Ngô Quốc Anh

May 30, 2010

Filed under: Giải Tích 6 (MA5205) — Tags: — Ngô Quốc Anh @ 20:29

Let us recall the following result

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 $\overline \Omega$.

The above result has been proven in this entry. If we chose $p=0$ then we would have $\displaystyle\int_\Omega dx =\int_0^M \left(\int_{\{u=t\}}\frac{1}{|\nabla u|}d\sigma\right)dt$.

If we now replace $\Omega=\{u>t\}$, we get $\displaystyle\int_{\{u>t\}}dx =\int_t^M \left(\int_{\{u=t\}}\frac{1}{|\nabla u|}d\sigma\right)dt$.

By differentiating with respect to $t$, we arrive at $\displaystyle \frac{d}{dt}\int_{\{u>t\}}dx =-\int_{\{u=t\}}\frac{1}{|\nabla u|}d\sigma$.

In this entry, we shall prove the foregoing identity is actually true. This is the second interesting formula I have mentioned before. The proof, of course again, is based on a clever choice of test function.

Proof. Let $\varepsilon>0$. Define $\displaystyle f=-{\rm div}\left( \frac{\nabla u}{|\nabla u|^2+\varepsilon}\right)$.

Multiplying by $(u-t)^+$ and integrating by parts we get $\displaystyle \int_{\{u>t\}}\frac{|\nabla u|^2}{|\nabla u|^2+\varepsilon}dx = \int_{\{u>t\}}f(u-t)dx$.

Differentiating with respect to $t$ gives $\displaystyle -\frac{d}{dt}\int_{\{u>t\}}\frac{|\nabla u|^2}{|\nabla u|^2+\varepsilon}dx = \int_{\{u>t\}}fdx$.

For a sufficiently small $h$, we can integrate the previous relation from $t-h$ to $t$ to get $\displaystyle \int_{\{t-h\leqslant u\tau\}}fdx\right)d\tau=\int_{t-h}^t\left( \int_{\{u=\tau\}}\frac{|\nabla u|}{|\nabla u|^2+\varepsilon}d\sigma\right)d\tau$.

We are now in a position to apply the Dominated Convergence Theorem to pass to the limit as $\varepsilon\to 0$ to get $\displaystyle \int_{\{t-h\leqslant u.

Note that $\displaystyle \int_{\{t-h\leqslant ut-h\}}dx-\int_{\{u>t\}}dx$.

Thus, dividing by $h$ and taking the limit as $h\to 0$ yields the desired result.

Remark. The following is also true $\displaystyle \frac{d}{dt}\int_{\{u>t\}}dx =-\int_{\{u^\star=t\}}\frac{1}{|\nabla u^\star|}d\sigma$.