Is it true that, $|e^{x}-e^{y}|\leq C \cdot |x-y|$?

Question

Define $f:\mathbb R \to \mathbb R$ such that $f(x)= e^{x}-1:= \sum_{n=1}^{\infty} \frac{x^{n}}{n!};$ for $x\in \mathbb R.$

My Question: Can we expect $|f(x)-f(y)|\leq |x-y| \cdot C;$ where $C$ is constant, $x, y \in \mathbb R$; if yes what can we say about $C$ ?

Ruff Attempt: $$f(x)-f(y)= \sum_{n=1}^{\infty} \frac{1}{n!} (x^{n}-y^{n})= \sum_{n=1}^{\infty}\frac{1}{n!} (x-y)(x^{n-1}+x^{n-2}y+...+y^{n-1})= (x-y)\sum_{n=1}^{\infty} \frac{1}{n!}(x^{n-1}+x^{n-2}y+...+y^{n-1}) ;$$

I am not sure what I have done so far is legitimate; and also I have question; Is $$\sum_{n=1}^{\infty}\frac{1}{n!} (x^{n-1}+x^{n-2}y+...+y^{n-1})$$ is converges ? Please correct me if I have done some thing wrong here;

Answer 1

No, there is no such $C$. Assume that there is and take $y = 0$. Then $|e^x - 1| \le C|x|$ for all $x$ which would for example imply that $e^x/x^2 \to 0$ as $x \to \infty$ and that is false.

On the other hand, if $x$ and $y$ are restricted to a compact set, then the mean value theorem implies the existence of such a $C$.

In other words, $f$ is locally Lipschitz, but not Lipschitz on $\mathbb{R}$.

Answer 2

WLOG: $x<y$.

As you probably know, for a continuously differentiable function, you have $$f(x) - f(y) = f'(\zeta)(x-y)$$ for some value $\zeta\in[x,y]$.

This means, of course, that $|e^x - e^y| = |e^\zeta||x-y|$ for some $\zeta\in[x,y]$. Since this means that $e^\zeta \geq e^x$, this also means that

$$|e^x - e^y| \geq e^x |x-y|,$$ and this means that the constant $C$ you are looking for does not exist.