Vtyjkyuk

I am trying to determine if the following function is Lipschitz or not $$f(x)=sqrt1-x^2 textfor -1leq xleq 1$$

My attempt:

Suppose $f$ is Lipschitz on $[-1,1]$. This implies $exists LinmathbbR$ such that $forall x,yin [-1,1]$
beginalign
|sqrt1-x^2-sqrt1-y^2|&leq L|x-y| \
L&geqfracsqrt1-x^2-sqrt1-y^2geqfrac \
endalign
If we set $y=0$ and take $xrightarrow 0$, then $Lrightarrowinfty$.

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.

Are you sure about$$
fracsqrt1-x^2-sqrt1-y^2geqfrac$$

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.

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 $sqrt2n-1leq$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 $sqrt2n-1$ is unbounded

As already pointed out by Saucy O'Path, the inequality $$leftlvert sqrt1-x^2-sqrt1-y^2rightrvertgeleftlvert -sqrt1-y^2rightrvert$$ 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 $x=1-frac1n$ and $y=1$ with $ninmathbbN^+$. Then $x,yin[-1,1]$ and
$$|sqrt1-x^2-sqrt1-y^2|leq L|x-y|$$
implies that
$$|sqrt1-1+frac2n-frac1n^2-sqrt1-1|leq Lleft|1-frac1n-1right|$$
that is
$$sqrt2n-1leq L$$
which is a contradiction because the sequence $sqrt2n-1_ninmathbbN^+$ is unbounded.

You could observe that for all $xne y$ in $[-1,1]$ such that either $x^2ne 1$ or $y^2ne 1$ $$fracsqrt1-x^2-sqrt1-y^2x-y=frac1-x^2-1+y^2(x-y)left(sqrt1-x^2+sqrt1-y^2right)=-fracx+ysqrt1-x^2+sqrt1-y^2$$

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