For an analysis exercise, I had to show that the function $\sqrt{1-x^2}$ was uniformly continuous, but not lipschitz continuous on the interval $[-1,1]$. I was able to show it was uniformly continuous, however I keep running into problems showing that it is not lipschitz.
Any help is appreciated.
Best Answer
Following the definition of Lipschitz condition for $f$, we need to show that for some constant $K>0 $ one has $$ |f(x) - f(y)| \leq K |x- y| \text{ for all } x,y \in [-1,1]. $$ Now take $y = 1$, then $f(y) = 0$ and the above becomes $$ \tag{1} \sqrt{1 - x^2} \leq K |x - 1|, \text{ for all } x \in [-1,1]. $$ However, $$ \lim\limits_{x\to 1-}\frac{\sqrt{1 - x^2}}{1-x} = \lim\limits_{x\to 1-}\frac{\sqrt{1 + x}}{\sqrt{1 - x}} = + \infty, $$ hence no $K$ satisfies $(1)$.