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It is known that if a real continuous function $f(x)$ satisfies a local $\alpha$-Hölder condition on a closed interval $[a,b]$, the box dimension of the graph of $f(x)$ on $[a,b]$ will be not greater than $2-\alpha$.

But if the proposition above is taken inversely, that is, if the box dimension of the graph of a real continuous function $f(x)$ on $[a,b]$ will be not greater than $2-\alpha$, how much can we guarantee that $f(x)$ satisfies a local $\alpha-$Hölder condition on a closed interval $[a,b]$?

For instance, the box dimension of the graph of $\sqrt{x}$ is one on $[0,1]$, and $\sqrt{x}$ satisfies a local $1$-Hölder condition on $[a,1]$ given any small positive $a$. Here we may say that $\sqrt{x}$ does not satisfy a local $1$-Hölder condition only at $0$, and one point of exception may not affect the value of the box dimension. So how much is such exception allowed that it does not affect the value of the box dimension?

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The box dimension is not countably stable, so even one exceptional point could drive it up. For example, given $\epsilon>0$, the function $f(x)=x^{1/2} \sin(1/x)$ on $[0,1]$, with $f(0)=0$, is locally Lipschitz on $(0,1]$, yet the graph of $f$ has box dimension at least $1.25$. Indeed, for $j<k$, to cover the graph above $[1/(2\pi(k+j)), 1/(2\pi(k+j+1))]$, at least $ck^{3/2}$ squares of side length $1/(4k^2)$ are needed. Summing over $j<k$ we get $c_1 k^{5/2}$ squares of side length $1/(4k^2)$.

You can get a result of the type you want by working with the Hausdorff or packing dimension of the graph, as these notions are stable under countable union.

In order to prove the packing dimension of the graph of $f$ is at most $2-\alpha$, it suffices to assume that the exceptional set (where the local $\alpha$-Hölder condition fails to hold) has packing dimension at most $1-\alpha$. [This condition is sharp, unless you make further assumptions on the behavior of $f$ on the exceptional set.]

To see this, partition the graph of $f$ to the part over the exceptional set (that clearly has packing dimension at most $2-\alpha$, and for each integer $m$, the part $A_m$ where the $\alpha$-Hölder condition holds with constant $m$. The usual argument you referred to shows that for each $m$, the box dimension of $A_m$ is at most $2-\alpha$. Then use the connection between packing dimension and Box dimension (see, e.g. [1] [2] or [3]) to conclude.

[1] Falconer, Kenneth. Fractal geometry: mathematical foundations and applications. John Wiley & Sons, 2004.

[2] Mattila, Pertti. Geometry of sets and measures in Euclidean spaces: fractals and rectifiability. No. 44. Cambridge university press, 1999.

[3] Bishop, Christopher J., and Yuval Peres. Fractals in probability and analysis. Vol. 162. Cambridge University Press, 2017.

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  • $\begingroup$ Thank you. But how may I prove this proposition? $\endgroup$
    – Watheophy
    Oct 20, 2021 at 8:32
  • $\begingroup$ I corrected my answer and included a few details. If this is unclear, I can add more. $\endgroup$ Oct 21, 2021 at 19:17
  • $\begingroup$ How may I prove the packing dimension of the exceptional set must be at most $1-\alpha$? Thanks. $\endgroup$
    – Watheophy
    Nov 4, 2021 at 9:18
  • $\begingroup$ I did not say it must be at most $1-\alpha$. I just wrote that it suffices. $\endgroup$ Nov 7, 2021 at 0:58

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