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$\newcommand\P{\mathcal P}$A "partition" $P$ (of the interval $[0,1]$) is a finite sequence $(t_0,\dots,t_n)$ such that $0=t_0<\cdots<t_n=1$; then the mesh of $P$ is $\|P\|:=\max_{1\le j\le n}(t_j-t_{j-1})$.

Fix any sequence $\P:=(P_k)$ of "partitions" $P_k=(t_{k,0},\dots,t_{k,n_k})$ such that $\|P_k\|\to0$ (as $k\to\infty$).

For a real-valued function $f$ on $[0,1]$, define its quadratic variation (with respect to $\P$) by the formula $$[f]_\P:=\limsup_{k\to\infty}\sum_{j=1}^{n_k}(f(t_{k,j})-f(t_{k,j-1}))^2.$$

Suppose now that $f$ is differentiable and $[f]_\P<\infty$. Does it then necessarily follow that $[f]_\P=0$?

This question is a modification of the (now answered, positively) previous question A dichotomy for the quadratic variation of differentiable functions?. The difference is that now the sequence $\mathcal P$ is fixed.

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    $\begingroup$ Just to make sure I understand correctly: the difference is mainly that you are taking a fixed sequence $\mathcal{P}$ so that using the notation of your previous question, it may be that $[f] = \infty$? $\endgroup$ Jul 23 at 17:30
  • $\begingroup$ @WillieWong : Thank you for your comment. Yes, the difference is that now the sequence $\mathcal P$ is fixed. I have added this comment to the question. $\endgroup$ Jul 23 at 17:40
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Using your previous example of $f(x) = x^2 \cos(x^{-4})$. Note that on any interval $[\epsilon,1]$ the function is continuously differentiable, and hence has quadratic variation 0.

Construct $P_k$ so that the following points are contained in the partition:

  1. 0
  2. $\frac{1}{\sqrt[4]{\pi}} \ell^{-1/4}$ for natural $\ell$ between $k$ and $2k$.
  3. choose a partition of $[(k\pi)^{-1/4}, 1]$ such that the quadratic variation of $f$ on that interval is less than 1/10, and such that the points are no further than $1/k^{1/4}$ apart.

With this the sum relative to $P_k$ of $$ \sum (f(t_{k,n}) - f(t_{k,n-1}))^2 $$ evaluates to approximately $\ln(2)$.

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