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  1. MHB Master
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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 5: Continuous Functions ...

    I need help in fully understanding an aspect of Example 5.4.6 (b) ...


    Example 5.4.6 (b) ... ... reads as follows:






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

    "... ... However, there is no number $ \displaystyle K \gt 0$ such that $ \displaystyle \lvert g(x) \lvert \le K \lvert x \lvert $ for all $ \displaystyle x \in I$. ... ... "


    Can someone please explain why the above quoted statement holds true ...

    Peter


    *** EDIT 1 ***


    Just noticed that for $ \displaystyle x$ less than $ \displaystyle 1$ we have $ \displaystyle \sqrt{x}$ is larger than $ \displaystyle x$ ... ...

    e.g. $ \displaystyle \sqrt{0.0004}$ is $ \displaystyle 0.02$ ... ... and then we require $ \displaystyle K$ such that ...

    $ \displaystyle \lvert 0.02 \lvert \le K \lvert 0.0004 \lvert$

    ... so a large $ \displaystyle K$ is required ... ... and the required number will get larger and larger without bound as $ \displaystyle x$ gets smaller ...



    Is the above the correct explanation for $ \displaystyle f$ not being Lipschitz on $ \displaystyle I$ ... ... ?


    Peter



    *** EDIT 2 ***


    It may be helpful for readers of the above post to have access to B&S's definition of the Lipschitz function/condition ... ... so I am providing the following text from Bartle and Sherbert ...







    Note that in the above example B&S take $ \displaystyle u $ as the point $ \displaystyle u = 0$ ... ...


    Peter
    Last edited by Peter; September 12th, 2017 at 00:35.

  2. MHB Craftsman

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    #2
    Hi Peter,

    Your reasoning in Edit 1 is correct. Symbolically you could note that for $x\in (0,2]$, $x^{-1/2}\leq K$, verifying your claim that $K$ grows without bound as $x\rightarrow 0^{+}$.

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