LEMMA Suppose $g \in L_{\text {loc }}^{1}\left(\mathbb{R}^{n}\right)$ is nonnegative. Then for any $u \in C(\bar{\Omega}) \cap$ $C^{2}(\Omega)$ there holds $$ \int_{\tilde{M}_{(0)}} g \leq \int_{\Gamma^{+}} g(D u)\left|\operatorname{det} D^{2} u\right| $$

where $\Gamma^{+}$is the upper contact set of $u$ ( that is $\Gamma^{+}=\{y \in \Omega: u(x) \leq u(y)+D u(y) \cdot(x-y) \text { for any } x \in \Omega\} \text {. }$) and $\tilde{M}=\left(\sup _{\Omega} u-\sup _{\partial \Omega} u^{+}\right) / d$ with $d=\operatorname{diam}(\Omega)$

PROOF OF LEMMA : Without loss of generality we assume $u \leq 0$ on $\partial \Omega$(this is because that the relation it's preserved by the translation.). Set $\Omega^{+}=\{u>0\}$. By the area formula for $D u$ in $\Gamma^{+} \cap \Omega^{+} \subset \Omega$, we have $$ \int_{D u\left(\Gamma^{+} \cap \Omega^{+}\right)} g \leq \int_{\Gamma^{+} \cap \Omega^{+}} g(D u)\left|\operatorname{det}\left(D^{2} u\right)\right|, \tag{1} $$ where $\left|\operatorname{det}\left(D^{2} u\right)\right|$ is the Jacobian of the map $D u: \Omega \rightarrow \mathbb{R}^{n}$. In fact, we may consider $\chi_{\varepsilon}=D u-\varepsilon$ Id $: \Omega \rightarrow \mathbb{R}^{n}$. Then $D \chi_{\varepsilon}=D^{2} u-\varepsilon I$, which is negative definite in $\Gamma^{+}$. Hence by the change-of-variable formula we have $$ \int_{\chi_{\varepsilon}\left(\Gamma^{+} \cap \Omega^{+}\right)} g=\int_{\Gamma^{+} \cap \Omega^{+}} g\left(\chi_{\varepsilon}\right)\left|\operatorname{det}\left(D^{2} u-\varepsilon I\right)\right|, \tag{2} $$ which implies (1) if we let $\varepsilon \rightarrow 0$. ...

I can't work out the proof of this lemma, there are two question:

  1. why the change of variable formula holds? I found a post here.However it's unclear $\Gamma^+ \cap \Omega^+$ is convex or not?
  2. why after taking the limit the equality (2) becomes inequality (1)?

1 Answer 1


I'll try to speak to your stated questions 1 and 2, assuming you are okay with $D\chi_{\epsilon}$ being negative definite in $\Gamma^{+}$, I've also taken the liberty to include a calculation at the end of this answer trying to justify this.

$\textbf{For question 1}$:

I believe we can conclude $\chi_{\epsilon}$ is injective in our domain of interest as follows:

Suppose that $y_1,y_2 \in \Gamma^{+}$ are such that $\chi_{\epsilon}(y_1) = \chi_{\epsilon}(y_2)$, which is equivalent to saying that $Du \restriction_{y_1} - \epsilon y_1 = Du \restriction_{y_2} - \epsilon y_2$.

Now this implies $Du \restriction_{y_1} \cdot (y_2 - y_1) - \epsilon y_1 \cdot (y_2 - y_1) = Du \restriction_{y_2} \cdot (y_2 - y_1) - \epsilon y_2 \cdot (y_2-y_1)$, rearranging further this gives $Du \restriction_{y_1} \cdot(y_2 - y_1) + Du \restriction_{y_2} \cdot (y_1-y_2) = -\epsilon ||y_1 - y_2||^2$ .

The LHS of the last line in the above paragraph is $\geq u(y_2) - u(y_1) + u(y_1) - u(y_2) = 0$ by definition of $\Gamma^{+}$, but this implies that $-\epsilon ||y_1 - y_2||^2 \geq 0$ for $\epsilon > 0$, so $||y_1-y_2|| = 0$, i.e. $y_1 = y_2$.

Thus $\chi_{\epsilon}$ is a diffeomorphism onto its image which gives equality (2).

$\textbf{For question 2}$:

I believe the equality becomes the desired inequality by a direct application of Fatou's lemma, since everything is positive.

$\textbf{Calculation justifying negative definiteness}$

It is sufficient to show that $\Delta(h) = \frac{h^t D^2u \restriction_{y} h}{h^t h} < 0$ for all $h$ sufficiently small, and for any $y \in \Gamma^{+}$.

To this end, write $Du \restriction_{y+h} - Du \restriction_{y} = (D^2 u \restriction_{y})(h) + \underline{o}(||h||)$ (the underlined term is just supposed to indicate it is really a vector)

and $u(y+h) - u(y) - Du \restriction_{y+h} \cdot h = o(||h||^2)$ where the latter follows by a Taylor expansion and the former by definition of derivative.

Now substitute the expression for $Du \restriction_{y+h}$ obtained via a Taylor expansion into the first expression, and then apply the fact that $y \in \Gamma^{+}$ and the Cauchy-Schwarz inequality, to derive an inequality of the form $\Delta(h) \leq -\frac{o(||h||^2)}{||h||^2}$, which yields the result.

  • $\begingroup$ Thank you the proof of the injectivity of $\chi_\epsilon$ looks a bit tricky, it's hard for me to come up with this idea, do you have some intuition? or is there some machinery? $\endgroup$
    – yi li
    Jun 23 at 11:27
  • 1
    $\begingroup$ I can't really say, I basically stumbled upon it by trying to make use of the definition of $\Gamma^{+}$ and manipulating the equations. I had a little more intuition for the negative definiteness justification, I knew that I needed to make use of $u \in C^2$ somehow, and it turned out that what I needed was an $o(||h||^2)$ error term in the first-order difference quotient of $u$ to get the desired inequality, so I basically 'worked backwards' from the final inequality. $\endgroup$ Jun 23 at 11:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.