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!
Asked
Active
Viewed 80 times
2 Answers
3
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 $y=1-2h$: $$\frac{|\sqrt{1 -x} -\sqrt{1-y}|}{|x-y|}=\frac{1}{\sqrt{1 -x} +\sqrt{1-y}}=\frac{1}{(1+\sqrt 2)\sqrt h}\rightarrow +\infty$$
Stefan Lafon
- 13,728
1
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 $\max(x,y)$!).
dcolazin
- 2,262