Prove that the function $ g $ satisfying $g(g(x))=2g(x)-x $ is strictly monotonic.

Question

Let $ g :\Bbb R \to \Bbb R$ a continuous function such that $$(\forall x\in \Bbb R)\; g(g(x))=2g(x)-x$$

Prove that $ g $ is injective and strictly monotonic.

I took $ x,y\in \Bbb R $. $$g(x)=g(y)\implies g(g(x))=g(g(y))$$ $$\implies 2g(x)-x=2g(y)-y $$ $$\implies x=y$$

but to show that it is strictly monotonic, I didn't find a simple answer. Any help is appreciated.

@hamam_Abdallah I don't think it has a name but it is not hard to prove. Injectivity implies that that it is locally monotonic, continuity ensures that it isn't a crooked step function that changes direction on certain intervals. See if you can make that description precise. — CyclotomicField, Jan 09 '22 at 14:29
See for example https://math.stackexchange.com/q/170147/42969 — Martin R, Jan 09 '22 at 14:50

score 1 · Accepted Answer · answered Jan 09 '22 at 14:52

A Continuous $1-1$ function $f:\Bbb{R}\to \Bbb{R} $ is strictly monotone.

Hint:(Use Intermediate value theorem)

Strategy:

Assume $f$ is not strictly monotone.

Then, there exists $ x,y,z\in \Bbb{R}$ with $ x<y<z $ such that either:

$f(x)\le f(y)\ge f(z)$

Or

$f(x)\ge f(y) \le f(z) $

Now ignore the equality between all of them otherwise it fails to be one-to-one.

Now use intermediate value theorem to contradict that $f$ is one -to-one.

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