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  1. MHB Master
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    #1
    I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ...

    I am currently reading Chapter 3: Continuous Functions on Intervals and am currently focused on Section 3.1 Limits and Continuity ... ...

    I need some help with the proof of Corollary 3.13 ...


    Corollary 3.13 reads as follows:





    Can someone help me to prove that if $ \displaystyle f$ is continuous then $ \displaystyle f^+ = \text{max} (f, 0)$ is continuous ...


    My thoughts are as follows:


    If $ \displaystyle c$ belongs to an interval where $ \displaystyle f$ is positive then $ \displaystyle f^+$ is continuous since $ \displaystyle f$ is continuous ... further, if $ \displaystyle c$ belongs to an interval where $ \displaystyle f$ is negative then $ \displaystyle f^+$ is continuous since $ \displaystyle g(x) = 0$ is continuous ... but how do we construct a proof for those points where $ \displaystyle f(x)$ crosses the $ \displaystyle x$-axis ... ..




    Help will be much appreciated ...

    Peter
    Last edited by Peter; January 17th, 2020 at 05:00.

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  3. MHB Master
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    #2
    To prove $ \displaystyle f^+$ is continuous at $ \displaystyle x_0$ consider 3 cases:
    1) $ \displaystyle f(x_0)> 0$.
    2) $ \displaystyle f(x_0)= 0$.
    3) $ \displaystyle f(x_0)< 0$.

    Since f is continuous, in case (1) there exist an interval around $ \displaystyle x_0$ such that f(x)> 0
    and $ \displaystyle f^+(x)= f(x)$
    for all x in the interval

    Since f is continuous, in case (1) there exist an interval around $ \displaystyle x_0$ such that f(x)< 0 and $ \displaystyle f^+(x)= 0$ for all x in the interval.

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    #3 Thread Author
    Quote Originally Posted by HallsofIvy View Post
    To prove $ \displaystyle f^+$ is continuous at $ \displaystyle x_0$ consider 3 cases:
    1) $ \displaystyle f(x_0)> 0$.
    2) $ \displaystyle f(x_0)= 0$.
    3) $ \displaystyle f(x_0)< 0$.

    Since f is continuous, in case (1) there exist an interval around $ \displaystyle x_0$ such that f(x)> 0
    and $ \displaystyle f^+(x)= f(x)$
    for all x in the interval

    Since f is continuous, in case (1) there exist an interval around $ \displaystyle x_0$ such that f(x)< 0 and $ \displaystyle f^+(x)= 0$ for all x in the interval.


    Thanks for the help ...

    BUT ... you do not describe what to do in case 2 ... and as I indicated it is when f(x) = 0, specifically when f crosses the x-axis that I am having trouble dealing with ...

    Peter

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    #4 Thread Author
    Quote Originally Posted by Peter View Post
    Thanks for the help ...

    BUT ... you do not describe what to do in case 2 ... and as I indicated it is when f(x) = 0, specifically when f crosses the x-axis that I am having trouble dealing with ...

    Peter



    After reflecting on this problem for some time ... here is my proof for the situation where the point investigated is a point where $ \displaystyle f$ crosses the x-axis ...


    I think it will suffice to prove that $ \displaystyle f^+$ is continuous for the case where a point $ \displaystyle c_1 \in \mathbb{R}$ is such that for $ \displaystyle x \lt c_1, \ f(x) = f^+(x)$ is positive and for $ \displaystyle x \gt c_1, \ f^+(x) = 0$ ... ... while for some point $ \displaystyle c_2 \gt c_1$ we have that $ \displaystyle f^+(x) = 0$ for $ \displaystyle x \lt c_2$ and $ \displaystyle f(x) = f^+(x)$ is positive for $ \displaystyle x \gt c_2$ ... ...


    ... see Figure 1 below ...










    Now consider an (open) neighbourhood $ \displaystyle V$ of $ \displaystyle f^+(c_1)$ where...



    $ \displaystyle V= \{ f^+(x) \ : \ -f^+(a_1) \lt f^+(c_1) \lt f^+(a_1)$ for some $ \displaystyle a_1 \in \mathbb{R} \}

    $

    so ...


    $ \displaystyle V= \{ f^+(x) \ : \ -f^+(a_1) \lt 0 \lt f^+(a_1)$ for some $ \displaystyle a_1 \in \mathbb{R} \}$ ...




    Then ... (see Figure 1) ...



    $ \displaystyle (f^+)^{ -1 } (V) = \{ a_1 \lt x \lt a_2 \}$ which is an open set as required ....



    Further crossings of the x-axis by $ \displaystyle f$ just lead to further sets of the nature $ \displaystyle \{ a_{ n-1 } \lt x \lt a_n \}$ which are also open ... so ...




    $ \displaystyle (f^+)^{ -1 } (V) = \{ a_1 \lt x \lt a_2 \} \cup \{ a_3 \lt x \lt a_4 \} \cup \ldots \cup \{ a_{n-1} \lt x \lt a_n \}$



    which being a union of open sets is also an open set ...




    The proof is similar if $ \displaystyle f$ first crosses the x-axis from below ...







    Is that correct?


    Peter
    Last edited by Peter; January 18th, 2020 at 20:58.

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    #5
    Quote Originally Posted by HallsofIvy View Post
    To prove $ \displaystyle f^+$ is continuous at $ \displaystyle x_0$ consider 3 cases:
    1) $ \displaystyle f(x_0)> 0$.
    2) $ \displaystyle f(x_0)= 0$.
    3) $ \displaystyle f(x_0)< 0$.

    Since f is continuous, in case (1) there exist an interval around $ \displaystyle x_0$ such that f(x)> 0
    and $ \displaystyle f^+(x)= f(x)$
    for all x in the interval

    Since f is continuous, in case (1) there exist an interval around $ \displaystyle x_0$ such that f(x)< 0 and $ \displaystyle f^+(x)= 0$ for all x in the interval.
    TYPO: in the last sentence "case (1)" should have been "case (2)".

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