Currently I'm reading a paper titled "Fixed Point Theorems for Multivalued $\alpha$-Function" and I have a problem on proofing one of its function's properties.
Let $J \subseteq [0,\infty)$ that contains 0. A function $\psi : J \rightarrow J$ is gauge function of order $r \geq 1$ if (i) $\psi(\lambda t) \leq \lambda^r \psi(t)$, for all $\lambda \in (0,1)$ and $t\in J$, and (ii) $\psi(t) < t$, for all $t \neq 0$.
How can I proof this?
If $\phi : J \rightarrow J$ is gauge function of order $r > 1$ and there exist a nondecreasing and non-negative function $\varphi : J\rightarrow J$ such that $\phi(t) = t \varphi(t)$, for all $t \in J$, then show that $\varphi(0) = 0$.
Firstly, as $\varphi$ is non-decreasing, and non-negative, it is continuous, and $\varphi(t) \ge 0$
Now, we have $$\phi(\lambda t) = \lambda t\varphi(\lambda t)$$ and \begin{align} \phi(\lambda t) &\le \lambda^r \phi(t) \\ &= \lambda^rt\varphi(t) \end{align}
So, $$0\le \varphi(\lambda t)\le \lambda^{r-1}\varphi(t)$$
Taking the limit $\lambda \to 0$, as $r>0$, $\lambda^{r-1} \to 0$. Thus, $\lim_{\lambda\to 0} \varphi(\lambda t) =0$
As $\varphi$ is continuous, $\lim_{\lambda\to 0} \varphi(\lambda t) = \varphi(0)$. So, the result is proved.