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

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


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.

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Hint: Hölder inequality. – Did Nov 9 '12 at 6:17
@did Thanks. Do you have any tips on how to progress furtheR? – Jase Nov 9 '12 at 8:04
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 $\Pi$. – Did Nov 9 '12 at 10:26

This is more like a comment but there are a few typos in the above.

It should read

$\lim (\sum ((X_{t_i} - X_{t_{i-1}})^p)^a)^{1/a} + \lim (\sum ((X_{t_i} - X_{t_{i-1}})^{q-p})^b)^{1/b}$ if Holder's inequality were to be applied properly.

But if one were to set $ap=q$ then $b = \frac{q}{q-p}$ whereas if $b(q-p)=q$ then we would be in the same position as before. This value of $a$ and $b$ does not render the RHS to be in the form that Did mentioned in the above comment - the RHS has a factor of a power of the $q$-variation and a factor of a power of the $p$-variation.

Am I mistaken or was I not clever enough?

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