$p$-variation of a continuous local martingale

Question

Here are two interesting statements stated from a book which I do not know how to prove: Let $\{V^{n}_t \}$ be a sequence of processes defined by $$ V^{n}_t = \sum_{k=0}^{\lceil 2^n t \rceil -1} \big| X_{(k+1)2^{-n}} -X_{k2^{-n}} \big|^p, $$ where $p>1$, $X$ is a continuous local martingale in some filtered probability space, with $X_0=0$.

FACT $1$:

If $p>2$, then $V^n_t \rightarrow 0$ uniformly on compacts in probability.

FACT $2$:

If $1<p<2$ and $\limsup_{n \rightarrow \infty} V^n_t < +\infty$ for some $t \geq 0$, then $X$ is indistinguishable from $0$ on $[0,t]$.

Most standard textbooks only dwell on the case when $p=2$. I wonder whether anyone could help me understand these statements.

Answer 1

Recall that the following statements are equivalent:

1. $Y_n \to Y$ in probability
2. For any subsequence $(Y_{n(k)})_k$ of $(Y_n)_n$ there exists a subsequence which converges almost surely to $Y$.

Whenever confusion may arise, we write $V_t^n(p)$ for the $p$-variation of $(X_t)_{t \geq 0}$. For simplicity of notation, we consider the case that $(X_t)_{t \geq 0}$ is a continuous $L^2$-martingale. (Else: Use localizing techniques.)

Fact 1: Let $(V_t^{n(k)})_{k}$ be an arbitrary subsequence. For simplicity of notation, we relabel the subsequence and write again $(V_t^n)_{n}$. We know that $V_t^n(2) \to \langle X \rangle_t$ uniformly on compact sets in probability, therefore there exists a subsequence $V_t^{n(j)}$ such that $$\sup_{s \in [0,t]} |V_s^{n(j)}(2)-\langle X \rangle_s| \to 0 \qquad \text{as} \, \, j \to \infty. \tag{1}$$ Using

$$\sum_k |X_{(k+1) 2^{-n}} - X_{k 2^{-n}}|^p \leq \max_{k} |X_{(k+1) 2^{-n}}-X_{k 2^{-n}}|^{p-2} \sum_k |X_{(k+1)2^{-n}}-X_{k2^{-n}}|^2$$

and that the sample paths are uniformly continuous on $[0,t]$, we get

$$\begin{align*} V_s^{n(j)}(p) &\leq \max_{|u-v| \leq 2^{-n(j)},u,v \leq t} |X_u-X_v|^{p-2} V_s^{n(j)}(2) \\ &\to 0 \cdot \langle X \rangle_s = 0 \end{align*}$$

almost surely uniformly for $s \in [0,t]$. Using again the statement from the beginning of this answer, we conclude $V_t^n \to 0$ uniformly on compact sets.

Fact 2: Mind that the statement only holds true if $X_0=0$. (If e.g. $X=c$ for some constant $c$, then all assumptions are satisfied, but $X$ is indistuinguishable from $0$.)

As in the first proof, we choose a subsequence such that $(1)$ holds. Now

$$\sum_k |X_{(k+1)2^{-n}} -X_{k2^{-n}}|^{(2-p)+p} \leq \max_k |X_{(k+1)2^{-n}}-X_{k2^{-n}}|^{2-p} \sum_k |X_{(k+1)2^{-n}} -X_{k2^{-n}}|^p$$

implies

$$V_s^{n(j)}(2) \leq \max_{|u-v| \leq 2^{-n(j)},u,v \leq t} |X_u-X_v|^{2-p} V_s^{n(j)}(p) \tag{2}$$

for any $s \leq t$. Since $\limsup_{n \to \infty} V_s^n(p)< \infty$, the uniform continuity of the sample paths entails

$$\langle X \rangle_s = \limsup_{j \to \infty} V_s^{n(j)}(2) = 0.$$

Hence, $\mathbb{E}(X_s^2) = \mathbb{E}(X_0^2)=0$. Finally, the continuity of the sample paths implies

$$\sup_{s \in [0,t]} |X_s|=0$$

almost surely.