How to show that the natural logarithm is Lipschitz on $[\beta, \infty)$

Question

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?

Answer 1

The derivative, being $1/x$ is upper bounded by $1/\beta$ and also monotone decreasing on this domain.

Answer 2

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 $\ln$).

Answer 3

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.