Timeline for adjoint of this closed (?) operator
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Jan 24, 2015 at 22:10 | comment | added | Christian Remling | @AntonioRapallino: This is correct. It's easy to verify, from the definition of the adjoint (use that the bounded operator is defined everywhere). | |
| Jan 24, 2015 at 22:04 | comment | added | Antonio Rapallino | you used that $T^* = gS^*$. Does this mean that for a bounded and s.a. operator $f$ and a closed operator $S$ we have in general $(S\circ f)^*= f \circ S^*$? | |
| Jan 24, 2015 at 17:04 | history | edited | Christian Remling | CC BY-SA 3.0 |
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| Jan 23, 2015 at 23:13 | comment | added | Christian Remling | @AntonioRapallino: That is correct. Since $f'$ can be an arbitrary $L^2$ function here, the limit need not exist. | |
| Jan 23, 2015 at 23:12 | comment | added | Antonio Rapallino | not sure if I get this right. if $f$ is continuous, then I would know that $gf \rightarrow 0$ at the boundary, but how do I deduce this for $gf'$? | |
| Jan 23, 2015 at 22:56 | vote | accept | Antonio Rapallino | ||
| Jan 23, 2015 at 22:56 | comment | added | Antonio Rapallino | ah thank you, but the limit that I wanted to exist for functions $f \in D(T^*)$ does not necessarily have to exist, right? I am talking about the $\lim_{x \rightarrow \pm 2 \pi} g(x)f'(x)=0$. | |
| Jan 23, 2015 at 22:41 | history | answered | Christian Remling | CC BY-SA 3.0 |