Timeline for Positive semigroups and convex operator
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
Commonmark migration
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| Jan 21, 2014 at 16:42 | vote | accept | Shinning Star | ||
| Jan 15, 2014 at 8:58 | comment | added | András Bátkai | I suggest you to ask a separate question where you formulate clearly what is your question. In this thread you already asked three questions by editing, and it gets confusing. | |
| Jan 15, 2014 at 8:55 | comment | added | András Bátkai | $Zv\neq v$. The linear polynomial is never considered to be an identity element in these spaces, only if you make an algebra by composition. | |
| Jan 15, 2014 at 8:42 | comment | added | András Bátkai | $1$ is the constant one function, not $Z$.... | |
| Jan 15, 2014 at 7:03 | comment | added | András Bátkai | Why do you think so? You mix up something again... | |
| Jan 8, 2014 at 8:18 | vote | accept | Shinning Star | ||
| Jan 21, 2014 at 12:00 | |||||
| Jan 7, 2014 at 21:07 | history | edited | András Bátkai | CC BY-SA 3.0 |
fixed typo
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| Jan 7, 2014 at 20:38 | comment | added | András Bátkai | I deleted my comments and added a detailed example. You should consider real valued functions if you want to preserve positivity. Good day to you. | |
| Jan 7, 2014 at 20:34 | history | edited | András Bátkai | CC BY-SA 3.0 |
Expanded as an answer to the demands of OP.
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| Jan 6, 2014 at 9:40 | vote | accept | Shinning Star | ||
| Jan 6, 2014 at 10:49 | |||||
| Jan 6, 2014 at 9:17 | comment | added | Shinning Star | Sir, for $X=C(\Gamma)$ (continuous functions on unit circle), rotation group is defined as $$(T(t)f)(s)=f(e^{it}.z),\quad f\in X, z\in\Gamma$$ Let $\phi$ be mirroring along $x$-axis (which means to take conjugate) i.e $\phi(z)=\overline{z},\quad z\in\Gamma.$ If I take $f(z)=z^2$, for $z\in\Gamma$, then f is continuous. $(T(t)f)(z)=f(e^{it}.z)=z^2e^{2it}$, $\phi[(T(t)f)(z)]=\overline{z}^2e^{-2it}$, $\phi(f(z))=\overline{z}^2$, $T(t)[\phi(f(z))]=\overline{e^{it}.z}^2=\overline{z}^2.e^{-2it}$ Sir this is how I tried. It may be wrong. | |
| Jan 5, 2014 at 15:26 | history | edited | András Bátkai | CC BY-SA 3.0 |
expanded answer.
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| Jan 2, 2014 at 23:05 | history | answered | András Bátkai | CC BY-SA 3.0 |