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Let $$V=\{ u \in W^{1,2}(\mathbb{R}) : \int_{\mathbb{R}}{\vert u(x)\vert^2 x^2dx }\lt \infty\}$$ with the scalar product $$\langle u,v\rangle = \int_{\mathbb{R}}{ u(x)v(x) (1+x^2)dx}+\int_{\mathbb{R}}{ u'(x)v'(x)dx}$$

Let $B_1(0)^V$ denote the unit sphere with respect to the above norm and $I_N =[-N,N] \subset \mathbb{R}$

For $u \in B_1(0)^V$ . Show that for every $\epsilon$ >0, there exists an $N>0$ such that $\Vert u \Vert_{L^2(\mathbb{R^n} \setminus I_N)}\le \epsilon$ and that the embedding $V \hookrightarrow L^2(\mathbb{R})$ is compact.

I am already stuck on the first part of this question. Doesn't $u \in V$ imply that $u$ vanishes for $x \rightarrow \infty$ ? How can I use this to prove the first part of the question. Also on a side note, for the second part I was given the hint that for compact intervals $I$ the embedding $W^{1,2}(I) \hookrightarrow L^2(I)$ is compact.

Would appreciate any help on this question.

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  • $\begingroup$ For the second part I don't know, but for the first one, you have \begin{align} \|u\|_{L^2(\mathbb R \backslash I_N)} &= \left(\int_{\mathbb R \backslash I_N} |u^2(x)| x^2 x^{-2}\right)^{1/2}\\ &\le \|x ~u \|_{L^2(\mathbb R \backslash I_N)} \|1/x\|_{L^2(\mathbb R \backslash I_N)}\\ &= \|1/x\|_{L^2(\mathbb R \backslash I_N)} \to 0 \end{align} for $N \to \infty$. $\endgroup$
    – Falcon
    Jan 8, 2022 at 19:41

1 Answer 1

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For the first, note that $$ \| u\|_{L^2(\mathbb{R}\setminus I_N)}^2 = \int_{\mathbb{R}\setminus I_N} |u(x)|^2\, dx \leq \int_{\mathbb{R}\setminus I_N} |u(x)|^2 \dfrac{x^2}{N^2}\, dx \leq N^{-2}\| u\|_{V}^2. $$

For the second part, look here for an argument along the same lines.

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