Is $f(x)=\sqrt{1-x^2}$ Lipschitz?

Question

I am trying to determine if the following function is Lipschitz or not $$f(x)=\sqrt{1-x^2} \ \ \ \text{for} \ -1\leq x\leq 1$$

My attempt:

Suppose $f$ is Lipschitz on $[-1,1]$. This implies $\exists L\in\mathbb{R}$ such that $\forall x,y\in [-1,1]$ \begin{align} |\sqrt{1-x^2}-\sqrt{1-y^2}|&\leq L|x-y| \\ L&\geq\frac{|\sqrt{1-x^2}-\sqrt{1-y^2}|}{|x-y|}\geq\frac{|-\sqrt{1-y^2}|}{|x-y|}=\frac{\sqrt{1-y^2}}{|x-y|} \\ \end{align} If we set $y=0$ and take $x\rightarrow 0$, then $L\rightarrow\infty$.

This is a contradiction, as $L$ is finite. Hence $f$ is not Lipschitz.

I'm wondering if the my logic is correct, particularly my second line of working. Any advice would be greatly appreciated.

Answer 1

As already pointed out by Saucy O'Path, the inequality $$\left\lvert \sqrt{1-x^2}-\sqrt{1-y^2}\right\rvert\ge\left\lvert -\sqrt{1-y^2}\right\rvert$$ does not hold in $[-1,1]$ (take for example $-1<x<1$ and $y=0$).

On the other hand, you may modify your approch by taking $x=1-\frac{1}{n}$ and $y=1$ with $n\in\mathbb{N}^+$. Then $x,y\in[-1,1]$ and $$|\sqrt{1-x^2}-\sqrt{1-y^2}|\leq L|x-y|$$ implies that $$|\sqrt{1-1+\frac{2}{n}-\frac{1}{n^2}}-\sqrt{1-1}|\leq L\left|1-\frac{1}{n}-1\right|$$ that is $$\sqrt{2n-1}\leq L$$ which is a contradiction because the sequence $\{\sqrt{2n-1}\}_{n\in\mathbb{N}^+}$ is unbounded.

Answer 2

Are you sure about$$ \frac{|\sqrt{1-x^2}-\sqrt{1-y^2}|}{|x-y|}\geq\frac{|-\sqrt{1-y^2}|}{|x-y|}$$

Note that for two positive numbers, $a$ and $b$ we do not necessarily have $$|a-b|\ge |b|$$ For example $|5-3|=2 < 3$

Note that the absolute value of the derivative near boundary points gets very large.

I suggest that you focus at the endpoints of the domain and apply mean value theorem.

Answer 3

as Robert mentioned taking $x=1-1/n$, n is natural and $y=1$..but i want to add this thing that if f is lipschitz then $\sqrt{2n-1}\leq$L must hold for all n in natural because $x=1-1/n$ lies in [-1,1] for all n in natural but that cant be possible because $\sqrt{2n-1}$ is unbounded

Answer 4

You could observe that for all $x\ne y$ in $[-1,1]$ such that either $x^2\ne 1$ or $y^2\ne 1$ $$\frac{\sqrt{1-x^2}-\sqrt{1-y^2}}{x-y}=\frac{1-x^2-1+y^2}{(x-y)\left(\sqrt{1-x^2}+\sqrt{1-y^2}\right)}=-\frac{x+y}{\sqrt{1-x^2}+\sqrt{1-y^2}}$$

Which diverges to $-\infty$ when $(x,y)\to (1,1)$ and to $\infty$ when $(x,y)\to(-1,-1)$.