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Math Help - Sets of Monotonic Functions

  1. #1
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    Sets of Monotonic Functions

    Let  \mathcal{M} be the set of montonic functions  \mathbb{R} \ \rightarrow \ \mathbb{R} . That is,  f \ \in \ \mathcal{M} if either

     f(x) \ \geq \ f(y) \ \forall \ x > y ,

    or

     f(x) \ \leq \ f(y) \ \forall \ x > y .

    Is  \mathcal{M} a subgroup of  \mathcal{F} under pointwise addition?

    Thank-you in advance for any help.
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  2. #2
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    Hi

    We have id_{\mathbb{R}}+(-id_{\mathbb{R}})=\theta , where \theta:\mathbb{R}\rightarrow\mathbb{R} is the zero map.
    Can you transform id_\mathbb{R} or/and -id_\mathbb{R} so their sum will become a function strictly decreasing on ]-\infty ,0] and strictly increasing on [0,+\infty[ ? Can \mathcal{M} be a subgroup?
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  3. #3
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    Brain Exploding

    Okay, Im really confused and in desparate need of help.

    I know that the set  \mathcal{F} of functions \mathbb{R}\rightarrow\mathbb{R} forms a group under addition.

    Now for  \mathcal{M} to be a subgroup of  \mathcal{F} the following conditions have to be met:

    (1) For every  x,y, \in \mathcal{F}, xy \in \mathcal {F}
    (2)  1 \in \mathcal{F}
    (3) For every  z \in \mathcal{F}, z^{-1} \in \mathcal{F}

    But I thought that that the set  \mathcal{F} of functions \mathbb{R}\rightarrow\mathbb{R} does not form a group under multiplication, which to me, seems to conflict with condition (1) from above.

    Ahhhhhhhhhhh , I think im getting things so confused
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  4. #4
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    (In your post, I think some occurences of \mathcal{F} should be replaced by \mathcal{M})

    I guess you think that \mathcal{M} doesn't satisfy (1), and it is true.

    I was trying to give you something easy to visualize in my last post.
    For example, let f and g be the functions defined by:

    f(x)=\begin{cases}x & x\leq 0 \\ 2x & x\geq 0\\\end{cases}

    g(x)=\begin{cases}-2x & x\leq 0 \\ -x & x\geq 0 \\\end{cases}

    Are they in \mathcal{M} ? And their pointwise addition?
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