prove that e^x is not Lipschitz on R

Question

I know that the definition of Lipschitz function is: If A⊆R and f:A→ R is function. Then, f is said to be Lipschitz on A if there exists a constant K>0 such that,|f(x)−f(y)|≤K|x−y| for all x,y∈A. but how to prove that e^x is not Lipschitz on R?

Answer 1

$e^x$ is not uniformly continuous and so is not Lipschitz. Take for example $x_n=\ln n,\ $ $y_n=\ln(n+1)$ then $y_n-x_n\to 0$ but $e^{y_n}-e^{x_n}\to 1$. Roughly speaking $e^x$ diverges to $+\infty$ with big velocity, bigger than a linear function

Answer 2

Because the derivative isn't bounded. Convexity makes it very straighforward, though still true while being not convex. $$f(y) - f(x) \geq f'(x) (y-x) $$

Answer 3

$\newcommand{\R}{\mathbb{R}}$ This kind of problem can be easily tackled using proof by contradiction.

Suppose $e^x$ is Lipschitz on $\R$. Then, there is a constant $L > 0$ such that for all $x,y \in \R$, $|e^x - e^y| \leq L|x-y|$. But according to the mean value theorem, there is $c\in (x,y)$ such that $$ \frac{e^x - e^y}{x - y} = e^c $$ Therefore, if we raise both $x$ and $y$, it can be arbitrarily large and of course, it can be larger than any $L > 0$, which is a contradiction.