1
$\begingroup$

I have a question about Sobolev spaces.

$(1,2)$-Sobolev space on $\mathbb{R}^{d}$ (denoted by $H^{1,2}(\mathbb{R}^{d})$) is defined as follows: \begin{align*} H^{1,2}(\mathbb{R}^{d}):=\left\{f \in L^{2}(\mathbb{R}^{d};dx) \left| \frac{\partial f}{\partial x_{i}} \in L^{2}(\mathbb{R}^{d};dx), 1\leq i \leq d \right. \right\} \end{align*} with derivatives in the Schwartz distribution sence. It is kwnon that $H^{1,2}(\mathbb{R}^{d})$ is a Hilbert space with the inner product \begin{align*} (f,g):=\int_{\mathbb{R}^{d}}fg\,dx+\sum_{i=1}^{n} \int_{\mathbb{R}^{d}}\frac{\partial f}{\partial x_{i}}\frac{\partial g}{\partial x_{i}}dx \end{align*}

Question

Let $(E_{n})_{n \in \mathbb{N}}$ be an increasing Lebesgue measurable subsets such that $\mu(\mathbb{R}^{d} \setminus E_{n})\to 0$ as $n \to \infty$.

Suppose that $f_{n} \to f$ in $H^{1,2}(\mathbb{R}^{d})$

Then can we show that $\displaystyle \lim_{n \to \infty}\int_{\mathbb{R}^{d}\setminus E_{n}} (\nabla f_{n}, \nabla f_{n}) dx =0\cdots(1)$ ?

My Idea

Since $(\int_{\mathbb{R}^{d}} (\nabla f_{n}, \nabla f_{n}) dx)_{n \in \mathbb{N}}$ is convergent sequence, it is bounded. Therefore $(\int_{\mathbb{R}^{d}\setminus K_{n}} (\nabla f_{n}, \nabla f_{n}) dx)_{n \in \mathbb{N}}$ is bounded. Hence there exists a subsequence such that $ \left(\int_{\mathbb{R}^{d}\setminus K_{n_{j}}} (\nabla f_{n_{j}}, \nabla f_{n_{j}})dx\right)_{j \in \mathbb{N}} $ is a convergent sequence. Can we show $ \int_{\mathbb{R}^{d}\setminus K_{n_{j}}} (\nabla f_{n_{j}}, \nabla f_{n_{j}})dx \to 0 $ ?

Thank you in advance.

$\endgroup$
1
$\begingroup$

Notice that: \begin{align} \frac 12\int_{\mathbb{R}^d\setminus E_n}|\nabla f_n|^2 \le &\ \int_{\mathbb{R}^n\setminus E_n}|\nabla f_n - \nabla f|^2 + \int_{\mathbb{R}^n\setminus E_n}|\nabla f|^2 \tag 1\\ \le &\ \int_{\mathbb{R}^d}|\nabla f_n - \nabla f|^2 + \int_{\mathbb{R}^n\setminus E_n}|\nabla f|^2 \tag 2\\ \to &\ 0. \tag 3 \end{align}

  • $(1)$ follows from $(a + b)^2 \le 2a^2 + 2b^2$,
  • $(2)$ follows from monotonicity of the integral,
  • $(3)$ follows from convergence in $L^2$ of the gradients for the first term and monotone convergence theorem for the second term.
$\endgroup$
  • 1
    $\begingroup$ Thanks for your reply! I understood. Your proof is very easy to understand! $\endgroup$ – sharpe Sep 28 '15 at 5:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.