From a complete space, how do we find an uniformly convergent subsequence?

Question

I'm reading a proof of the Lévy-Ciesielski construction of the Brownian Motion, from the book by "Brownian Motion: An Introduction To Stochastic Processes" by Schilling. There, the authors state a similar reasoning to the following:

We have a sequence of continuous (in $t$) functions $\{W_n(t)\}$, such that $W_n(t,\omega):=\sum^{n-1}_{i=0} G_i(\omega)S_i(t)$, where $\omega$ is an outcome, $G_i\sim^{iid} N(0,1)$ and $S_i(t)$ is the $i$-th Schauder function at $t$, i.e.

With $N>n$ both in $\mathbb{N}$:

Why is that? What theorem are the authors using?

Edit: So I got a hold of a 2nd edition, and it seems that instead of lim inf, they use lim sup. So, I think we now have a Cauchy sequence, that in a complete space converges... Can we work something out from here? Yes, read the comment section in this question, specially Saz comments.

Answer 1

This is not an answer for your original problem. After reading the book you mentioned above, I can explain why $\{W_{2^j}(w,t)\}$ has a subsequence that uniformly converges.

Our task is to show there exists $\Omega_0$ with $\mathbb P(\Omega_0)=1$ such that for any $w\in\Omega_0$, the sample paths $\{W_{2^j}(w,t)\}$ has a subsequence that uniformly converges.

For clarity, we let $u_n(w,t):=W_{2^n}(w,t)$ and $X_{n,N}(w):=\sup_{t\in[0,1]}|u_n(w,t))-u_N(w,t)|$ . On Page 25 it has been shown that $$||X_{n,N}||_{L^4}\le3^{1/4}\sum_{j=n+1}^N2^{-j/4}\le 2^{-n/4}C,$$ where $C$ is an absolute constant. Let $Y_n=\liminf_{N\rightarrow\infty}X_{n,N}$. Then by Fatou's Lemma, we have that $$\mathbb{E}|Y_n|^4\le\liminf_{N\rightarrow\infty}\mathbb{E}|X_{n,N}|^4\le 2^{-n}C^4.$$ By Markov's inequality, we have that $$\mathbb{P}(Y_n>2^{-n/8})\le\mathbb{E}|Y_n|^4/2^{-n/2}\le2^{-n/2}C^4.$$ Let $A_n$ be the event that $Y_n>2^{-n/8}$. We see that $\sum_{n\ge1}\mathbb P(A_n)\le\sum_{n\ge1}2^{-n/2}C^4<\infty$. So by Borel-Cantelli Lemma, we can conclude that $$\mathbb P(\bigcap_{k\ge1}\bigcup_{n\ge k}A_n)=0$$ Let $\Omega_0=(\bigcap_{k\ge1}\bigcup_{n\ge k}A_n)^c=\bigcup_{k\ge1}\bigcap_{n\ge k}A_n^c$. Then $\mathbb P(\Omega_0)=1$.

Next we will show for any $w\in\Omega_0$, the sample paths $\{W_{2^j}(w,t)\}$ has a subsequence that uniformly converges.

From the definition of $\Omega_0$, we know that for any $w\in\Omega_0$, there exists $K=K(w)$ such that $Y_n(w)\le2^{-n/8}$ for all $n\ge K$. That is, \begin{equation} \liminf_{N\rightarrow\infty}X_{n,N}(w)\le2^{-n/8},\mathrm{for ~all~} n\ge K.~~~~~~~~(*) \end{equation} Let $n_1=\max\{1,K\}$, then $(*)$ tells us that there exists $n_2\ge\max\{2,K,n_1+1\}$ such that $$X_{n_1,n_2}(w)<2\cdot2^{-n_1/8}\le 2\cdot2^{-1/8}.$$ Iteratively, for every $j\ge3$, by $(*)$ there exists $n_j\ge\max\{j,K,n_{j-1}+1\}$ such that $$X_{n_{j-1},n_j}(w)<2\cdot2^{-n_{j-1}/8}\le 2\cdot2^{-(j-1)/8}.$$ In other words, we find an increasing integer-valued sequence $n_1,n_2,\ldots$ satisfying that for all $j\ge2$, $$\sup_{t\in[0,1]}|u_{n_{j-1}}(w,t)-u_{n_j}(w,t)|\le2\cdot2^{-(j-1)/8}.$$ This could imply that $\{u_{n_j}(w,t)\}$ uniformly converges in $t\in[0,1]$.

Answer 2

My guess is that the first edition meant something like "for all sufficiently large $N$, there exist arbitrarily large $n$ such that $\sup_t |\dots| < \epsilon$" and the liminf was a sloppy way to write it. Formally, something like: $$\forall \epsilon > 0\, \exists N'\, \forall N \ge N'\, \forall n' \ge N\, \exists n \ge n' \sup_{t \in [0,1]} |W_{2^N}(t,\omega) - W_{2^n}(t,\omega)| < \epsilon.$$ Then you could find a subsequence $n_k$ such that $|W_{2^{n_k}}(t,\omega) - W_{2^{n_{k+1}}}(t,\omega)| < 2^{-k}$ and it would follow that this subsequence was uniformly Cauchy, hence (by completeness) converges uniformly.

For the second edition, the authors probably realized that they could say something a little stronger (basically a limsup instead of liminf) and thus avoid the need for a subsequence.