Show that $t-t^{2}+s-s^{2}-2(s\wedge t-st)\leq C|t-s|^{r}$ for all $t,s\in\mathbb{R}_{\geq 0}$, and constant $C, r>0$.

Question

I am in the end of a proof, and I want to show that:

For all $s,t\in\mathbb{R}_{\geq 0}$, there exists $C,r>0$ such that $$t-t^{2}+s-s^{2}-2(s\wedge t-st)\leq C|t-s|^{r}.$$

My attempt so far is that for $t\leq s$, we have $$-t-t^{2}+s-s^{2}+2st\leq -t+s+2st\leq -t+s+2s^{2}$$ but then I cannot have any form of $|t-s|^{2}$.

So it turned out that $st$ is some error term, and I want it to disappear earlier: $$-t-t^{2}+s-s^{2}+2st\leq -t-t^{2}+s-s^{2}+2ss=-t-t^{2}+s+s^{2}=(s-t)+s^{2}-t^{2}$$ however $t\leq s$, so I cannot achieve further bound...

Is there any way to prove this? Thank you!

Answer 1

$C=r=1$ works fine. You need $t-t^{2}+s-s^{2}-2(s\wedge t-st) \leq |t-s|$. Suppose (w.l.o.g.) $s \leq t$. Then this becomes $t-t^{2}+s-s^{2}-2(s-st) \leq |t-s|$ or $t-s -(t-s)^{2} \leq t-s$ which is true becasue $(t-s)^{2} \geq 0$.