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Let $\Omega$ be the open unit ball in $\mathbb R^n$. Consider the set $\mathcal D$ of continuous functions $f:\Omega \to \mathbb R$ that are differentiable a.e, and with $|\nabla f| \leq 1$ wherever $f$ is differentiable.

Question: What is the value of

$$\frac{1}{\mu(\Omega)} \sup_{f \in \mathcal D} \inf_{g \in C^{\infty}(\Omega)} \int_{\Omega} |\nabla f - \nabla g| d\mu?$$

Remarks: The value of the expression is between $0$ to $1$ inclusive, as can be seen by taking $g$ to be a constant function. The “most likely” values to me are either $0$ or $1$, though I would be pleasantly surprised if it took an intermediate value!

Note: Here $\mu$ denotes the standard Lebesgue measure on $\Omega$.

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    $\begingroup$ My first idea would be using the devil's staircase. If you would have required $f$ and $g$ to have the same boundary values, then you could actually have an arbitrarily large value by using multiples of that. But the way you phrase it, e.g. in 2d you would have to alternate between staircases going up and ramps going down along a circle to achieve something that cannot be approximated by gradients of smooth functions. So while the answer is certainly not 0, I am not sure how close to 1 it can be easily pushed. $\endgroup$
    – mlk
    Commented Aug 30, 2021 at 12:39
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    $\begingroup$ When $n=1$ the answer is certainly 0. Indeed in this case $\nabla f=f'$ a. e., and by your condition $|f'|\leq 1$ a. e., so $f'\in L^1$, but $C_0^\infty$ is dense in $L^1$. $\endgroup$
    – Alexandre Eremenko
    Commented Aug 30, 2021 at 13:07
  • $\begingroup$ @Alexandre Eremenko Yes agreed there! Very nice. $\endgroup$
    – Nate River
    Commented Aug 30, 2021 at 13:12

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