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Let $B$ be a Banach space. Let $\{Y_{n}\}$ be a sequence of $B$ valued random variables.


  1. $P(\{Y_{n}\} \mbox{is bounded}) = 1$,

  2. fo every $\epsilon>0$, there exists a finite dimensional subspace $F$ such that $P(\limsup_{n} q_{F}(Y_{n})\leq \epsilon) = 1$.

Then show that $P(\{Y_{n}\} \mbox{ is relatively compact}) = 1$. Where $q_{F}(x) = d(x,F)$ is the distance of $x$ from $F$.

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its not HW....I came across this in the proof of Law of iterated logarithms for Banach Space valued RVS by Kuelbs....I know this would be easy to prove..... – user24367 Feb 18 '13 at 0:28
Which part can you not prove? What have you tried so far? See – Yemon Choi Feb 18 '13 at 2:03
Also, check your statement ... fix the conclusion, perhaps... – Gerald Edgar Feb 18 '13 at 14:26
up vote 0 down vote accepted

For $\omega\in\Omega$, define $S(\omega):=\(Y_n(\omega),n\in\mathbb N\)$. We have to show that $P(\omega\mid S(\omega)\mbox{ is relatively compact})=1$.

Considering $\varepsilon=1/j$, we can see that there is $\Omega'$ of probability $1$ such that for each $\omega\in \Omega'$,

  1. $\sup_n\lVert Y_n(\omega)\rVert<\infty$, and
  2. for each integer $j$, $\limsup_{n\to +\infty}d(Y_n(\omega),F_j)\leqslant \frac 1j$, where $F_j$ corresponds to the subspace for $\varepsilon=\frac 1j$.

We say that a set $S\subset B$ is flately concentrated if for each $ \varepsilon>0$, we can find a finite dimensional subspace $F$ of $B$ such that for all $x\in S$, $d(x,F)<\varepsilon$.

It's linked to relative compactness (using completeness) by the following:

A subset $S$ of the Banach space $B$ has a compact closure if and only $S$ is bounded and flately concentrated .

To conclude here, we have to check that $S(\omega)$ is flately concentrated for each $\omega\in\Omega'$. Argue by contradiction (the problem will come from infinitely many $n$).

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