Pointwise Hölder continuity of order $1-\varepsilon$

Question

For $\varepsilon > 0$, we say a function $f: \mathbb R^n \to \mathbb R^m$ is (pointwise) $(1 - \varepsilon)$-Hölder continuous at $x \in \mathbb R^n$ if

$$ \lim_{ y \to 0} \frac{f(x + y) - f(x)}{\lVert y \rVert^{1 - \varepsilon}} = 0.$$

Question: Suppose $f$ is $(1 - \varepsilon)$-Hölder continuous at all $x \in \mathbb R^n$ for every $0 < \varepsilon < 1$. Is it true that $f$ is differentiable everywhere?

Answer 1

One counter example is the Takagi function $\tau$, see Theorems 8.1 and 8.2 in the survey [2] (As noted yesterday at [4]).

More detail: Takagi [1] showed that his function $\tau$ is nowhere differentiable. On the other hand, It is easy to verify that for $0 \le x <x+h\le 1$, we have $$|\tau(x+h)-\tau(x)| \le 2 h\log_2(h)\,,$$ so $\tau$ is Holder continuous of order $\alpha$ for all $\alpha \in (0,1)$. See Theorem 8.1 page 15 in [2] (For refinements, see Kono [3].)

[1] T. Takagi, A simple example of the continuous function without derivative, from Proceedings of the Physico-Mathematical Society of Japan, ser II, Vol 1. 1903, pp 176-177. [Collected Papers of Teiji Takagi (S. Iyanaga, Ed), Springer Verlag, New York 1990].

[2] https://arxiv.org/pdf/1112.4205.pdf

[3] N. Kono, On generalized Takagi functions, Acta Math. Hung. 49 (1987), No 3-4, 315–324.

[4] https://www.facebook.com/groups/1923323131245618/posts/3414086925502557/?comment_id=3414118705499379&reply_comment_id=3414119175499332&notif_id=1647334661196681&notif_t=group_comment&ref=notif )