show the existence of a class $\mathcal{K}$ function

Question

A function $f: \mathbb{R}_+ \rightarrow \mathbb{R}_+$ is a class $\mathcal{K}$ function if it is continuous, strictly increasing and $f(0) = 0$. Now suppose $\lambda$ is a class $\mathcal{K}$ function, does there exist a class $\mathcal{K}$ function $\rho < \mathrm{id}$ such that for all $M \geq x \geq c$, $\lambda(x - c ) \leq \rho\circ\lambda(x)$

Answer 1

Since $\lambda$ is of class $\mathcal{K}$, then we have that $\lambda(x-c)<\lambda(x)$ for all $x\ge c$. Define $\sigma>0$ as

$$\sigma:=\max_{x\in[c,M]}\dfrac{\lambda(x-c)}{\lambda(x)}.$$

From the definition of a class $\mathcal{K}$ function, we have that $\sigma<1$. Therefore, one can take $\rho(x)=\sigma x$.

Example. $\lambda(x)=x^2$, then we have that $\sigma=(M-c)^2/M^2<1$.

We can verify the inequality $\lambda(x-c)\le\sigma\lambda(x)$ for all $x\in[c,M]$. This is equivalent to saying that

$$(-2Mc+c^2)x^2+2cM^2x-M^2c^2\ge0,\ x\in[c,M].$$

The case $c=M$ is immediate, so let us focus on the case $M>c$. Evaluating the left hand side at $x=c$ and $x=M$ yields

$$c^2(M-c)^2\ \mathrm{and}\ 0,\ \mathrm{respectively}.$$

Finally, observing that the left-hand side is strictly concave since $-2Mc+c^2<0$, allows to state that the left hand side is nonnegative for all $x\in[c,M]$.