1. 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.

2. 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?

3. 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

4. (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?