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Does there exist a real valued function on $[0, 1]$ that is differentiable everywhere, but for every $\alpha > 0$ is nowhere locally $\alpha$-Hölder continuous? That is, it is not $\alpha$-Hölder on any open subinterval.

I hope I am not overlooking something elementary, but to my utmost surprise I actually think this might be true…

The answer by user479223 here shows that such a function can be differentiable almost everywhere, in fact even increasing and continuous.

PS: Wouldn't it be funny if there was also a function that was Holder continuous of every order less than 1, but nowhere differentiable…

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    $\begingroup$ There is something that is buzzing in my head and says:- "Dini continuity is the key concept here", but I cannot figure out if this is true or it is just a buzz. $\endgroup$
    – Daniele Tampieri
    Commented Jun 8 at 5:21

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No, such a function does not exist. In general, there is no differentiable function $f:[0,1]\to\mathbb{R}$ with derivative unbounded in any open interval. If that was possible, we could construct by recursion two sequences of points $x_n<y_n$, for $n\in\mathbb{N}$, such that

  1. $[x_{n+1},y_{n+1}]\subseteq(x_n,y_n)$.
  2. $\lim_n\frac{|f(y_n)-f(x_n)|}{|y_n-x_n|}=\infty$.

Now let $p$ be the limit of the sequences $(x_n)_n$ and $(y_n)_n$. Note that $x_n<p<y_n$ for all $n$, and for all $n$, $\frac{|f(y_n)-f(x_n)|}{|y_n-x_n|}\leq\max\left(\frac{|f(p)-f(x_n)|}{|p-x_n|},\frac{|f(y_n)-f(p)|}{|y_n-p|}\right)$ (proof by picture). So by condition 2 above, $f$ cannot be differentiable at $p$.

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    $\begingroup$ Hmm, doesn’t $x^2 \sin (1/x^3)$ have unbounded derivative near 0? $\endgroup$
    – Nate River
    Commented Jun 8 at 14:19
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    $\begingroup$ Indeed, $x^2\sin(1/x^3)$ has unbounded derivative near $0$. In case that is the confusion, by "locally unbounded" I mean "unbounded in every open interval" $\endgroup$
    – Saúl RM
    Commented Jun 8 at 14:19
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    $\begingroup$ Ahh understood. $\endgroup$
    – Nate River
    Commented Jun 8 at 14:29
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    $\begingroup$ So the set of points $x\in[0,1]$ such that $f’$ is not locally bounded at $x$ is a closed set with empty interior, and in fact any closed set with empty interior $C\subset[0,1]$ is the set of points $x\in[0,1]$ such that $f’$ is not locally bounded at $x$ for some everywhere differentiable $f:[01,\to\mathbb R$ $\endgroup$
    – Pietro Majer
    Commented Jun 8 at 20:12
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    $\begingroup$ @PietroMajer yes, that is a nice statement. I guess you can obtain such $f$ by introducing one small copy of the function $x^2\sin(1/x^3)$ in each of the extremes of the (countably many) gaps of the set $C$ $\endgroup$
    – Saúl RM
    Commented Jun 8 at 20:24

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