Here is the exercise that I got.

Verify that the class $\mathcal{G} = \left\{ g: \mathbb{R} \to [0,1], g \text{ is increasing}\right\}$ is not totally bounded for the supremum norm on $\mathbb{R}$.

I am trying to prove this by constructing a counterexample (for a given $\epsilon > 0$ and some functions $g_1,\cdots,g_n$), but I don't know how. The rough idea I have is that, to construct $f$ such that $\|f - g_i\|_\infty > \epsilon$ for all $g_i$'s. So maybe only at one point $x_i$, $f$ and $g_i$ are far away. For example, maybe at point $x_1$, $f$ is only close to $\max_i g_i$ and far away from $\min_i g_i$, but at another point $x_2$, $f$ is only close to $\min_i g_i$ but far away from $\max_i g_i$, but I don't know how to formalize this, partly because I don't know how far is $\max_i g_i$ from $\min_i g_i$.

The picture in my head is that, if $g_i$ is like a increasing straight line, then I can consider $f = \frac{1}{2}$, so $f$ and $g$ will be far away when $x$ is big or small. If $g$ is quite flat, I can take $f$ to be an increasing line.

I am not sure if I am thinking correctly and how to proceed. Could someone give me a hint?

New contributor
TJnFvYLDu3 is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
up vote 2 down vote accepted

The below will work for functions into $[-\pi/2,\pi/2]$. You would just need to shift and stretch a little for your case. Recall that $\arctan(x)$ is strictly increasing and ranges in $(-\pi/2,\pi/2)$. Replacing $x$ by $\alpha x$ for $\alpha>1$ does a horizontal compression. So for large $\alpha$, $\arctan(\alpha x)$ will approach its asymptotes faster despite still starting off at $(0,0)$.

The sequence $f_n(x) = \arctan(nx)$, $n\geq 1$ is contained in $\mathcal{G}$ but can't be contained in finitely many balls of radius $\frac{\pi}{6}$, $B(g,\frac{\pi}{6})$. If they were, then all $f_n$'s would be within distance $\frac{\pi}{3}$ of the set $\{f_{n_1},\ldots,f_{n_k}\}$ for some finite indices $n_1 < \ldots < n_k$. But (feel free to check) that if you let $n >> n_k$, then the distance will approach $\pi/2$.

Your Answer

TJnFvYLDu3 is a new contributor. Be nice, and check out our Code of Conduct.
 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.