Timeline for Duality argument to get $L^\infty-L^2$ inequality
Current License: CC BY-SA 3.0
11 events
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May 13, 2022 at 2:49 | comment | added | Mr. Proof | Does the second norm need correction, namely to make it as follows? $$ \|T\|_{L^1\rightarrow L^2}=\|T^*\|_{L^2 \rightarrow L^\infty\}=\sup_{f,g}\frac{\langle Tf,g\rangle}{\|f\|_1\|g\|_2}\.$$ | |
Feb 9, 2016 at 17:29 | comment | added | Lior Silberman | Basically you differentiate both sides wrt $t$ and use the self-adjointness of the Laplace operator, but this brushes asides various minor technical issues. | |
Feb 9, 2016 at 16:27 | comment | added | FavorExistingPopularTags | @LiorSilberman or Nate: do you know how to prove that $(e^{-Ht})^* = e^{-Ht}$ using the weak formulation? It amounts to showing something like $(u(t),v(0))_{L^2} = (u(0),v(t))_{L^2}$, where $u$ and $v$ solutions of heat equation. I want to avoid this exponential representation | |
Feb 9, 2016 at 15:32 | comment | added | Lior Silberman | This is not obvious to me. I agree that $T=\exp(-Ht)\colon L^2\to L^2$ has $T^*=T$, but I don't see why that automatically shows something about the dual of $S=\exp(-Ht)\colon L^1\to L^2$. | |
Feb 9, 2016 at 15:09 | comment | added | Nate Eldredge | In the last step, all we really need is that the semigroup $e^{-Ht}$ is a bounded self-adjoint operator, which follows from the fact that $H$ is self-adjoint. We don't need to use the fact that a heat kernel exists, which is quite a bit more difficult to prove. | |
Feb 9, 2016 at 14:27 | history | edited | Denis Serre | CC BY-SA 3.0 |
added 156 characters in body
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Feb 9, 2016 at 13:48 | comment | added | Lior Silberman | In the second line too. Fixed now. | |
Feb 9, 2016 at 13:47 | history | edited | Lior Silberman | CC BY-SA 3.0 |
corrected the order of spaces
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Feb 9, 2016 at 13:32 | comment | added | FavorExistingPopularTags | Thanks. your domain and range are mixed in the first line btw. | |
Feb 9, 2016 at 13:27 | vote | accept | FavorExistingPopularTags | ||
Feb 9, 2016 at 13:13 | history | answered | Lior Silberman | CC BY-SA 3.0 |