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Is it true that the graph of a function $\varphi:\mathbb [0,1]\to\mathbb R$ which is discontinuous at each $x$, has lower box dimension strictly greater than one?

If not, what extra condition do we need? (Something about the "size" of its "jumps", possibly?)

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    $\begingroup$ The characteristic function of the rationals is discontinuous everywhere, but its graph is contained in the union of two line segments, so box dimension is 1. $\endgroup$
    – Vaughn Climenhaga
    Commented Dec 29, 2016 at 13:15
  • $\begingroup$ Indeed. So, we do need something extra to discard such obvious counterexamples. $\endgroup$
    – Nikita Sidorov
    Commented Dec 29, 2016 at 13:42
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    $\begingroup$ To generalize Vaughn's comment, let $f:[0,1]\to [\delta,\infty)$ be any function, and let $g$ agree with $f$ on the irrationals and be $0$ on the rationals. Then the lower box dimension of the graph of $g$ is smaller or equal than that of $f$, with equality if $f$ is continuous. So it seems to me that being discontinuous at all points, regardless of the size of jumps, in itself has no bearing on the box dimension of the graph. $\endgroup$
    – Pablo Shmerkin
    Commented Dec 29, 2016 at 14:25
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    $\begingroup$ To avoid this kind of constructions, one would need to assume something far stronger of the type: if $f|_D$ is continuous, then $D$ is nowhere dense. I have no intuition for what kind of functions satisfy this, but they might be wild enough for their graph to have large (even full?) box dimension. $\endgroup$
    – Pablo Shmerkin
    Commented Dec 29, 2016 at 14:34

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