# Bounding the difference of numbers between 0 and 1 with the same power

I would like to prove the following inequality (I guess it holds but I'm not able to formally do it). Consider two numbers $$x,y\in (0,1)$$ and the positive real number $$\alpha$$. Then, can I write $$|(1-x)^\alpha-(1-y)^\alpha|\leq c \cdot |x-y|$$ where $$c$$ is a constant depending on $$\alpha$$? Thank you!

Function $$f$$ defined by $$f(x)=(1-x)^\alpha$$ is continuously differentiable, therefore for all $$x, y$$ in $$[0, 1]$$, $$|f(x)-f(y)|\leq |x-y| \cdot \sup_{t\in[0,1]} |f^\prime(t)|$$ Since $$f^\prime(t)=-\alpha(1-t)^{1-\alpha}$$, the inequality you're seeking is true if $$\alpha \geq 1$$, with $$\boxed{|(1-x)^\alpha-(1-y)^\alpha|\leq \alpha \cdot |x-y|}$$ Note that if $$\alpha\in (0,1)$$, the inequality is not true. For instance, take $$\alpha=\frac 1 2$$, and $$x=1-h$$, and $$y=1-2h$$: $$\frac{|\sqrt{1 -x} -\sqrt{1-y}|}{|x-y|}=\frac{1}{\sqrt{1 -x} +\sqrt{1-y}}=\frac{1}{(1+\sqrt 2)\sqrt h}\rightarrow +\infty$$
Consider $$x,y\in[0,1]$$ and the function $$f(x)=(1-x)^\alpha$$. Then the inequality holds if $$f$$ is Lipschitz continuous. Because a derivable function is Lipshitz continuos (here an answer), if $$\alpha\geq1$$ $$f$$ is derivable and so the inequality holds.
But for $$0<\alpha<1$$, $$f$$ isn't continuously differentiable on $$[0,1]$$ (the derivative is $$-\infty$$ at $$x=1$$), but because $$x,y \neq 1$$, $$f$$ is derivable on $$[0,\max(x,y)]$$ and so the inequality holds only here ($$c$$ depends on $$\max(x,y)$$!).