I have faced the following problem, regarding to the Martingale Theory. Because this area far from my area I don't know whether this problem is in literature or this can be simple question for specialists of this area. If like this things in any literature can someone suggest me?

Let $$ (\mathbb{P}_{n}(t))_{n \in \mathbb{N}_{0}} = \left( \left\{ I^{n}_{j}(t),~ 1\leq j \leq q_{n}(t) \} \right\} \right)_{n \in \mathbb{N}_{0}} $$ be a sequence of partitions of $ [0,1)$ depending on parameter $t\in J\subset (0,1)$ such that $$ \underset{t\in J}{\sup}\underset{1\leq j\leq q_{n}(t)}{\max}|I^{n}_{j}(t)|\leq \lambda ^{n} $$ for some $\lambda \in (0,1)$ and for any $t\in J$ we have $q_{n}(t)\rightarrow \infty.$

Suppose that $ f $ is $ 1 $-periodic and that $ f \in {L^{p}}([0,1) $, where $ p > 1 $. Next, for any $t\in J,$ we define a sequence $ (G_{n}(\cdot, t): [0,1) \to \mathbb{C})_{n \in \mathbb{N}_{0}} $ of functions by $$ \forall n \in \mathbb{N}_{0}, ~ \forall x \in [0,1), ~ \forall j \in \{ 1,\ldots, q_{n}(t) \}: \\ {G_{n}}(x, t) \stackrel{\text{df}}{=} \frac{1}{\left| I^{n}_{j}(t) \right|} \int_{I^{n}_{j}(t)} f(s) ~ \mathrm{d}{s}, \quad \text{if $ x \in I^{n}_{j}(t) \in D_{n}(t) $}. $$ Assume that for any $t\in J$ the random value $G_{n}(\cdot, t)$ is a $L^{p}$ bounded martingale w.r.t $\mathbb{P}_{n}(t).$ It is well known (according to Doob's theorem) $$ G_{n}(\cdot, t)\rightarrow f, \,\,\,a.s. $$ and $$ E(|G_{n}(\cdot, t)-f|^{p})\rightarrow 0. $$

Question. Is it true $$ \underset{t\in J}{\sup}E(|G_{n}(\cdot, t)-f|^{p})\rightarrow 0? $$

Since the lengths of atoms of partitions uniformly tend to zero, I think it is true, but I do not know how to show it.

  • $\begingroup$ Does $E$ here simply mean integration over $[0,1]$ with Lebesgue measure? Intuitively, it seems to me that using martingale methods here is overkill; this looks like it may just need some clever calculus. For instance, if $f$ is continuous you get uniform convergence of $G_n$ to $f$, and the uniform distance between $G_n$ and $f$ depends only on the mesh size and the modulus of continuity of $f$. I don't see offhand how to do the $L^p$ case, though. $\endgroup$ – Nate Eldredge Nov 10 '14 at 15:14
  • $\begingroup$ @NateEldredge. $E$ is usual Lebesgue integral. Yes you are right, if the lengths of atoms of partition are same then it can be shown $$\underset{t\in J}{\sup}E(|G_{n}(\cdot, t)-f|^{p})\leq 2\underset{|h|\leq \lambda^{n}}{\sup}E(|f(x+ h)-f(x)|^{p})$$. But in the case of different lengths I have got some problem $\endgroup$ – Alex Nov 11 '14 at 0:53

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