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The Question: Let $X$ be a continuous process, and suppose $0 < p < q$.

Prove the case $V_t^p(X) < \infty \implies V_t^q(X) = 0$.

Definitions:

The standard setup.

$\Pi := \{t_0,t_1,...,t_N\}$ where $0 = t_0 \leq t_1 \leq ... \leq t_N = t$.

$||\Pi|| := \text{max}_{1 \leq i \leq N}|t_i - t_{i-1}|$

$V_t^a(X) := \lim_{||\Pi||\to 0}\sum_{i=1}^N |X_{t_i} - X_{t_{i-1}}|^a$

My Current Progress:

$V_t^q(X) = \lim_{||\Pi||\to 0}\sum_{i=1}^N |X_{t_i} - X_{t_{i-1}}|^p |X_{t_i} - X_{t_{i-1}}|^{q-p}$.

By the Holder Inequality, for $\frac1a + \frac1b = 1$:

$$V_t^q(X) = \lim_{||\Pi|| \to 0}\sum_{i=1}^N|X_{t_i}-X_{t_{i-1}}|^p|X_{t_i}-X_{t_{i-1}}|^{q-p} $$ $$ V_t^q(X) \leq \lim_{||\Pi|| \to 0}\Big[\sum_{i=1}^N(|X_{t_i}-X_{t_{i-1}}|^p)^a\Big]^\frac1b \Big[\sum_{i=1}^N(|X_{t_i}-X_{t_{i-1}}|^{p-q})^b\Big]^\frac1a $$ $$ = \lim_{||\Pi|| \to 0}\Big[\sum_{i=1}^N(|X_{t_i}-X_{t_{i-1}}|^p)^a\Big]^\frac1b \lim_{||\Pi|| \to 0}\Big[\sum_{i=1}^N(|X_{t_i}-X_{t_{i-1}}|^{p-q})^b\Big]^\frac1a $$ My Request:

Please help me to progress further on this interesting problem.

  • 1
    Hint: Hölder inequality.2012-11-09
  • 0
    @did Thanks. Do you have any tips on how to progress furtheR?2012-11-09
  • 2
    Yes: choose your $a$ and $b$ such that a factor of the RHS is a power of the $q$-variation. This allows to bound $V_t^q(X,\Pi)$ by a power of $V_t^p(X,\Pi)$, for every $\Pi$. Then take the supremum over $\Pi$.2012-11-09
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    @Did Assuming $ \lim_{||\Pi|| \to 0}\Big[\sum_{i=1}^N(|X_{t_i}-X_{t_{i-1}}|^p)^a\Big]^\frac1a \lim_{||\Pi|| \to 0}\Big[\sum_{i=1}^N(|X_{t_i}-X_{t_{i-1}}|^{p-q})^b\Big]^\frac1b$ How would one choose $a, b$ for the desired result? As user157279 pointed out, choosing $ap=q$ will result in the exact same equation and choosing $(q-p)b=p$ will result in $ \lim_{||\Pi|| \to 0}\Big[\sum_{i=1}^N|X_{t_i}-X_{t_{i-1}}|^{\frac{p^2}{2p-q}}\Big]^\frac{2p-q}{p} \lim_{||\Pi|| \to 0}\Big[\sum_{i=1}^N|X_{t_i}-X_{t_{i-1}}|^{p}\Big]^\frac{q-p}{p}$2017-10-30
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    The right factor is finite by assumption but we can't say anything about the other factor, can we?2017-10-30

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