Thread: Continous, unbounded function in metric spaces

1. Continous, unbounded function in metric spaces

Let $\displaystyle f:R \rightarrow R$ be a continuous, unbounded function. Show that there is a number $\displaystyle t_{0}$ for which $\displaystyle \{f(nt_{0}): n$ is an integer } is an unbounded set.

2. This is quite a difficult problem. I think the easiest way to attack it is to use a proof by contradiction.

So suppose that the result is false, and that for each real number t, the set $\displaystyle \{f(nt):n\in\mathbb{Z}\}$ is bounded. For each natural number m, let $\displaystyle K_m = \{t\in\mathbb{R}:|f_n(t)|\leqslant m \text{ for all }n\in\mathbb{Z}\}$. Then $\displaystyle K_m$ is closed (not too hard to prove), and $\displaystyle \bigcup_mK_m = \mathbb{R}$.

Now comes the hard part. Suppose that $\displaystyle K_m$ contains an interval $\displaystyle [t_1,t_2]$, where $\displaystyle t_1<t_2$. Then for all $\displaystyle s\in [t_1,t_2]$ and for all integers n, $\displaystyle |f(ns)|\leqslant m$. But as n increases, the intervals $\displaystyle [nt_1,nt_2]$ start to overlap. In fact, this will happen as soon as n is sufficiently large that $\displaystyle nt_2>(n+1)t_1$, or $\displaystyle n>\tfrac{t_1}{t_2-t_1}$. But that means that every sufficiently large real number x is in one of the intervals $\displaystyle [nt_1,nt_2]$ and so $\displaystyle |f(x)|\leqslant m$. It is easy to conclude from that that f is bounded. Since we are told that f is unbounded, it follows that $\displaystyle K_m$ cannot contain an interval. And since $\displaystyle K_m$ is closed, that means that $\displaystyle K_m$ is nowhere dense.

Thus $\displaystyle \{K_m: m\in\mathbb{N}\}$ is a countable collection of nowhere dense sets whose union is the whole of $\displaystyle \mathbb{R}$. But that is impossible by the Baire category theorem. That contradiction completes the proof.

Here is a very similar problem. Let f be a continuous function and suppose that for every t>0 the sequence f(t), f(2t), f(3t), ... tends to 0. Prove that f(x) tends to 0 as x→∞. The proof of that result also relies on the trick of using overlapping intervals and the Baire category theorem. There is a long and careful discussion of that problem in this entry in Tim Gowers's blog.