(Weak) Triangle Inequality for a class $\mathcal{K}$ function

Question

Let $\alpha$ be a class $\mathcal{K}$ function defined on $[0,a)$. Then \begin{equation} \alpha(r_1+r_2) \leq \alpha(2r_1) + \alpha(2r_2), \quad \forall r_1,\,r_2 \in [0,\,a/2). \end{equation}

Definition (class $\mathcal{K}$ function): A continuous function $\alpha: [0, \,a) \rightarrow [0,\,\infty)$ is a class $\mathcal{K}$ function if it is strictly increasing and $\alpha(0) = 0$.

I have seen this result in at least two resources and a proof was not provided in neither of them. The authors explained that it is a direct consequence of the increasing property of class $\mathcal{K}$ functions, but there is one particular case that is not so obvious to me.

Here's my thoughts on it:

• Equality holds when $r_1 = r_2 = 0$.
• When $r_1 = r_2 \neq 0$, the result is true because of its increasing nature.
• For $r_1 \neq r_2$, we know that $\alpha(r_1) < \alpha(2 r_1)$ and $\alpha(r_2) < \alpha(2 r_2)$ because they are increasing. Given that $\alpha(0) = 0$, if either $r_1$ or $r_2$ is $0$, then the inequality also holds.
• But for the case where $r_1 \neq r_2$ and both are not $0$, can we say anything about the relationship between $\alpha(r_1)+\alpha(r_2)$ and $\alpha(r_1+r_2)$? Or is there anything else to be applied for this last case?

Answer 1

Without loss of generality assume that $0 \le r_1 \le r_2$. Then $$ \begin{align} 0 \le 2 r_1 &\implies 0 = \alpha(0) \le \alpha(2r_1) \\ r_1 + r_2 \le 2 r_2 &\implies \alpha(r_1 +r_2) \le \alpha(2 r_2) \end{align} $$ because $\alpha$ is increasing. Adding these inequalities gives the desired estimate: $$ \alpha(r_1 +r_2) \le \alpha(2r_1) + \alpha(2 r_2) \, . $$

Since $\alpha$ is strictly increasing, equality holds only if $0 = r_1 = r_2$.

Answer 2

Since $\alpha$ is strictly increasing and since $r_1+r_2\leq2\max(r_1,r_2)\in\{2r_1,2r_2\}$ you get $\alpha(r_1+r_2)\leq\alpha(2\max(r_1,r_2))\leq\alpha(2r_1)+\alpha(2r_2)$.