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Let $H:=\left\{u\in L^2(R^N):\nabla u \in L^2(R^N)\right\}$ and a functional $$f(u)=\int_{R^N} |\nabla u|^2dx+\left(\int_{R^N} |\nabla u|^2dx\right)^2.$$

If $\{u_n\}\subset H$ is a sequence such that $u_n \rightharpoonup u$, is it true that

$$\liminf_{n\rightarrow +\infty}\left[\int_{R^N} |\nabla u_n|^2dx+\left(\int_{R^N} |\nabla u_n|^2dx\right)^2\right]\geq \int_{R^N} |\nabla u|^2dx+\left(\int_{R^N} |\nabla u|^2dx\right)^2 \;?$$

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2 Answers 2

Given any Hilbert space, if $u_n\rightharpoonup u$ weakly, then $\liminf_{n\to\infty}\|u_n\| \ge \|u\|$.

Proof: W.l.o.g. $\|u\| = 1$. $$ \|u\| = \langle u,u \rangle = \lim_{n\to\infty} \langle u,u_n \rangle $$ and $$ \langle u,u_n \rangle \le \|u_n\| $$ so $$ \lim_{n\to\infty} \langle u,u_n \rangle \le \liminf_{n\to\infty}\|u_n\|$$

Actually a similar argument works on any Banach space, and it is not just for Hilbert spaces.

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I guess the inner product is given by $$\langle u,v\rangle=\int_{\mathbb R^N}u(x)v(x)dx+\sum_{j=1}^N\int_{\mathbb R^N}\dfrac{\partial u}{\partial x_j}(x)\dfrac{\partial v}{\partial x_j}(x)dx.$$ Note that we can find a subsequence $\{u_{n_k}\}$ such that $\lim_{k\to\infty}f(u_{n_k})=\liminf_{n}f(u_n)$, and since the sequence $\{|\nabla u_{n_k}|\}$ and $\{u_{n_k}\}$ are bounded in $L^2(\mathbb R^N)$ , we can extract a subsequence such that $\{u_{\varphi(n)}\}$ and $\{\nabla u_{\varphi(n)}\}$ are weakly convergent in $L^2$. We denote by $v$ and $(v_1,\ldots,v_N)$ these limits. Then weak convergence shows that $\dfrac{\partial v}{\partial x_j}=v_j$, so $u=v$. We can conclude since in an Hilbert space, if a sequence $\{x_n\}$ converges weakly to $x$ then $\lVert x\rVert\leq \liminf_n \lVert x_n\rVert$: \begin{align*} \liminf_n\: f(u_n)&=\lim_{k\to \infty}f(u_{\varphi(k)})\\ &= \liminf_{k\to\infty}\:\left(\int_{\mathbb R^N}|\nabla u_{\varphi(k)}(x)|^2dx+\left(\int_{\mathbb R^N}|\nabla u_{\varphi(k)}(x)|^2dx\right)^2\right)\\ &\geq \liminf_{k\to\infty}\:\int_{\mathbb R^N}|\nabla u_{\varphi(k)}(x)|^2dx+ \left(\liminf_{k\to\infty}\int_{\mathbb R^N}|\nabla u_{\varphi(k)} (x)|^2dx\right)^2\\ &=\int_{\mathbb R^N}|\nabla u(x)|^2dx+\left(\int_{\mathbb R^N}|\nabla u(x)|^2dx\right)^2\\ &=f(u). \end{align*}

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Thank your answer, Dvaide Girudo! –  Jiahua Jin Dec 2 '11 at 9:17
    
But I cannot see that $u_n\rightharpoonup u$ guarantees finding a subsequence $\{u_{n_k}\}$ such that $lim_{k\rightarrow+\infty} f(u_{n_k})=\liminf_{n\rightarrow} f(u_{n_k} )$. –  Jiahua Jin Dec 2 '11 at 9:24
    
I have get your meaning, the limit of the weak convergance sequence is the unique and same as its any convergant subsequence. Your precise and detailed answer makes me clear. –  Jiahua Jin Dec 2 '11 at 13:28
    
@Davide Giraudo: I would like to ask your comments in the following question math.stackexchange.com/questions/482684/… Thank you for your kind help –  blindman Sep 4 '13 at 5:00

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