I was asked to produce a bounded sequence in $L^p[0,1]$ without a convergent subsequence. Here is what I came up with:
Define
- $f_1$ to be $1$ on $[0,1/2)$ and $0$ otherwise
- $f_2$ to be $1$ on $[0,1/4)$, $[1/2,1/2+1/4)$ and $0$ otherwise
- $f_3$ to be $1$ on $[0,1/8)$,$[1/4,1/4+1/8)$,$[1/2,1/2+1/8)$, $[1/2+1/4,1/2+1/4+1/8)$ and $0$ otherwise
and so on. Since, for $m\neq n$ we have $1/2=||f_n-f_{m}||_p$ (if $p=\infty$, this difference is just $1$), we conclude that no convergent subsequence can exist.
The downside of this, if correct, is that it is quite tedious to write down $f_n$ explicitly, and show that $||f_n-f_m||_p=1/2$ or $1$.
Do you have any other more explicit, simpler examples?
Best Answer
Here's an example that's easy to write down: For $n=1,2,\dots$ define $I_n$ to be the interval $(1/(n+1),1/n),$ and set
$$f_n = \mathbb 1_{I_n}\cdot (n(n+1))^{1/p}.$$
Then for $m\ne n,\,\|f_m-f_n\|_p=2^{1/p}.$