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*}


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.


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.
  • 1
    $\begingroup$ Thanks for your reply! I understood. Your proof is very easy to understand! $\endgroup$ – sharpe Sep 28 '15 at 5:10

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