Timeline for A specific mollified functions in the Sobolev space H^1(R)
Current License: CC BY-SA 3.0
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| Aug 15, 2016 at 10:23 | answer | added | D G | timeline score: 1 | |
| Jul 27, 2016 at 13:20 | comment | added | Willie Wong | These questions are extremely elementary and you should probably ask on Math.SE instead. But a sketch of (1): the cut-off function $\chi_{[-R,R]}$ is (implicitly in the argument above) smooth. In fact, you should know how to do this since that's half of the method you quoted in the question body itself. For (2) a hint: convolution by a non-negative bump function of integral one can only reduce the oscillation. | |
| Jul 27, 2016 at 10:37 | comment | added | papnass | Could you please send any references you might have | |
| Jul 27, 2016 at 10:19 | comment | added | papnass | Then I have two questions. First, why $\Vert \nabla v_{n}-\nabla u\Vert_{L^2(\mathbb{R})}\longrightarrow 0$ as $n\longrightarrow \infty$? And how can prove using the uniform continuity the fact that $|v_{n,\varepsilon}(x)|\leq u(x)$? Thanks | |
| Jul 27, 2016 at 10:19 | comment | added | papnass | Ok for convolution. Now, as well as I understand the goal is to show that $\forall \varepsilon'>0$, for $n$ large enough and $\varepsilon$ small enough we have $$\Vert v_{n,\varepsilon}-u\Vert_{H^1(\mathbb{R})}\leq \Vert v_{n,\varepsilon}-v_{n}\Vert_{H^1(\mathbb{R})}+\Vert v_{n}-u\Vert_{H^1(\mathbb{R})}\leq \frac{\varepsilon'}{2}+\frac{\varepsilon'}{2}\leq\varepsilon'.$$ | |
| Jul 26, 2016 at 18:36 | comment | added | Willie Wong | Do the usual thing by convolving with a smooth bump function; call the smoothed versions $v_{n,\epsilon}$. Since $v_{n,\epsilon} \to v_n$ in $H^1$ as $\epsilon \to 0$, and $H^1(\mathbb{R})$ embeds in $L^\infty$, the $\epsilon_n$ room you left in step one is big enough. // You can be even more precise by using uniform continuity to show that as long as your smooth bump function has sufficiently small support this will work. | |
| Jul 26, 2016 at 18:32 | comment | added | papnass | Can you please explain more how to transform the functions $v_n$ into a functions in $C^{\infty}_{0}(\mathbb{R})$ which converge toward to $u$ using the $H^{1}$ norm and maintain the condition $|v_n(x)|\leq u(x)$ a.e. | |
| Jul 26, 2016 at 18:17 | comment | added | papnass | Thank for your answer, it is not necessary to have a convolution against mollifier. I just want a sequence of $C^{\infty}(\mathbb{R})$ with compact support and the condition $|u_n(x)|\leq u(x)$ a.e. | |
| Jul 26, 2016 at 15:15 | comment | added | Willie Wong | Rough sketch: since $u > 0$, you can find a sequence $R_n \nearrow \infty$ and $\epsilon_n \searrow 0$ such that the functions $$ v_n = \chi_{[-R_n, R_n]}\cdot (u - \epsilon_n) $$ is non-negative and converges to $u$ in $H^1$. For each $v_n$ you can mollify. Then you can diagonalize and use the fact that $u\in H^1 \implies u$ is uniformly continuous to ensure the smoothed versions of $v_n$ remains below $u$. | |
| Jul 26, 2016 at 15:05 | review | First posts | |||
| Jul 26, 2016 at 15:10 | |||||
| Jul 26, 2016 at 15:00 | comment | added | Willie Wong | If you insist on constructing it from convolution against a mollifier: no. But if you just want a sequence of smooth approximations: yes. | |
| Jul 26, 2016 at 14:55 | history | asked | papnass | CC BY-SA 3.0 |