# Ngô Quốc Anh

## May 24, 2010

Filed under: Giải Tích 6 (MA5205) — Tags: — Ngô Quốc Anh @ 18:55

I found two interesting formulas related to co-area formula while reading some tricks done by Talenti regarding to the best constant of the Sobolev inequality. The first result is to derive a representation of

$\displaystyle \int_\Omega |\nabla u|^pdx$

and the second result is to deal with differentiation of level sets. Having all these stuffs, I will derive a very short and beautiful proof concerning the lower bound of $\int \exp(u) dx$ where $u$, a positive solution to the following PDE

$\displaystyle-\Delta u = e^u$

This proof I firstly learned from a paper published in Duke Math. J. in 1991 by W. Cheng and C. Li [here].

Co-area formula. Suppose that $\Omega$ is an open set in $\mathbb R^n$, and $u$ is a real-valued Lipschitz function on $\Omega$. Then, for an integrable function $g$

$\displaystyle\int_\Omega g(x) |\nabla u(x)| dx = \int_{-\infty}^\infty \left(\int_{\{u=t\}}g(x) dH_{n-1}(x)\right)dt$

where $H^{n-1}$ is the $(n-1)$-dimensional Hausdorff measure.

The Sard theorem. Let $f :\mathbb{R}^n \rightarrow \mathbb{R}^m$ be $C^k$, $k$ times continuously differentiable, where $k \geqslant \max\{n-m+1, 1\}$. Let $X$ be the critical set of $f$, the set of points $x$ in $\mathbb R^n$ at which the Jacobian matrix of $f$ has ${\rm rank} < m$. Then $f(X)$ has Lebesgue measure $0$ in $\mathbb R^m$.

The Sard theorem has some useful applications. For example, if $u \in \mathcal D(\Omega)$ the space of test functions where $\Omega \subset \mathbb R^n$, then for almost every $t$ in the range of $u$, we have that $|\nabla u|\ne 0$ on the level set $\{u=t\}$. Thus that level set will be an $(n-1)$-dimensional surface. Furthermore

$\{u=t\}=\partial \{u>t\}$

and

$|\{u=t\}|=0$.

Theorem. Let $\Omega \subset \mathbb R^n$ be an open set and $u \in \mathcal D(\Omega)$. If $u \geqslant 0$ then for any $1 \leqslant p<\infty$, we have

$\displaystyle\int_\Omega|\nabla u|^p dx =\int_0^M \left(\int_{\{u=t\}}|\nabla u|^{p-1}d\sigma\right)dt$

where $M=\sup u$ over $\overline \Omega$.

Proof. Two cases are possible

If $2 \leqslant p<\infty$. Define

$\displaystyle f=-{\rm div}(|\nabla u|^{p-2}\nabla u)$

then for all $v \in W_0^{1,p}(\Omega)$ we have

$\displaystyle \int_\Omega |\nabla u|^{p-2}\nabla u \cdot \nabla vdx = \int_\Omega f v dx$.

Here is the trick. Let $t>0$. Set

$\displaystyle v=(u-t)^+ \in W_o^{1,p}$.

Using this test function and observing that $v$ is supported on the set $\{u>t\}$ and that on this set $v=u-t$, we get

$\displaystyle \int_{\{u>t\}} |\nabla u|^pdx = \int_{\{u>t\}}f(u-t)dx$.

Thus, differentiating with respect to $t$, we get

$\displaystyle -\frac{d}{dt}\int_{\{u>t\}} |\nabla u|^pdx = \int_{\{u>t\}}f dx$.

Integrating this relation over the range of $u$, i.e. on $[0,M]$ we get

$\displaystyle\int_\Omega|\nabla u|^p dx =\int_0^M \left(\int_{\{u>t\}}fdx\right)dt$.

For $t$ such that $|\nabla u| \ne 0$ on the level set $\{u=t\}$, the Green theorem implies

$\displaystyle \int_{\{u>t\}}fdx=-\int_{\{u=t\}}|\nabla u|^{p-2}\nabla u \cdot \nu d\sigma=\int_{\{u>t\}}|\nabla u|^{p-1}d\sigma$

since on the level set $\{u=t\}$ the tangential derivatives of $u$ vanish and since $u>t$ inside this surface, we have

$\displaystyle -\nabla u \cdot \nu = |\nabla u|$.

