L
Lv 7
L 發問時間： 科學數學 · 1 0 年前

# [Sobolev space] 不等式證明 #3

Fix α > 0 and let U = the open ball B_1(0) of |R^n.

Show there exists a constant C = C(n,α) such that

∫_U u^2 dx <= C *∫_U |Du|^2 dx

provided |{x in U | u(x) = 0}| >= α , u in H^1(U).

### 2 個解答

• Eric
Lv 6
1 0 年前
最佳解答

|u - (u)|^2 = u^2 - 2u(u) ＋ (u)^2

Let V = the volume of B(0,1)

∫|u - (u)|^2 = ∫u^2 - 2∫u(u) ＋ ∫(u)^2

= ∫u^2 - 2(u)^2 V ＋ (u)^2 V = ∫u^2 - (u)^2 V

2007-06-22 10:30:02 補充：

\documentclass[12pt]{article}

\usepackage{amsmath,amsthm,amsfonts}

\pagestyle{empty}

\begin{document}

\begin{proof}

Let us mimic the proof of the Poincar\'e inequality. Suppose the claim is false; then there exists, for each $k \in \mathbb{N}$, a function $u_k \in H^1(U)$ with $\|u_k\|_{L^2(U)} = 1$ (after normalizing), such that

$\|Du_k\|_{L^2(U)} < \frac{1}{k}.$

In light of the Rellich-Kondrachov compactness theorem, there exists a subsequence $u_{k_j}$ and $u \in L^2(U)$ such that $u_{k_j} \rightarrow u$ in $L^2(U)$. It follows that $\|u\|_{L^2(U)} = 1$.

On the other hand, since $\|Du_{k_j}\|_{L^2(U)} < 1/k_j$, we have, for each $i = 1,\ldots,n$ and $\phi \in C^\infty_\mathrm{c}(U)$,

$\int_U u\phi_{x_i} \, dx = \lim_{j \rightarrow \infty} \int_U u_{k_j} \phi_{x_i} \, dx = - \lim_{j \rightarrow \infty}\int_U \partial_{x_i}u_{k_j} \phi = 0,$

and hence $u \in H^1(U)$ and $Du = 0$ a.e. Since $U$ is connected, $u$ must be constant a.e.; but $u$ is zero on a set of positive measure, so $u = 0$ a.e., contradicting the fact that $\|u\|_{L^2(U)} = 1$.

\end{proof}

\end{document}

• L
Lv 7
1 0 年前

用 Poincare's inequality with p=2 兩次再相乘可得

∫_U |u - (u)_U|^2 <= C * ∫_U |Du|^2

其中 C = C(n) , (u)_U = (∫_U u)/|U|

2007-06-21 14:56:55 補充：

∫_U u^2 <= ∫_U |u - (u)_U|^2 ??

2007-06-21 17:03:02 補充：

So∫_U u^2 > ∫_U |u - (u)_U|^2 ...

{u=0} 不會太小的條件不知怎麼用上去