Is ${\log(x)}^2$ uniformly continuous on $(1, \infty)$?

Question

Is $f(x)={\log(x)}^2$ uniformly continuous on $(1,\infty)$?

I have found the above question in an old exam and was wondering how to solve it. Actually, a few friends and I have found 2 different solutions, one showing that the function is uniformly continuous, while the other showing that it isn't, and we can't figure out which one is wrong.

Here is the first one:

Let $\varepsilon = 1$ and $\delta>0$ be given. Let $x = \max\{\frac{\delta}{2},e\} > 1$, and $y = x + \frac{\delta}{2} > 1$. Then we have $$|x - y| = \left|x - \left(x + \frac{\delta}{2}\right)\right| = \frac{\delta}{2} < \delta$$ and \begin{align*} |f(x)-f(y)| &= \left|{\log(x)}^2 - {\log(y)}^2 \right| \\[0.5em] &= \left|(\log x + \log y)(\log x - \log y) \right| \\[0.5em] &\geq |(\log e)(\log x - \log y)| \\[0.5em] &= |\log x - \log y\:| \\[0.5em] &= \left| \log \left(x+\frac{\delta}{2}\right) - \log x\right| \\[0.5em] &= \left| \log \left( 1+ \frac{\delta}{2x} \right)\right| \\[0.5em] &\geq 1 \\ &= \varepsilon, \end{align*} so $f(x)$ is not uniformly continuous.

On the other hand, intuitively $f(x)$ "looks" uniformly continuous. I believe that since $f \in \mathcal{C}^1$ in $(1, \infty)$ and the derivative is bounded, $f(x)$ should also be uniformly continuous, but obviously not both can be true, so I must be missing something.

Answer 1

If $f : [a,\infty) \longrightarrow \mathbb R$ has bounded (not necessarily continuous) derivative on its domain, then it is uniformly continuous. Indeed, the mean value theorem tells you that for any $x < y$ you have that $$|f(x)-f(y)| = |f'(\xi)(x-y)| \leqslant M |x-y|$$ for some absolute constant depending on $f'$. In other words, your function is Lipschitz continuous, which is a strong form of uniform continuity.

Answer 2

You already pointed it out. $f'$ is bounded on $(1, \infty)$, whence $\lvert f(x) - f(y) \rvert \leq \sup_{\xi \in (1, \infty)}\lvert f'(\xi) \rvert \lvert x -y \rvert$ for all $x, y \in (1, \infty)$. This is uniform continuity (Lipschitz continuity even).

Concerning your first argument: The second line already is wrong. There is no way that $\tfrac{\delta}{2} >\delta$.
Moreover, it is logically flawed. To "disprove" uniform continuity, you need to show that there is $\varepsilon >0$ such that for every $\delta> 0$ there is a pair $x_\delta, y_\delta \in (1, \infty)$ such that $\lvert x_\delta - y_\delta \rvert < \delta$ and $\lvert f(x_\delta) - f(y_\delta)\rvert \geq \varepsilon$. The second line really blurs whether you are attempting to show this.

And as Martin R. correctly points out in the comments: For the proof to work, $\delta$ has to be allowed to become arbitrarily small. But then $\left \lvert \log\left( 1+ \tfrac{\delta}{2x} \right) \right \rvert$ approaches $0$.