For a real-valued function $f$ on $[0,1]$, define its quadratic variation by the formula $$[f]:=\limsup\sum_{j=1}^n(f(t_j)-f(t_{j-1}))^2,$$ where the $\limsup$ is taken over all "partitions" $0=t_0<\cdots<t_n=1$ of $[0,1]$ as $\max_{1\le j\le n}(t_j-t_{j-1})\to0$.

If $f$ is continuously differentiable or, more generally, uniformly Hölder continuous, then, of course, $[f]=0$. On the other hand, if $f(x)=x^2\cos(1/x^4)$ for $x\ne0$ and $f(0)=0$, then $f$ is differentiable, but $[f]=\infty$.

Suppose now that $f$ is differentiable and $[f]<\infty$. Does it then necessarily follow that $$\sum_{j=1}^n\Big(f\Big(\frac jn\Big)-f\Big(\frac{j-1}n\Big)\Big)^2\to0$$ as $n\to\infty$ (or maybe even that $[f]=0$)?

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    $\begingroup$ Hopefully I didn’t calculate wrongly, but I think $f(x) := x^2 \sin(5\pi/2x^2)$ is differentiable but $[f] = 2\pi^2/6$. $\endgroup$ – Nate River Jul 17 at 1:25
  • $\begingroup$ @NateRiver : I had examples like this in mind. However, for such examples we have $[f]=0$, I think. I think you missed the condition that the mesh must go to $0$. If I am mistaken here, please detail your calculation in an answer. $\endgroup$ – Iosif Pinelis Jul 18 at 2:15
  • $\begingroup$ Okay, I will write it up later today! $\endgroup$ – Nate River Jul 18 at 2:50
  • $\begingroup$ Oh, actually you are right - the quadratic variation will be the tail end of $\sum 1/n^2$, which is $0$ as you say. Very interesting question... $\endgroup$ – Nate River Jul 18 at 3:00

The paper linked formulates quadratic variation in a measure-theoretic framework. The references therein may also be of interest. As a disclaimer, I did not read this paper very closely, nor is this a research area I am familiar with. I imagine being able to access measure-theoretic tools might offer some interesting approaches to solving this problem.

In the paper, the authors assign positive finite (Hausdorff) measures to functions of bounded quadratic variation. Their Theorem 20 proves that for every positive finite measure on $[0,1]$, there is a function of finite quadratic variation that generates this measure (and I believe has quadratic variation related to the variation norm of the measure. So a nonzero finite measure would correspond with a nonzero but finite quadratic variation function).

The paper: https://www.ams.org/journals/tran/2011-363-08/S0002-9947-2011-05209-8/S0002-9947-2011-05209-8.pdf



Abstract: To each function $f$ of bounded quadratic variation we associate a Hausdorff measure $\mu_f$ . We show that the map $f \mapsto \mu_f$ is locally Lipschitz and onto the positive cone of $\mathcal{M}[0,1]$. We use the measures $\{\mu_f : f \in V_2\}$ to determine the structure of the subspaces of $V_2^0$ which either contain $c_0$ or the 2 square stopping time space $S^2$.

Chapter 3 is of most interest I think.


So I think I totally misunderstood Theorem 22 of the paper (I have removed my incorrect interpretation). The theorem states that the set $X:= \{x \in [0,1]: f \text{ is differentiable at } x\}$ has measure 0 under the quadratic variation measure $\mu_f$ (assuming $f$ has finite quadratic variation to begin with). If $f$ is differentiable on $[0,1]$ then we have $X = [0,1]$ and therefore $\mu_f([0,1]) = 0$. This in particular implies the quadratic variation of $f$ is zero (since $\mu_f([0,1])$ equals the quadratic variation). I believe this is Corollary 23 of the paper, which actually proves the stronger claim that if a function of finite quadratic variation is continuous and has only countably many points of non-differentiability then the function has quadratic variation zero. So it looks like the answer is indeed affirmative.

  • $\begingroup$ Thank you for this informative answer. The paper you linked seems indeed highly relevant. I will have to think more about this. $\endgroup$ – Iosif Pinelis Jul 22 at 1:50
  • $\begingroup$ I have added an edit to my response. I believe corollary 23 of the referenced paper states that any continuous function of finite quadratic variation with only countably many nondifferentiable points has quadratic variation zero. If this is indeed what it says then the answer to this question is affirmative. $\endgroup$ – Lars van der Laan Jul 22 at 3:37
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    $\begingroup$ This is great news! $\endgroup$ – Iosif Pinelis Jul 22 at 3:48
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    $\begingroup$ This is a great find, to both OP and the answerer. $\endgroup$ – Nate River Jul 22 at 5:07

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