Timeline for If $A$ is a dissipative self-adjoint operator with spectral decomposition $(H_λ)$, then $e^{tA}x$ tends to the projection of $x$ onto $H_0$ as $t→∞$
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 19, 2019 at 6:57 | vote | accept | 0xbadf00d | ||
| Apr 1, 2019 at 20:15 | review | Close votes | |||
| Apr 8, 2019 at 13:01 | |||||
| Apr 1, 2019 at 19:46 | comment | added | Christian Remling | To elaborate on Nik's comment, your claim follows from functional calculus (+ dominated convergence), since $\|e^{tA}x-E_0x\|^2=\int_{(-\infty, 0]} |e^{ts}-\chi_{\{ 0\} }(s)|^2\, d\rho(s)$ for some finite measure $\rho$. (And, indeed, this is certainly not research level.) | |
| Apr 1, 2019 at 17:25 | comment | added | Jochen Glueck | It might be worthwhile to add that the following much more general result is true: If $(T(t))_{t\ge 0}$ is a bounded $C_0$-semigroup with generator $A$ on a reflexive Banach space (say, over $\mathbb{C}$) and if $\sigma(A) \cap i\mathbb{R} \subseteq \{0\}$, then $T(t)$ converges strongly as $t \to \infty$. This is (one version of) the so-called ABLV theorem. | |
| Apr 1, 2019 at 14:03 | answer | added | 0xbadf00d | timeline score: 0 | |
| Apr 1, 2019 at 11:31 | comment | added | Nik Weaver | I don't think it's research level ... in the multiplication operator picture $e^{tA}$ is multiplication by $e^{-tx}$ on $[0, \infty)$ and $E_0$ is multiplication by $1_{\{0\}}$. Yes, the former converges strongly to the latter. | |
| Apr 1, 2019 at 10:50 | history | asked | 0xbadf00d | CC BY-SA 4.0 |