Determining if a function is Lipschitz

Question

In a certain exercise I am asked to show that a certain function is Lipschitz. This function is $f(x) = arcsin(x), f:(-1,1) \rightarrow \mathbb{R}$. My attempt consisted on trying to bound the derivative, which is $f'(x) = \frac{1}{\sqrt{(1-x^2)}}$. However, it is easy to see that this derivative cannot be bounded when $x$ tends to $1$ or $-1$, although if one chooses a compact interval contained in $(-1,1)$, let it be $(a,b)$, the derivative can be bounded. Does it mean that is only Lipschitz in any interval strictly contained in $(-1,1)$?
Not only that, but I was wondering if that argument of bounding the derivative on a certain interval is valid for any function. For example, if $f(x) = x^2$, this function is not globally Lipschitz, but it is locally if we consider a closed interval (and I think that even an open one, or none of these options). In this same exercise, there is another paragraph which asks me to say if it is Lipschitz or locally Lipschitz this same function of $arcsin(x)$, but considering the closed interval $[-1,1]$. Is there any significant difference?

Answer 1

Part 1: I agree with your assessment, and the function is only Lipschitz on any compact subset of $(-1, 1)$ (referred to as locally Lipschitz). This is because the derivative will be bounded on those compact subsets, giving you a finite constant.

Part 2: You are referring to the distinction between Lipschitz and locally Lipschitz, which are not the same, as pointed out in your question. However, bounding the Lipschitz constant by the $\sup |f'|$ is nonetheless valid on any interval of your consideration.

Part 3: In a closed interval Lipschitz means the same as locally Lipschitz, since the interval is compact already. Dinstinguish:

• On $(-1, 1)$, the function is locally Lipschitz because you can find an open interval around any point $x\in (-1, 1)$ in which the function is Lipschitz. Note that this explicitly lets you stay clear of the 'problematic' points $x=\pm 1$.
• On $[-1, 1]$, to be locally Lipschitz, you would need $f$ to be Lipschitz also on some open interval around $x=-1$ and $x=1$ respectively, e.g. some interval $(1-\epsilon, 1]$, which does contain the problematic boundary point with infinite slope.

Note that in the metric space $[-1, 1]$ an open interval around $1$ is of the form $(a, 1]$, i.e. it being closed on the right is no accident.

See the Properties-Section of Lipschitz continuity

Answer 2

DominikS has already answered most of your concerns, but without proofs, so I'll prove that $\arcsin$ is locally Lipschitz on $(-1,1)$ but not Lipschitz on $(-1,1)$. To see that $\arcsin$ is locally Lipschitz on $(-1,1)$, it suffices to show that it is locally Lipschitz on each compact subinterval of $(-1,1)$. Let $[a,b]\subseteq(-1,1)$. As $\arcsin'(x)=\frac{1}{\sqrt{1-x^2}}$ defines a continuous function on $[a,b]$, it is also bounded by some $L$. Then, for any $x,y\in[a,b]$ with $x<y$, find some $\xi\in(a,b)$ such that

$$\frac{\arcsin(x)-\arcsin(y)}{x-y}=\arcsin'(\xi).$$

Then

$$\lvert \arcsin(x)-\arcsin(y)\rvert=\lvert\arcsin'(\xi)\rvert\lvert x-y\rvert\leq L\rvert\lvert x-y\rvert.$$

It follows that $\arcsin$ is Lipschitz on $[a,b]$ with a Lipschitz constant $L$.

As for $\arcsin$ not being Lipschitz on $(-1,1)$, it suffices to show that the function $h$ defined by

$$h(x,y)=\frac{\arcsin(x)-\arcsin(y)}{x-y}$$

for $x,y\in(0,1)$ with $x\neq y$ is unbounded. Indeed this is enough, as if it was Lipschitz with Lipschitz constant, then we would have the bound

$$\lvert h(x,y)\rvert\leq L,$$

contradicting the unboundednes. Now to see that $h$ is unbounded, take two sequences $\{x_j\}_{j\in\mathbb{N}}$, $\{y_j\}_{j\in\mathbb{N}}$ in $(0,1)$ with $x_j<y_j$ for all $j\in\mathbb{N}$ such that $x_j\to1$ as $j\to\infty$. For each $j\in\mathbb{N}$, find some $\xi_j\in(x_j,y_j)$ such that

$$h(x_j,y_j)=\arcsin'(\xi_j)=\frac{1}{\sqrt{1-\xi_j^2}}.$$

As $x_j\to1$, and $x_j<\xi_j<1$ for all $j\in\mathbb{N}$, it follows that $\xi_j\to1$ as $j\to\infty$, and so

$$\lim_{j\to\infty}h(x_j,y_j)=\lim_{j\to\infty}\frac{1}{\sqrt{1-\xi_j^2}}=\infty.$$

But this implies that $h$ is unbounded, as for any $L>0$, we can find a $j\in\mathbb{N}$ such that

$$h(x_j,y_j)>L.$$

It follows that $\arcsin$ is not Lipschitz on $(-1,1)$.