# Showing function is not Lipschitz continuous

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.

• What is the definition of Lipschitz continuity? That is a good starting point Jan 14 '19 at 19:58
• consider the Lipschitz condition near $x=1$, when one of the points is set to $1$.
– Hayk
Jan 14 '19 at 20:02
• It is that there exists $K>0$ such that for $x,y \in [-1,1]$, $|f(x)-f(y)|<K|x-y|$. I know I want to show that such a $K$ cannot exist, but I am lost as to how I can do this. Jan 14 '19 at 20:04

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 $$(1)$$.