Topology-Continuos Functions

  • #1
108
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Homework Statement


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.


Homework 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] ...

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!
 

Answers and Replies

  • #2
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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
 
  • #3
352
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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?
 
  • #4
108
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That one set contains all the squares of the other one?
This is excatly what is missing in the way to soloution...

:(

TNX
 
  • #5
352
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No, that's not right.

Write it out and look at it carefully. What is squared is not what you think is squared.
 
  • #6
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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
 
  • #7
108
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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:
  • #8
352
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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
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.
 
  • #9
108
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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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