# question about the proof for the Key Lemma for Alexandroff maximum principle

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)?

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.

• 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? Jun 23 at 11:27
• 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. Jun 23 at 11:33