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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.

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  • $\begingroup$ Not true. The function can increase dramatically over a short interval. $\endgroup$ – Don Thousand Jan 17 at 15:54
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Consider $$ f(x) = \begin{cases} x, & x > 1 \\ x^2 & x \in [0,1] \end{cases} $$ around $x \approx 0.9$.

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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 $\psi(2\cdot 1000)=500>200=2\cdot \psi(1000)$.

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    $\begingroup$ $\psi(2.2000)$ is not 500,+ and has two values at the same time $\endgroup$ – Motaka Jan 17 at 16:27

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