Timeline for Application of uniform boundedness (Banach-Steinhaus) principle
Current License: CC BY-SA 4.0
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| Nov 18, 2020 at 10:46 | history | edited | katago | CC BY-SA 4.0 |
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| Nov 18, 2020 at 10:30 | history | edited | katago | CC BY-SA 4.0 |
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| Nov 18, 2020 at 9:12 | history | edited | katago | CC BY-SA 4.0 |
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| Nov 18, 2020 at 4:12 | comment | added | katago | we can use convolution of $\phi$ with a approximation to identity or use step function to approximate and then convolution with some approximation to identity(there are some other way to prove this of course) to prove testfuntion is dense in distribution if the underly space is separable. | |
| Nov 18, 2020 at 4:07 | comment | added | katago | @Fozz, if the underly space is separable, there is $R^3$ of course it is. then the test function space is dense in distribution, and we can use the limit to define the action of $u_{R}$ on a general distribution $\phi$ by $\int \phi_{k} u_{R} d x \rightarrow \int \phi u_{R} d x$ where $ \phi_{k}$ is a sequence of function weak converage to $\phi$ in distribution, we only need to prove it is well-defined(do not rely on the choice of sequence) and it is linear, this is easy. | |
| Nov 18, 2020 at 4:03 | comment | added | katago | @Fozz, in fact in your case because of priori estimate and application of Leray–Schauder principle we can prove $u_{R} \in \dot{H}^{1}$, so the argument can all live in $\dot{H}^{1} \subset H^ {-1}$ and the limit is also in $\dot{H}^{1}$ | |
| Nov 18, 2020 at 4:01 | comment | added | Fozz | For $\phi\in\mathcal D$, the map $\phi\mapsto\int\phi u_R\,dx$ IS a linear operator, but how do we know it's continuous with respect to $\dot{H^1}$? I guess what I mean is, if we approximate $\phi\in \dot{H^1}$ by $\phi_k\in\mathcal D$ in the space $\dot{H^1}$, how does it follow that $\int \phi_k u_R\,dx\to\int\phi u_R\,dx$ as well? | |
| Nov 18, 2020 at 3:50 | comment | added | katago | from $\dot H^1 \to R$, test function space is dense in $\dot H^1$, so there is a natural extension to make $u_R$ to be a linear operator on $\dot H^1$ | |
| Nov 18, 2020 at 3:44 | comment | added | Fozz | Are you viewing $u_R$ as a linear operator from $\dot{H^1}$ to $\mathbb{R}$ ($\phi\mapsto\int\phi u_R\,dx$) or from $\dot{H^1}$ to $H^1$ ($\phi\mapsto \phi u_R$)? | |
| Nov 18, 2020 at 3:43 | comment | added | Fozz | The space $\dot{H^1}$ is the space of distributions $f$ such that $\int_{\mathbb{R}^3}|\xi|^2|\mathcal{F}f(\xi)|^2\,d\xi<\infty$ (here $\mathcal{F}$ is the Fourier transform). Basically, it's the space of distributions whose gradient is in $L^2$. | |
| Nov 18, 2020 at 3:24 | history | edited | katago | CC BY-SA 4.0 |
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| Nov 18, 2020 at 3:18 | history | edited | katago | CC BY-SA 4.0 |
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| Nov 18, 2020 at 3:11 | history | answered | katago | CC BY-SA 4.0 |