Metric space compactness

Question

Consider the space $C([0,1])$ equipped with the uniform norm. Find a sequence of functions $\{g_n\}$ in $C([0,1])$ so that $\overline{\{g_n\}}$ is compact, but $g_n$ does not converge uniformly.

I'm confused about $\overline{\{g_n\}}$ and how to prove its compactness

Thanks!

Would $|sin(kx)|$, $k \in \mathbb{N}$ fit in your problem? – Sahil Silare Mar 24 '19 at 04:42 — Sahil Silare, Mar 24 '19 at 04:42

score 1 · Answer 1 · answered Mar 24 '19 at 05:00

1

$g_n\equiv 1$ for $n$ even and $g_n\equiv 0$ for $n$ odd.

answered Mar 24 '19 at 05:00

Kavi Rama Murthy

311,013

Boring yet effective.. – Henno Brandsma Mar 24 '19 at 06:59
@HennoBrandsma Is $g_n(x)=exp(exp(sin(cos(artan(exp(x))))))$ for $n$ even and $g_n(x)=x^{216}-34x^675+130$ for $n$ odd less boring? -:) – Kavi Rama Murthy Mar 24 '19 at 11:26
then you’d need more of an argument for continuity. – Henno Brandsma Mar 24 '19 at 11:37

Gary Moon · Answer 2 · 2019-03-24T05:45:39.677

To prove that for some given $g_n$, $\overline{\{g_n\}}$ is compact, we can show that the $g_n$ are uniformly bounded and equicontinuous. Arzela-Ascoli then implies that $\{g_n\}$ is precompact in $C([0,1])$, which gives what you want (i.e., compactness of the closure). So, the work will be in finding a sequence of such $g_n$ which do not converge.

Edit: I suppose I used "work" rather loosely above. Combine the given argument with the choice of $g_n$'s suggested and that should do it.

Metric space compactness

2 Answers2