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Math Help - Isometric

  1. #1
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    Isometric

    Show that C[0,1] and C[a,b] are isometric.
    Solution:
    A mapping T of X into X' is said to be isometric if for all x, y in X
    d'(Tx,Ty)=d(x,y)
    C is continous [0,1] and [a,b] closed interval..
    The distance on Function space C[a,b] is d(x,y)=max|x(t)-y(t)|

    I couldnt procced from here....i need some help....

    thanks
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kinkong View Post
    Show that C[0,1] and C[a,b] are isometric.
    Solution:
    A mapping T of X into X' is said to be isometric if for all x, y in X
    d'(Tx,Ty)=d(x,y)
    C is continous [0,1] and [a,b] closed interval..
    The distance on Function space C[a,b] is d(x,y)=max|x(t)-y(t)|

    I couldnt procced from here....i need some help....

    thanks
    What if you did a mapping that bijectively took [a,b]\to[0,1] in an increasing way, call this g, and you considered T:C[a,b]\to C[0,1] defined by Tf(x)=f(g(x))?
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  3. #3
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    Thank u very much...
    But how can we prove that the map is bijective...i mean how can we see that it is one-to-one and onto...
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kinkong View Post
    Thank u very much...
    But how can we prove that the map is bijective...i mean how can we see that it is one-to-one and onto...
    Have you tried to prove it?
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  5. #5
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    yes i thought about it..but i couldnt find a way...please can u help me...
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  6. #6
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    Re: Isometric

    This is what I did:

    You are looking for a function, say F from C[a,b] into C[0,1], which is injective.
    Choose F such that F(u)=u(b*t-(1-t)*a). Here you take a function u which is an element of C[a,b] and, using F this u becomes an element of C[0,1].

    This function clearly preserves the distance propierty (isometric) which you are requiered. Indeed:

    Lets take u,w \in C[a,b] and F allows to take F(u),F(w) \in C[0,1], now
    d_{C[a,b]}(u,w)=\max_{t\in[a,b]} |u(t)-w(t)|=\max_{t\in[0,1]}|u(bt-(1-t)a)-u(bt-(1-t)a)|=d_{C[0,1]}(F(u),F(w)).

    Finally  F is bijective from the form it was selected.
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  7. #7
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    Re: Isometric

    Quote Originally Posted by kinkong View Post
    Thank u very much...
    But how can we prove that the map is bijective...i mean how can we see that it is one-to-one and onto...
    You were told to find a mapping from [a, b] to [0, 1]. There are many such mappings, some are bijective and some are not. Which did you use?

    To prove that a mapping, f, is one-to-one, you need to show that "if f(x)= f(y) then x= y". HOW you do that depends upon exactly what f is. To prove that a mapping, from [a, b] to [0, 1] is onto, you need to show that "if y is in [0, 1] then there exist an x in [a, b] such that f(x)= y.
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