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.

  • $\begingroup$ What is the definition of Lipschitz continuity? That is a good starting point $\endgroup$ Jan 14 '19 at 19:58
  • 1
    $\begingroup$ consider the Lipschitz condition near $x=1$, when one of the points is set to $1$. $\endgroup$
    – Hayk
    Jan 14 '19 at 20:02
  • $\begingroup$ 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. $\endgroup$
    – Tyler6
    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 $(1)$.


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