# Let $\alpha >1.$ Then $\forall x\gt 0: \psi(\alpha x)\leq \alpha \psi( x)\;.$ True or False?

Let $$\psi$$ be a function satisfying :

• $$\psi: \mathbb{R}^+\rightarrow \mathbb{R}^+$$ .

• $$\psi$$ is non-decreasing.

• $$\psi (x)< x, \forall x> 0$$.

I want to know if the following statement is true: $$\text{Let } \alpha >1. \text{ Then }\, \forall x> 0: \psi(\alpha x)\leq \alpha \psi( x)\;.$$

If not, can you give me a counter example please.

• Not true. The function can increase dramatically over a short interval. – Don Thousand Jan 17 at 15:54

Consider $$f(x) = \begin{cases} x, & x > 1 \\ x^2 & x \in [0,1] \end{cases}$$ around $$x \approx 0.9$$.
There is an easy counterexample: imagine the function $$\psi(x)$$ defined by parts as
$$\psi(x)=\begin{cases}\dfrac{x}{10} &x\leq 1000\\ \dfrac{2x}{5}+60 &x>1000\end{cases}$$
Clearly $$\psi(x)\leq x$$ for all $$x$$, it is non-decreasing, but for $$x=1000$$ and $$\alpha=2>1$$ we have $$\psi(2\cdot 1000)=500>200=2\cdot \psi(1000)$$.
• $\psi(2.2000)$ is not 500,+ and has two values at the same time – Motaka Jan 17 at 16:27