Is $\ln(x)$ uniformly continuous?

Question

Let $x\in[1,\infty)$. Is $\ln x$ uniformly continuous? I took this function to be continuous and wrote the following proof which I'm not entirely sure of.

Let $\varepsilon>0 $, $x,y\in[1, ∞)$ and $x>y$. Then, $\ln x< x$ and $\ln y< y$ and this follows that $0<|\ln x-\ln y|<|x-y|$ since $x> y$. Choose $δ=ϵ$. Now suppose $|x-y|< δ$. Then, $|\ln x-\ln y|<|x-y|<\varepsilon$

It would be much appreciated if someone could validate my proof

Answer 1

You can prove something more general:

PROP Suppose $f:[a,\infty)\to\Bbb R$ has bounded derivative. Then $f$ is uniformly continuous on its domain.

P Pick $x,y\in[a,\infty)$ arbitrarily. By the mean value theorem, we can write $$|f(x)-f(y)|=|f'(\xi)||x-y|$$

Let $M=\sup\limits_{x\in[a,\infty)}|f'(x)|$. Then $$|f(x)-f(y)|\leq M|x-y|$$

Thus, for any $\epsilon$ we may take $\delta=\frac{\epsilon}{2M}$. Note that in your case $M=1$. I only divide by $2$ to turn $\leq$ into $<$.

ADD This means, for example, that $\log x$ (over $[a,\infty)$, $a>0$), $\sin x$, $\cos x$, $x$, and similar functions are all uniformly continuous. Note, for example, that $\sin(x^2)$ is not uniformly continuous. Note that we actually prove $f$ is $1$-Lipschitz with constant $M$, so this might be of interest.

Answer 2

An easier argument is to note that the derivative of $\ln x$ is bounded by 1 on the interval $[1,\infty)$. Therefore $\ln x$ is Lipschitz and in particular uniformly continuous.

Answer 3

Alternatively, you can prove a function is uniformly continuous based off the following idea:

$f$ is uniformly continuous if and only if for any sequence $\{a_n\},\{b_n\}$

$$ \lim\left(a_n-b_n\right)=0 \Rightarrow \lim\left(f\circ a_n-f\circ b_n\right)=0. $$

Let $\{a_n\}, \{b_n\}$ satisfy our hypothesis ($\lim\left(a_n-b_n\right)=0$), then we have $\lim a_n = \lim b_n$ and so

$$ \lim\left(f\circ a_n-f\circ b_n\right) =\lim\left(\ln(a_n)-\ln(b_n)\right)=\lim\ln\left(\frac{a_n}{b_n}\right) = \ln(1) = 0. $$

Answer 4

No, your proof has a problem. If $f(x)<x$ for all $x\in[1,\infty)$, it does not follow that $|f(x)-f(y)|<|x-y|$ for all $x$ and $y$.

You have $x>y$, and using the fact that $\ln$ is increasing, $|\ln x -\ln y|=\ln x - \ln y$. But how do you conclude that this is less than $x-y=|x-y|$? We know that $\ln x<x$, which gives $\ln x - \ln y <x-\ln y$. But $\ln y<y$ applied to the last expression gives $x-\ln y>x-y$, which doesn't help. Replacing $\ln x$ with $x$ makes the expression bigger, while replacing $\ln y$ with $y$ makes the expression smaller. To ensure that the net result is bigger, you need to know that $x-\ln x > y- \ln y$. But this is just a rearrangement of the inequality that you want to prove.

In summary: The conclusion that $|\ln x -\ln y|\leq |x-y|$ for all $x,y\geq 1$ is true, but more is needed to show it. Some methods to complete the proof are given in the other answers.

Answer 5

Assume $x>y>1$, Then by triangle inequality and the fact that $y>1$:

$$\frac{x}{y} < \frac{|x-y|}{|y|} + 1 < |x-y|+1$$

Let $\epsilon > 0$, choose $\delta = e^{\epsilon}-1$, then we have

$$|\ln(x)- \ln(y)| = \ln(\frac{x}{y}) < \ln(|x-y|+1) < \ln(e^{\epsilon} - 1+1) = \epsilon$$