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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
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
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