Find a function that satisfies the condition.

Question

Let $\epsilon > 0$ be fixed and $t$ a variable that takes values in the universal covering space of ${\mathbb{C} \setminus \{0\}}$.

Find a continuous function $f(s$) such that $$|t \log t| = |t| \sqrt{(\ln |t|)^2 + (\arg t)^2} < \epsilon$$ whenever $|t| < f(\arg t)$.

Note: Basically, since $t$ is in the universal covering space of ${\mathbb{C} \setminus \{0\}}$, $\log {t}$ and $\arg {t}$ are well-defined functions and not multi-valued like they usually are.

Answer 1

New strategy: Replace $\lvert t \rvert$ and $\arg t$ by real variables $r>0$ and $\theta$, respectively. Define $g_\theta \colon \mathbb R_{>0} \to \mathbb R_{>0}$ by $g_{\theta}(r) = r^2((\ln r)^2 + \theta^2)$; note that this last expression is essentially the square of the expression we are interested in. Construct a continuous function $h \colon \mathbb R \to \mathbb R_{>0}$ so that $g_\theta(r) \leq h(\theta) \cdot r$ for all $r \leq 1$ and $\theta \in \mathbb R$.

Edit (2014-06-21): Since the asker has been inactive for almost a month now, I'm just going to write down the full solution and put it in the spoiler block below. Hover your cursor over to unhide the text.

It suffices to find a continuous function $f \colon \mathbb R \to \mathbb R_{>0}$ such that for $r>0$ and $\theta$ real, we have $$r^2\bigl((\ln r)^2 + \theta^2\bigr) < \epsilon^2$$ whenever $r < f(\theta)$. Note first that since $\lim_{r \to 0} r (\ln r)^2 = 0$, we can use compactness to argue that the function $r \mapsto r(\ln r)^2$ is bounded on $(0,1]$. Pick an upper bound $\alpha$ (one can use calculus to show that $\alpha = 4/e^2$ is the smallest possible choice). If we now define $f$ by $$f(\theta) = \min\left\{1, \frac{\epsilon^2}{\alpha + \theta^2}\right\},$$ then when $r < f(\theta)$, we obtain $$r^2\bigl((\ln r)^2 + \theta^2\bigr) = r\bigl(r(\ln r)^2 + r \theta^2\bigr) < r\bigl(\alpha + \theta^2\bigr) < \epsilon^2$$ as desired.

Answer 2

A good strategy I've learned throughout undergraduate mathematics courses is to always simplify, and reword. It usually helps see a problem from a different perspective. Going off this notion, I'll make a few points

1. $|t\log{t}| = |t||logt| = |t|\sqrt{ (ln|t|)^2} + (\arg t)^2)$ So essentially the important parts of this equality is that $|logt| = |t|\sqrt{ (ln|t|)}^2 + (\arg t)^2)$
2. Rewording the problem, your problem is finding f(s) so that

$|t| < f(arg t) \implies |\log(t)| = |t|\sqrt{ (ln|t|)^2} + (\arg t)^2)$

From here, I suggest maybe isolating $\arg t$ so that your new statement is simplified. I think this will provide more insight into solving the problem. Hope this helped, if even a little. And sorry about the poor formatting, still getting used to posting answers. Good luck!