I'm preparing for the functional analysis test and I have a doubt on $L^{p}$ spaces. Kolmogorov criterion characterizes precompactness in $L^{p}(\mathbb{R}^d )$ for any $p \neq +\infty$, and says the following:
$H\subseteq L^{p}(\mathbb{R}^{d})$ is precompact if and only if:
$H$ is bounded
$\lim\limits_{R\rightarrow 0}\hspace{2mm}\sup\limits_{f\in H}\int\limits_{|x|\geq R}|f(x)|^{p}dx=0$
$\lim\limits_{|a|\rightarrow0}\hspace{2mm}\sup\limits_{f\in H}\|T_{a}f-f\|_{p}=0$
Our professors told us that usually this criterion fails on the third point, but so far I did a lot of exercises and it always worked (up to mistakes). Since it can't always be like that, as any arbitrary collection of functions can't be precompact, I'm asking:
Can you give me an example where such criterion fails? Is there a general idea one should check to see "at first glance" when this criterion indeed fails?
One simple example of a collection which satisfies $1$ and $2$ but not $3$ is $B_\Omega$, the unit ball in $L^p(\Omega)$ for some bounded set $\Omega$, thought of as a subset of $L^p(\mathbb{R}^d)$. These are the functions in $L^p(\mathbb{R}^d)$ with norm at most $1$ and that are $0$ outside of $\Omega$. It is clear that $B_\Omega$ satisfies $1$ and $2$ but it is not compact as the unit ball in an infinite dimensional Banach space. Therefore $3$ cannot hold, which you may also verify directly by considering functions supported on very thin rectangles.
In general, given any collection of vectors it is usually relatively obvious if 1 and 2 hold and it is easy to see that a collection without one of these properties cannot be precompact. So, as I think about it, the real insight of the criterion is that among the collections which do not fail to be precompact in some "obvious" way, the precompact ones are exactly the ones satisfying 3.