, 0.25, 1 }


    Let's check another example: f(x)=2x+4 , g(x)=x -> f(x^2)=2x^2+4 , g(x^2)=x^2
    f(t)-g(t)={4, 4.25, 4.5 , 5}
    But f(t^2)-g(t^2)={4, 4.0625, 4.25, 5 }

    My assumtion was from the beginning that the distances are equal...
    Becuase, for example, if the supermum is in 0.5, then the supermum of the squares is in
    sqrt(0.5)... So the distances must be equal...Am I right this time?
     
    Last edited: Dec 10, 2009
  9. Dec 10, 2009 #8
    No; [tex]f(x^2) \neq f(x)^2[/tex]. You are not squaring the value of [tex]f[/tex]; you are measuring the value of [tex]f[/tex] at a different place.
     
  10. Dec 11, 2009 #9
    Yes...But as you can see in my message above you, I am almost sure that if the supermum of f is in x, then the supermum of f(x^2) is in sqrt(x)...It's only because it's f(x^2) and not f(x)^2 of course...
     
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