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

Assume

  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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Cross posted: mathoverflow.net/questions/122114/… –  Byron Schmuland Feb 18 '13 at 0:45
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1 Answer

up vote 1 down vote accepted

I will assume something stronger and remove the condition later: assume instead of $(2)$ that $P(\sup_n q_F(Y_n) < \epsilon)= 1$ for all $\epsilon > 0$ and a finite-dimensional subspace $F$. Let $F_k$ be the subspace obtained from the $\epsilon = 1/k$ challenge; it follows that $$ P(\sup_n q_{F_k}(Y_n) < 1/k) = 1 $$ by continuity from above. From now on we'll argue pointwise on this almost-sure set.

For each fixed $k$, let $Z_n^k \in F_k$ be chosen so that $d(Y_n,F_k) \geq \|Y_n - Z_n^k\| - 1/(2 k)$. Note that for each $k$, $\{Z_n^k\}$ is relatively compact by 1 and so possesses a convergent subsequence $n^{(k)}$. Moreover we can arrange things so that $n^{(k+1)} \subset n^{(k)}$ (by which I mean that the $(k+1)$-th sequence is a subsequence of the $k$-th. Take the diagonal subsequence of this array, which we'll denote by $\tilde{Y}_n$ (approximated on $F_k$ by $\tilde{Z}_n^k$ to within $1/(2k)$ by our construction).

Almost surely, $\{\tilde{Z}_n^k\}$ is Cauchy on $F_k$ by our construction. We have, for any $k$, $$ \|\tilde{Y}_n - \tilde{Y}_m\| \leq \|\tilde{Y}_n - \tilde{Z}_n^k\| + \|\tilde{Y}_m - \tilde{Z}_m^k\| + \|\tilde{Z}_n^k - \tilde{Z}_m^k\| \\ \Rightarrow \lim_{n \rightarrow \infty} \sup_{m \geq n} \|\tilde{Y}_n - \tilde{Y}_m\| \leq 1/k $$ and therefore $\{\tilde{Y}_n\}$ is a Cauchy sequence in $B$, hence relatively compact by the completeness of $B$.

I claim that my assumption incurs no loss of generality; indeed we can simply sample our subsequences along farther-out stretches of the sequence to obtain the same result.

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I dont get how continuity from above is used –  user24367 Feb 18 '13 at 8:58
    
To be more precise, we have a sequence of almost sure sets in response to the $\epsilon = 1/k$ challenge. What I meant is that their intersection is also almost sure, which I suppose "follows" from continuity from above. –  A Blumenthal Feb 18 '13 at 18:39
    
I didnot get why you can construct such sequences as $n^{(k+1)}\subset n^{(k)}$.....also the diagonalization argument... –  user24367 Feb 18 '13 at 23:19
    
Well we're certainly at liberty to select $n^{(1)}$ so that $\{Z_{n^{(1)}}^1\}$ is Cauchy. Then $n^{(1)}$ is infinite and so there is a further subsequence $n^{(2)}$ for which $Z_{n^{(2)}}^2$ is Cauchy. Iterating this process we get the desired nested subsequences. 'Taking the diagonal subsequence' means proscribing the subsequence $n^{(n)}$ (i.e. take the first one to be the first member of $n^{(1)}$, then the second to be the second member of $n^{(2)}$ and so on). –  A Blumenthal Feb 18 '13 at 23:34
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