Prove that $\ln(x)$ is not uniformly continuous on $(0, \infty)$
I have written a proof of this but I am not sure whether or not it's formally correct.
Assume that this function is uniformly continuous. Let's put $\epsilon = 1$. Now, choose $x$ and $y$ so, that $|x-y| < \delta$. Now, set $x= \delta$ and $y = \frac{\delta}{1000}$. Now, by definition of uniform continuity:
$$|\ln(x) – \ln(y)| < 1 \iff |\ln(\frac{x}{y})| < 1 \iff \ln(1000) < 1$$
Contradiction.
Could you tell me whether the method I used is correct?
Best Answer
You're not quite there yet, be a bit more precise:
Suppose $f(x) = \ln(x)$ is uniformly continuous. Then, taking $\varepsilon=1$ there must be some $\delta > 0$ such that
$$\forall x,y \in (0,\infty): |x-y| < \delta \rightarrow |f(x) - f(y)| = |\ln(\frac{x}{y})| < 1$$
Then define $x = \delta>0$ and $y = \frac{\delta}{100}>0$ and note that this obeys $|x-y| = \delta\frac{99}{100} < \delta$ while $\frac{x}{y} = \frac{\delta}{\frac{\delta}{100}} = 100$ and $\ln(100)> 1$ contradiction, so no such $\delta$ can exist.