4
$\begingroup$

Let $f \in W^{1, p} (\mathbb R^n)$. A classical approximation theorem (see for instance, the book by Evans and Gariepy) says that we can approximate $f$ by Lipschitz functions, in the sense that for every $\varepsilon > 0$, there exists some Lipschitz continuous $u_\varepsilon$ on $\mathbb R^n$ such that

$$\mu(\{f \neq u_\varepsilon\}) \leq \varepsilon.$$

Such a function $u_\varepsilon$ will be called an $\varepsilon$-Lipschitz approximation to $f$.

Motivated by this, we want to characterise membership in $W^{1, p}$ by the best Lipschitz constants of these approximations.

Given any measurable $f$, we define the $\varepsilon$-Lipschitz modulus of continuity $\omega_{\varepsilon} (f)$ by

$$\omega_{\varepsilon} f = \inf \{L \geq 0 \, | \, \text{There exists an }\varepsilon\text{-Lipschitz approximation to }f\text{ with Lipschitz constant }L\}$$

where by convention we set the infimum to be infinite if the set is empty.

Question: Let $1 \leq p \leq \infty$, and $f \in L^{1}_{\text{loc}} (\mathbb R^n)$. Is it true that $f \in W^{1, p}(\mathbb R^n)$ if and only if

$$\omega_{\varepsilon} f = O(\varepsilon^{-1/p})?$$

Here the $O$ notation is as $\varepsilon \to 0_+$, and we allow the implied constant to depend on $f$.

$\endgroup$
6
  • $\begingroup$ Hm, “only if” should be true by the Kirzbaum-Valentine theorem… $\endgroup$
    – Nate River
    Commented Oct 19 at 10:29
  • $\begingroup$ “If” direction seems much harder. $\endgroup$
    – Nate River
    Commented Oct 19 at 10:34
  • 4
    $\begingroup$ I think there is a 50% chance this is true. $\endgroup$
    – JJJ
    Commented Oct 27 at 2:07
  • $\begingroup$ The (concave) modulus of continuity $\omega(t)$ of a uniformly continuous function $f$ on a bounded convex domain $C$ can be reconstructed from the distance from the $k$-Lipschitz functions $\delta(k):=\inf_{g\in \text{Lip}_k(C)}\|f-g\|_{\infty,C}$ (they are related by a Legendre transform). Maybe there is an analogous fact for integral norms? $\endgroup$ Commented Nov 8 at 18:21
  • 1
    $\begingroup$ Something similar should be true. As a general principle, the quicker the approximation of a function by regular functions converges, the more regular is the function, and conversely $\endgroup$ Commented Nov 9 at 6:02

1 Answer 1

1
+250
$\begingroup$

The condition that $\omega_\epsilon f = O(\epsilon^{-1/p})$ does not imply that $f \in W^{1, p}$. It does imply that $\nabla f \in L^p_{weak}$, but we obtain a counterexample by looking for $\nabla f$ in $L^p_{weak} \setminus L^p$. Indeed if the domain is $[0, 1]$, we can use $f(x) = \frac{p}{p-1} x^{1-1/p}$. Since $f$ is $\epsilon^{-1/p}$-Lipschitz on $[\epsilon, 1]$, $\omega_\epsilon(f) \leq \epsilon^{-1/p}$ for all $\epsilon \in (0, 1)$. (When $p=1$, you can just use $\log$.)

If $f \in W^{1, p}$, then $\omega_\epsilon f = O(\epsilon^{-1/p})$, and this can be seen directly in the proof of theorem 6.14 in Evans-Gariepy. In particular, for every $\lambda > 0$, they construct a set $R^\lambda$, and prove that (1) $f|_{R^\lambda}$ is $C \lambda$-Lipschitz and (2) $|(R^\lambda)^c| = o(\lambda^{-p})$ as $\lambda \to \infty$.

$\endgroup$
1
  • $\begingroup$ Ah, so too good to be true! $\endgroup$
    – Nate River
    Commented Nov 9 at 3:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .