Timeline for Convergence in unbounded domains
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jan 23, 2020 at 10:04 | comment | added | Skeeve | weak convergence in $W_0^{1,2}(\mathbb R^N)$ of $f_n$ to zero means that $\int_{\mathbb R^N}(f_n g + \nabla f_n \cdot \nabla g) \, dx \to 0$ as $n\to \infty$ for any fixed $g\in W_0^{1,2}(\mathbb R^N)$. Once this is established, the absence of strong convergence is not difficult because $\|f(\cdot - n)\| = \|f\|$ does not converge to zero. You may give a look at the book Sobolev Spaces by Robert A. Adams. | |
| Jan 17, 2020 at 14:55 | comment | added | M.A | How can I verify that $\phi:x\mapsto \phi(x-n)$ converges weakly to $0$ but not strongly (excuse but I'm not familiar with functional analysis properties)? | |
| Jan 17, 2020 at 14:48 | history | edited | M.A | CC BY-SA 4.0 |
deleted 1109 characters in body; edited title
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| Jan 12, 2020 at 12:28 | history | edited | Todd Trimble | CC BY-SA 4.0 |
placed text in a self-"answer" into the question
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| Jan 12, 2020 at 10:48 | history | edited | M.A | CC BY-SA 4.0 |
edited title
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| Dec 10, 2019 at 19:57 | comment | added | Skeeve | but what if for some smooth compactly supported nonzero $\phi\colon \mathbb R^N \to \mathbb R$ we consider a sequence of the form $\phi(x-n)$, $x\in \mathbb R^N$? It should converge weakly to 0 (at least for $p=2$), but not strongly. | |
| Dec 10, 2019 at 16:06 | comment | added | M.A | $\mu$ is not singular with respect to Lebesgue measure. | |
| Dec 10, 2019 at 15:53 | comment | added | M.A | $\Omega$ is a bounded open subset of $\mathbb{R}^{N}$, $\mathcal{M}(\Omega)$ is the space space of absolutely continuous Radon measures and $1<p\leq N$. The bounded case concerns the case where $\Omega$ have finite measure ($\mu(\Omega)<\infty$ where $\mu\in\mathcal{M}(\Omega)$). | |
| Dec 10, 2019 at 14:30 | comment | added | Skeeve | What is denoted by $\Omega$ and $\mathcal M(\Omega)$? What is $p$? How can convergence in $L^2(\Omega)$ make sense if $\mu$ is mutually singular with respect to Lebesgue measure? Why the statement should hold in the bounded case (and what do you mean by bounded case)? | |
| Dec 10, 2019 at 9:05 | history | asked | M.A | CC BY-SA 4.0 |