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It is of course well-known that there are plenty of functions from $\mathbb R$ into itself which are continuous and nowhere differentiable. Although the Baire Category Theorem is enough to prove the existence (and genericity) of such functions, it is also possible (and instructive) to construct directly some examples: my favorite example is given in a paper by John McCarthy, An everywhere continuous nowhere differentiable function, Amer. Math. Monthly 60 (1953), 709 [MR 0057955]. The main reason for which I believe that the McCarthy example is the best is that it is possible to describe it completely in less than one page.

$\bullet$ Q1 $\bullet$ My first question is the following one: let $f$ be the function constructed in the reference above. Did anybody ever calculate the Hausdorff dimension of the graph of $f$, i.e. of the set $$ C_f=\{(x,f(x))\}_{x\in \mathbb R}\subset \mathbb R^2\quad? \tag{$\sharp$}$$ I would guess that $ \mathcal H^\alpha(C_f) $ is positive and finite for some $\alpha\in (1,2)$, but I do not know that $\alpha$. As a final remark, I would say that the intuitive descriptions of continuity, yielding sentences like "a function is continuous if you can draw its graph without lifting your pencil from the sheet" looks rather inappropriate in view of this $\mathcal H^1(C_f)=+\infty$.

$\bullet$ Q2 $\bullet$ I have a second question. Let $g$ be a continuous nowhere differentiable real-valued function defined on $\mathbb R$ and let $C_g$ be defined by ($\sharp$). Is it true that $ \mathcal H^1(C_g)=+\infty $?

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  • $\begingroup$ My guess is that McCafthy's example $f$ has Hausdorff dimension $2$. That would mean $\mathcal H^s(C_f) = +\infty$ for $0<s<2$. Also $\mathcal H^2(C_f) = 0$ (by Fubini). $\endgroup$ Commented Oct 24, 2023 at 12:18
  • $\begingroup$ Thanks, that means also that it is the most oscillatory situation. Is your guess grounded on some homogeneity property of some sort? Do you know if there are some examples such that there exists $\alpha$ as in my question Q1? $\endgroup$
    – Bazin
    Commented Oct 24, 2023 at 16:11
  • $\begingroup$ @Basin, I would guess :) the guess is based on the fact that McCarthy's curve is not $\alpha$-Hölder for any positive $\alpha$. Weierstrass' example is however Hölder continuous, so its graph has Hausdorff dimension strictly below 2. I don't know about its Hausdorff measure though. $\endgroup$
    – Kostya_I
    Commented Oct 24, 2023 at 16:45

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Answer to Q2 is yes.

If $\mathcal{H}^1(C_g)<\infty$, then the graph has finite length (it is known that for a one-to-one curve $\mathcal{H}^1$ coincides with the length). However that implies that $g$ has bounded variation (Theorem 97 in [H]) and functions of bounded variation are differentiable a.e.

[H] P. Hajłasz, Measure Theory

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