# convergence exercice

I have a question please, thanks to help me. Let $\Omega$ an open bounded, connexe and regular Let $(v_n)$ an sequence in $H^1(\Omega)$ and let $v \in H^1(\Omega)$ such that $v_n$ converge weakly in $L^2(\Omega)$ to $v$. How we can compute the limits $$\int_{\Omega} A |\nabla v_n|^2 dx$$ and $$\int_{\Omega} v_n dx$$ when $n$ tends to $+\infty$

($A$ is such that $\exists \alpha > 0, A(x) \xi \xi \geq \alpha |\xi|^2, \forall \xi \in \mathbb{R}^n$ and $\exists \beta > 0, |A(x) \xi| \leq \beta |\xi|, \forall \xi \in \mathbb{R}^n$) Thanks for the help.

• The first expression does not seem well-defined: $A$ is apparently a matrix from your footnote, but you have written it as acting on a scalar. Do you mean $|A \nabla v_n|^2$? – Anthony Carapetis Aug 7 '13 at 11:28
• the first expression is $$\int_{\Omega} A \nabla v \cdot \nabla v dx$$there is a problem in this expression? – jijii Aug 7 '13 at 13:00
• That makes more sense - you should be interpreting it as $(A \nabla v) \cdot \nabla v$, since the other grouping makes no sense! – Anthony Carapetis Aug 7 '13 at 13:12
• ok, so how we compute $$\lim_n \int_{\Omega} (A \nabla v_n) \cdot \nabla v_n dx$$? – jijii Aug 7 '13 at 13:16
• I don't think you have sufficient information to determine whether or not that converges - in particular the weak $L^2(\Omega)$ convergence tells you nothing about the behaviour of $\nabla v_n$. Consider $\Omega = (0,2 \pi)$, $v_n(x) = \sin (n x)$ and $A$ the identity; then $v_n \rightharpoonup 0$ weakly in $L^2$ but $\int A \nabla v_n \cdot \nabla v_n dx = n^2 \pi$ does not converge. – Anthony Carapetis Aug 7 '13 at 13:26

## 2 Answers

The limit of the second sequence should be obvious from the definition of weak convergence once you notice that $$\int_\Omega v_n dx = (v_n, 1)_{L^2(\Omega)}$$ where $1$ is the constant function on $\Omega$ with value 1.

Let $\Omega = (-\pi,\pi)$. Let $v_n(x) = \dfrac{\sin(n^2x)}{n}$. Let $A\xi = \xi$. It is clear that $v_n \rightharpoonup 0$ in $L^2(\Omega)$, since in fact $v_n$ converges uniformly. Now, $\nabla v_n(x) = n\cos(n^2 x)$ and thus $(A\nabla v_n) \cdot \nabla v_n = n^2 \cos^2(n^2 x)$. You get $$\int_{-\pi}^\pi (A\nabla v_n) \cdot \nabla v_n \, dx = \int_{-\pi}^\pi n^2\cos^2(n^2 x) \, dx = n^2 \pi$$ for all $n$. In this case the limit is infinite. You can modify this example in a trivial way to get a finite limit or a zero limit, or even a divergent sequence.

• any one can help me please – jijii Aug 9 '13 at 18:07