2
$\begingroup$

Suppose $\Omega$ is a bounded domain in $\mathbb{R}^n.$ Let $w\in H^{1,2}$ (standard Sobolev space, order 1, integrability 2) and $L>0$ be given. Is it then true that the function $w_L:=\min (L,w)$ is also in $H^{1,2}?$

I found this assertion in the book "Riemannian geometry and geometric analysis" of Jost, in the section concerning higher regularity of harmonic maps. There, one already knows that $f$ (the continuous weakly harmonic map) is in $H_{loc}^{1,4} \cap H^{2,2}_{loc}$ and considers $w:=|Df|^2.$

Now, the function $w_L$ can also be written as $x\mapsto w(x)\chi_L(w(x)),$ where $\chi_L$ denotes the characteristic function of the set consisting of all x s.t. $x\le L$. Considering the distributional derivative of $w_L$ gives you something involving a delta function(al), which is not in $L^2.$

Any suggestions? What am I doing wrong?

Improve this question
$\endgroup$

2 Answers 2

Reset to default
7
$\begingroup$

Note that $\min(w,L) = \frac{1}{2} (w+L) - \frac{1}{2} |w-L|$, so the main issue is to establish that the map $w \mapsto |w|$ is bounded on $H^{1,2}$. But this follows from the diamagnetic inequality $|\nabla |w|| \leq |\nabla w|$ (in the sense of distributions), which is obvious formally, but can be established rigorously by approximating the absolute value $|x|$ by the smoothed variant $(|x|^2+\varepsilon^2)^{1/2}$ and then letting $\varepsilon$ go to zero. (The diamagnetic inequality can be found for instance in the text of Lieb and Loss; it is more commonly applied in the context of covariant differentiation, but already has usefully non-trivial content for ordinary differentiation.)

More generally, composition with Lipschitz functions will preserve all $W^{s,p}(\Omega)$ spaces for $0 \leq s \leq 1$ and $1 < p < \infty$ by the chain rule (if $s=1$) or fractional chain rule (if $s<1$), using regularisation arguments as necessary to make the argument rigorous. (See for instance Taylor's book "Tools for PDE" for this sort of thing.)

Improve this answer
$\endgroup$
7
  • $\begingroup$ I never knew that inequality has a name. Do you happen to know the reason behind calling it the "diamagnetic inequality"? $\endgroup$
    – Willie Wong
    Commented Apr 3, 2011 at 0:27
  • $\begingroup$ I think in differential geometry it is sometimes called something like Kato's inequality? $\endgroup$
    – Deane Yang
    Commented Apr 3, 2011 at 1:15
  • 2
    $\begingroup$ Hmm, there seem to be conflicting attributions in the literature. I've seen the name "Kato's inequality" both for $|\nabla |w|| \leq |\nabla w|$ and for the related inequality $\Delta |w| \geq \hbox{sgn}(w) \Delta w$. The name "diamagnetic inequality" usually refers to the generalisation $|\nabla|w|| \leq |(\nabla+iA)w|$, which among other things implies that the presence of a magnetic field always serves to increase the ground state energy of a Schrodinger operator. This presumably helps explain the physical phenomenon of diamagnetism, though I haven't looked at the details. $\endgroup$
    – Terry Tao
    Commented Apr 3, 2011 at 4:16
  • 1
    $\begingroup$ Related to the above two inequalities is also Simon's diamagnetic inequality $|e^{tH}w|≤|e^{tH_0}|w||$, where $H$ is a magnetic Schrodinger operator and $H_0$ is the unmagnetised version. $\endgroup$
    – Terry Tao
    Commented Apr 3, 2011 at 4:27
  • $\begingroup$ Excuse me, but isn't the expression you wrote the expression for the maximum of w and L, rather than the minimum? $\endgroup$
    – Orbicular
    Commented Apr 3, 2011 at 9:54
1
$\begingroup$

Your formula for $w_L$ is incorrect. It should be

$$ x\mapsto w(x) \chi_L(w(x)) + L [1-\chi_L(w(x))] $$

Note that your expression is equal to 0 where $w > L$, in contrast of your earlier description of the function. Now you can see that the "delta function" "cancels out". I'll leave it as an exercise on how to make the argument rigorous.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Yeah, right! Stupid me. I'd prove the statement for $C^1$ functions, s.t. $L$ is a regular value and then use an approximation argument. This should work, right? Or did you have something different in mind? $\endgroup$
    – Orbicular
    Commented Apr 2, 2011 at 21:38
  • $\begingroup$ That, or you can look at difference quotients. $\endgroup$
    – Willie Wong
    Commented Apr 2, 2011 at 22:52

You must log in to answer this question.

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