If $1\leqslant p<2$. Due to the presence of singularity, we need to use an approximation technique. Let $\varepsilon>0$. Define

$\displaystyle f_\varepsilon=-{\rm div}\left((|\nabla u|^2+\varepsilon)^\frac{p-2}{2}\nabla u\right)$

then for all $v \in W_0^{1,p}(\Omega)$ we have

$\displaystyle \int_\Omega (|\nabla u|^2+\varepsilon)^\frac{p-2}{2}\nabla u \cdot \nabla vdx = \int_\Omega f_\varepsilon v dx$.

Choose the same test function as in the previous case we arrive at

$\displaystyle \int_{\{u>t\}} (|\nabla u|^2+\varepsilon)^\frac{p-2}{2} |\nabla u|^2 dx =\int_0^M \left(\int_{\{u=t\}} (|\nabla u|^2+\varepsilon)^\frac{p-2}{2} |\nabla u| d\sigma\right)dt$.

We now pass the limit as $\varepsilon \to 0$ on both sides. On the LHS, the integrand coverges pointwise to $|\nabla u|^p$; further, since $1 \leqslant p<2$, we have

$\displaystyle (|\nabla u|^2+\varepsilon)^\frac{p-2}{2} |\nabla u|^2 \leqslant |\nabla u|^p$

which is integrable. Thus, by the Dominated Convergence Theorem, we get

$\displaystyle \lim_{\varepsilon \to 0} \int_\Omega (|\nabla u|^2+\varepsilon)^\frac{p-2}{2} |\nabla u|^2 dx = \int_\Omega |\nabla u|^pdx$.

On the RHS, one again observes that

$\displaystyle (|\nabla u|^2+\varepsilon)^\frac{p-2}{2} |\nabla u| \leqslant |\nabla u|^{p-1}$

which is integrable on the set $\{u=t\}$. So again by the Dominated Convergence Theorem,

$\displaystyle \int_{\{u=t\}} (|\nabla u|^2+\varepsilon)^\frac{p-2}{2} |\nabla u| d\sigma \to \int_{\{u=t\}}|\nabla u|^{p-1} d\sigma$

as $\varepsilon \to 0$.

Further,

$\displaystyle \int_{\{u=t\}} (|\nabla u|^2+\varepsilon)^\frac{p-2}{2} |\nabla u| d\sigma \leqslant \int_{\{u=t\}}|\nabla u|^{p-1} d\sigma$

and

$\displaystyle \int_0^M \left(\int_{\{u=t\}} |\nabla u|^{p-1} d\sigma\right)dt < \infty$

since $u \in \mathcal D(\Omega)$. Thus another application of the Dominated Convergence Theorem yields

$\displaystyle \lim_{\varepsilon \to 0}\int_0^M \left(\int_{\{u=t\}} (|\nabla u|^2+\varepsilon)^\frac{p-2}{2} |\nabla u| d\sigma\right)dt = \int_0^M \left(\int_{\{u=t\}} |\nabla u|^{p-1} d\sigma\right)dt$.

The proof follows.

Remark. Concerning the Schwarz symmetrization, the following

$\displaystyle\int_\Omega|\nabla u^\star|^p dx \leqslant\int_0^M \left(\int_{\{u^\star=t\}}|\nabla u^\star|^{p-1}d\sigma\right)dt$

holds.

### Co-area formula

1. In case 1, the second from the last equation, the right most integral should be $\int_{u=t}$, right?

Comment by Chao — August 15, 2010 @ 19:13

• You are right, Chao. The correct one reads as follows

$\displaystyle \int_{\{u>t\}}fdx=-\int_{\{u=t\}}|\nabla u|^{p-2}\nabla u \cdot \nu d\sigma=\int_{\{u=t\}}|\nabla u|^{p-1}d\sigma$.

Thanks for pointing out this misprint.

Comment by Ngô Quốc Anh — August 15, 2010 @ 19:21

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