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    #1
    I am reading "Introduction to Real Analysis" (Fourth Edition) by Robert G Bartle and Donald R Sherbert ...

    I am focused on Chapter 6: Differentiation ...

    I need help in fully understanding an aspect of the proof of Theorem 6.1.6 ...


    Theorem 6.1.6 and its proof ... ... reads as follows:






    In the above text from Bartle and Sherbert we read the following:

    "... Since the function $ \displaystyle ( \psi \circ f ) \cdot \phi$ is continuous at $ \displaystyle c$, and its value at $ \displaystyle c$ is $ \displaystyle g' (f (c) ) \cdot f'(c)$ , Caratheodory's Theorem gives (11) ...


    Could someone please explain exactly how Caratheodory's Theorem gives (11) ...?


    Peter



    *** NOTE ***

    The post above mentions Caratheodory's Theorem ... so I am proving the text of the theorem statement ... as follows:


    Last edited by Peter; September 13th, 2017 at 02:31.

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    Quote Originally Posted by Peter View Post
    I am reading "Introduction to Real Analysis" (Fourth Edition) by Robert G Bartle and Donald R Sherbert ...

    I am focused on Chapter 6: Differentiation ...

    I need help in fully understanding an aspect of the proof of Theorem 6.1.6 ...


    Theorem 6.1.6 and its proof ... ... reads as follows:






    In the above text from Bartle and Sherbert we read the following:

    "... Since the function $ \displaystyle ( \psi \circ f ) \cdot \phi$ is continuous at $ \displaystyle c$, and its value at $ \displaystyle c$ is $ \displaystyle g' (f (c) ) \cdot f'(c)$ , Caratheodory's Theorem gives (11) ...


    Could someone please explain exactly how Caratheodory's Theorem gives (11) ...?


    Peter



    *** NOTE ***

    The post above mentions Caratheodory's Theorem ... so I am proving the text of the theorem statement ... as follows:



    After a little reflection I think the answer to my question is along the following lines ...

    In the proof of the Chain Rule (Theorem 6.1.6) B&S establish that:

    $ \displaystyle g(f(x)) - g(f(c)) = [ ( \psi \circ f(x) ) \cdot \phi (x) ] ( x - c )$

    Matching this with (10) in Caratheodory's Theorem and noting that (in the notation of that theorem) $ \displaystyle \phi (c) = f'(c)$, we then have (in Theorem 6.1.6) that

    $ \displaystyle [ ( \psi \circ f ) \cdot \phi ] (c) = ( g \cdot f)' (c) $


    Is that a correct interpretation ... ?

    Peter

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    #3
    Yes, it's correct. In my opinion, there is nothing here besides matching a general theorem statement with a concrete situation.

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    Quote Originally Posted by Evgeny.Makarov View Post
    Yes, it's correct. In my opinion, there is nothing here besides matching a general theorem statement with a concrete situation.



    Thanks Evgeny ...

    Peter

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