Let $\alpha_0: \mathbb{R}_{\geq 0} \to \mathbb{R}_{\geq 0}$ be a $\mathcal{K}$-function, i.e., a strictly increasing function such that $\alpha_0(0) = 0$, and $b \geq 0$ a given constant. Is there any way to lower bound what follow
$$ \alpha_0(x) + b \geq \alpha(x + b) $$
with a $\mathcal{K}$-function $\alpha$? I guess this should be a well-known result, any suggestions for some related theorems?
You're looking for a $\mathcal K$-function $\alpha$ such that $\alpha(x+b)\le\alpha_0(x)+b$.
Notice that $\alpha_0(x)+b$ is a strictly increasing function. So one would think of letting $\alpha(x+b)=\alpha_0(x)+b$. In other words, we take $\alpha(x)=\alpha_0(x-b)+b$ for $x\ge b$.
We need to define $\alpha$ on $[0,b]$ as well! Well notice that $\alpha(b)=b$, so any strictly increasing function such that $\alpha(0)=0$ and $\alpha(b)=b$ will do.