TITLE: Proving a Lipschitz constant does not exist.
QUESTION [7 upvotes]: So I wish to find for each of these functions a Lipschitz constant or prove that none exists. So my definition for a function to be Lipschitz is:
A function $f:[a,b] \rightarrow \mathbb{R}$ is Lipschitz if there exists a $L$ such that $|f(x) - f(y)| \leq L|x-y|$ for all $x,y \in [a,b]$.

$f(x) = \frac{1}{x}$ for $x \in (0, 1]$ 
$f(x) = e^x$ for $x \in \mathbb{R}$
$f(x) = \sqrt{1-x^2}$ for $x \in [-1,1]$

My attempt for 1) to prove $f$ is not Lipschitz is via contradiction. Suppose that $x, y \in (0,1]$ and $f$ is Lipschitz. Then there exists a $L$ such that,
$$|\frac{1}{x} -\frac{1}{y}| = \frac{|x-y|}{|xy|} \leq L |x-y|$$ for all $x,y \in (0, 1]$. This would imply that, $\frac{1}{|xy|} \leq L$ for all $x,y \in (0, 1]$. But such a $L$ cannot exist since we can make $x, y$ as small as we like, the fraction will grow.
So in conclusion I attempted 1) but not sure if I am correct. I am stuck on 2 and 3. Anybody can give me some hints? I'm thinking of using the Mean Value Theorem on question 2. Other than that I have no idea how to start.

REPLY [5 votes]: You are right concerning 1).
Concerning 2): assume that there exist $L$ such that bla-bla-bla. Then $|\Bbb e ^ x - 1| \le L |x - 0|$, or $\frac {\Bbb e ^ x - 1} x \le L$ for $x>0$. What happens for $x \to \infty$?
For 3): note that if a continuously derivable function is Lipschitz, then $|f(x) - f(x_0)| \le L|x - x_0|$, so $|\frac {f(x) - f(x_0)} {x-x_0} | \le L$, so $|f'(x_0)| \le L$. Now show that $(\sqrt{1-x^2}) '$ is not bounded on $(-1,1)$ by studying its behaviour toward $\pm 1$, which shows that no such $L$ can be found, therefore the function is not Lipschitz.