Let $y \in [0,M]$ where $M$ is a positive constant and $\alpha > 0$.
Since $y \rightarrow \exp(y)$ is $C^1$ on the compact domain $[0,M]$, then it is Lipchitz and we can write that $$ |\exp(y_1) - \exp(y_2)| \leq L |y_1-y_2| \forall y_1, y_2 \in [0,M]$$ Now, I'm wondering if we could say a similar thing for the function $y \rightarrow \exp(y/\alpha)$, that is $$ |\exp(y_1/\alpha) - \exp(y_2/\alpha)| \leq L_{\alpha} |y_1-y_2| \forall y_1, y_2 \in [0,M]$$ When $\alpha \geq 1$, I know that $(y_1/\alpha)$ and $(y_2/\alpha)$ belong to $[0,M]$, then we have $$ |\exp(y_1/\alpha) - \exp(y_2/\alpha)| \leq L |y_1-y_2| \forall y_1, y_2 \in [0,M]$$
I'm trying to see if this holds for the case when $\alpha < 1$. Intuitively, when $\alpha \rightarrow 0$, I think that $|\exp(y_1/\alpha) - \exp(y_2/\alpha)| \sim 0$ since the arguments will be dominated by the term $1/\alpha$ (not quite sure though), so in this case, I think the inequality should also hold.
However, if $\alpha <1$ but not very small (i.e. approaching zero), can we say something?