# Show that f is strictly increasing in RR ?

## $f : \mathbb{R} \to \mathbb{R}$ differentiable with $f '$ continuous in $\mathbb{R}$, $f \left(0\right) = 0$ , $f \left(1\right) = 1$ $f \left(f \left(x\right)\right) + f \left(x\right) = 2 x$ , $\forall$$x$$\in$$\mathbb{R}$ Show that $f$ is strictly increasing in $\mathbb{R}$

May 24, 2018

#### Explanation:

$f$ is differentiable in $\mathbb{R}$ and the property is true $\forall x$$\in$$\mathbb{R}$ so by differentiating both parts in the given property we get

$f ' \left(f \left(x\right)\right) f ' \left(x\right) + f ' \left(x\right) = 2$ (1)

If $\exists {x}_{0}$$\in$$\mathbb{R} : f ' \left({x}_{0}\right) = 0$ then for $x = {x}_{0}$ in (1) we get

$f ' \left(f \left({x}_{0}\right)\right) {\cancel{f ' \left({x}_{0}\right)}}^{0} + {\cancel{f ' \left({x}_{0}\right)}}^{0} = 2$ $\iff$

$0 = 2$ $\to$ Impossible

Hence, $f ' \left(x\right) \ne 0$ $\forall$$x$$\in$$\mathbb{R}$

• $f '$ is continuous in $\mathbb{R}$
• $f ' \left(x\right) \ne 0$ $\forall$$x$$\in$$\mathbb{R}$

$\to$ $\left\{\begin{matrix}f ' \left(x\right) > 0 \text{ & " \\ f'(x)<0" & }\end{matrix}\right.$ $x$$\in$$\mathbb{R}$

If $f ' \left(x\right) < 0$ then $f$ would be strictly decreasing

But we have $0 < 1$ ${\iff}^{f \downarrow}$ $\iff$ $f \left(0\right) > f \left(1\right)$ $\iff$

$0 > 1$ $\to$ Impossible

Therefore, $f ' \left(x\right) > 0$, $\forall$$x$$\in$$\mathbb{R}$ so $f$ is strictly increasing in $\mathbb{R}$