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Let $\Omega$ be an open, bounded, convex subset of $\mathbb R^n$. Given a Lipschitz function $f$ on $\Omega$, we denote by $D(f)$ the subset of $\Omega$ on which the classical derivative of $f$ exists.

Consider the set $\mathscr{D}$ of Lipschitz functions $f: \Omega \to \mathbb R$ whose derivative is everywhere discontinuous as a function on $D(f)$.

Question: Is $\mathscr{D}$ a residual subset of the set of Lipschitz functions under the Lipschitz norm?

Note: We use the Lipschitz norm

$$\|f\|_{\text{Lip}} :=\sup_{\Omega} |f| + \sup_{y \neq x}\frac{|f(y)–f(x)|}{|y–x|}·$$

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  • $\begingroup$ How is the Lipschitz norm defined? $\endgroup$ Commented 14 hours ago
  • $\begingroup$ @SaúlRM Ah, its $\sup |f| + \sup_{x \neq y}\frac{|f(y)–f(x)|}{|y–x|}$. $\endgroup$ Commented 13 hours ago

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Is seems $\mathscr{D}_\Omega$ contains a dense open subset of the family $\mathscr{L}_\Omega$ of Lipschitz functions $\Omega\to\mathbb{R}$.

For $\varepsilon>0$, let $\mathscr{L}_\Omega^\varepsilon$ be the set of functions $f\in\mathscr{L}_\Omega$ such that, for all open sets $B\subseteq\Omega$, there are two points $x,y\in B$ and $v\in\mathbb{R}^n\setminus\{0\}$ such that $x+v,y+v\in B$ and $$\|(f(x+v)-f(x))-(f(y+v)-f(y))\|>\varepsilon\|v\|.$$

Note that $\mathscr{L}_\Omega^\varepsilon\subseteq\mathscr{D}_\Omega$ for all $\varepsilon>0$, as any $f\in\mathscr{L}_\Omega^\varepsilon$ cannot satisfy for any open set $B\subseteq\Omega$ that, for all $x,y\in B$ such that $D_xf,D_yf$ are defined, we have $\|D_xf-D_yf\|_\infty<\frac{\varepsilon}{2}$.

Also, for all $0<\delta<\varepsilon$, $\mathscr{L}_\Omega^{\varepsilon-\delta}$ contains the $\frac{\delta}{10}$-neighborhood of $\mathscr{L}_\Omega^\varepsilon$ in the Lipschitz norm. Meaning that, if we prove that $\cup_{\varepsilon>0}\mathscr{L}_\Omega^\varepsilon$ is dense in $\mathscr{L}_\Omega$, we will conclude that $\mathscr{D}_\Omega$ contains a dense open subset of $\mathscr{L}_\Omega$.

So let $f\in \mathscr{L}_\Omega$ and $\varepsilon>0$. We want to find $\delta>0$ and a function $g\in\mathscr{L}_\Omega^\delta$ such that $\|g-f\|<\varepsilon$.

Claim: For any ball $B\subseteq\mathbb{R}^n$ of radius $r\leq1$, there is a Lipschitz function $h=h_B:\mathbb{R}^n\to\mathbb{R}$ such that $\|h\|_{\text{Lip}}\leq\frac{\varepsilon}{2}$, $h(x)=0$ for $x\not\in B$, and for all open subsets $B_1$ of $B$ there exist $x,y\in B_1$ such that $D_xh,D_yh$ are defined and $\|D_xh-D_yh\|_\infty=\varepsilon$.

Proof: We may assume $B$ is centered at $0$. Let $h(x)=r\frac{\varepsilon}{2}u\left(\left\|\frac{x}{r}\right\|\right)$, where $u$ is the pathological function defined in this MO answer. $\square$

Now, let $\delta=\varepsilon/10$, and let $\Omega_0$ be the open set of points $x_0\in\Omega$ such that for some nhood $U$ of $x_0$ we have that, for all $x,y\in U$ and $v\in\mathbb{R}^n\setminus\{0\}$ such that $x+h,y+h\in B$, $$\|(f(x+h)-f(x))-(f(y+h)-f(y))\|<\delta\|h\|.$$ So $\Omega_0$ is, intuitively, the set of points where $Df$ has no discontinuities with jump $>\delta$.

Now consider a countable family of disjoint balls $(B_n)$ of radii $r_n$ contained in $\Omega_0$ and dense in $\Omega_0$. We let $g=f+\sum_nh_{B_n}$; we then have $\|g-f\|=\left\|\sum_nh_{B_n}\right\|=\frac{\varepsilon}{2}$, and $g\in\mathscr{L}_\Omega^\delta$.

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  • $\begingroup$ I will revise/look for more typos tomorrow. If any part of the proof needs additional detail, please let me know. $\endgroup$ Commented 6 hours ago
  • $\begingroup$ Just got done skimming it, awesome that it had to use the pathological $\{0, 1\}$ valued gradient norm function to avoid inadvertently creating continuity points. Will check more carefully for subtle issues later, meanwhile I will also post my own approach which is more functional analytic. $\endgroup$ Commented 2 hours ago
  • $\begingroup$ Ah, it is more or less equivalent to yours… $\endgroup$ Commented 1 hour ago

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