How to show that the natural logarithm is Lipschitz on $[\beta, \infty)$ I want to show the following result:

Let $\ln(x)$ have domain $D = [\beta, \infty)$ then $|\ln(x) - \ln(y)|
 \leq \dfrac{1}{\beta} |x-y|, \forall x,y \in D$

I am confused as to how to prove this seemingly simple looking claim.
Can someone please help?
 A: The derivative, being $1/x$ is upper bounded by $1/\beta$ and also monotone decreasing on this domain.
A: Since we are dealing with the logarithm, we can use its properties. Assume without loss of generality that $\beta \leq x \leq y$:
$$
|\ln x - \ln y| = \ln\frac{y}{x} = \ln\!\left(1+\left(\frac{y}{x}-1\right)\right)
\leq \frac{y}{x}-1 = \frac{1}{x}\left(y-x\right) \leq \frac{1}{\beta}\left(y-x\right)
 = \frac{1}{\beta}|y-x|
$$
where we only used the inequality $\ln(1+u)\leq u$ for all $u > -1$. (Which is standard, and can be proven e.g. by concavity of $\ln$).
A: In addition to the above answers, one can also use the mean value theorem here. Since f(x) = ln(x) is concave, we have that it is monotonically increasing on $(0, \infty)$, and that $f''(x) < 0$. Therefore:
$$\begin{align}
\frac{\|log(x) - log(y)\|}{\|x-y\|} &\leq \sup_{x \in [\beta, \infty)}\frac{d}{dx}log(x) \\
&= \sup_{x \in [\beta, \infty)} \frac{1}{x} \\
\end{align}$$
Now since $f'(x)$ is monotonically decreasing on $[\beta, \infty$), we have that the supremum will be at $x=\beta$. Therefore we can conclude:
$$\frac{\|log(x) - log(y)\|}{\|x-y\|} \leq \frac{1}{\beta}$$
and the result follows.
