A narrower dichotomy for the quadratic variation of differentiable functions?

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

• 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 $[f] = \infty$? Jul 23 at 17:30
• @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. Jul 23 at 17:40

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:
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 $$\ln(2)$$.