Skip to main content
Antonio Rapallino's user avatar
Antonio Rapallino's user avatar
Antonio Rapallino's user avatar
Antonio Rapallino
Unregistered
  • Member for 9 years, 10 months
  • Last seen more than 9 years ago
comment
adjoint of this closed (?) operator
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 $(S\circ f)^*= f \circ S^*$?
comment
adjoint of this closed (?) operator
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 $gf'$?
awarded
accepted
comment
adjoint of this closed (?) operator
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 $\lim_{x \rightarrow \pm 2 \pi} g(x)f'(x)=0$.
awarded
comment
adjoint of this closed (?) operator
@NikWeaver so you know how to do this?-could you explain your approach?
comment
adjoint of this closed (?) operator
@ChristianRemling ah yes thank you, I forgot this.
revised
adjoint of this closed (?) operator
added 24 characters in body
Loading…
awarded
revised
Loading…
asked
Loading…