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I am currently dealing with an unbounded operator

$T:\{f \in L^2(-2\pi,2\pi); f \in AC((-2\pi,2\pi)), T(f) \in L^2, \lim_{x \rightarrow \pm 2 \pi} f(x)g(x)=0\} \subset L^2(-2\pi,2\pi)\rightarrow L^2(-2\pi,2\pi)$

$g \in C^{\infty}: g(-2\pi) = g(2\pi)=0$ and $g|_{(-2\pi,2\pi)} >0.$

Then I want to show that $T(f) = -i(gf)'$ is closed and has an adjoint operator $T^*(f) = -igf'$ defined on the domain $\{f \in L^2;f \in AC((-2\pi,2\pi)), T^*(f) \in L^2, \lim_{x \rightarrow \pm 2 \pi}g(x) f'(x)=0\}.$

Unfortunately, this is not my normal field of interest and so I thought that some people here might immediately know how to approach this problem.

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    $\begingroup$ I think this question belongs on math.stackexchange. $\endgroup$
    – Nik Weaver
    Commented Jan 23, 2015 at 21:21
  • $\begingroup$ You definitely need to add the condition (on $f$) that $(gf)'\in L^2$ to obtain an operator that maps to $L^2$ again. $\endgroup$
    – Christian Remling
    Commented Jan 23, 2015 at 21:22
  • $\begingroup$ @ChristianRemling ah yes thank you, I forgot this. $\endgroup$
    – Antonio Rapallino
    Commented Jan 23, 2015 at 21:50
  • $\begingroup$ @NikWeaver so you know how to do this?-could you explain your approach? $\endgroup$
    – Antonio Rapallino
    Commented Jan 23, 2015 at 21:50

1 Answer 1

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Let's first of all recall that the minimal operator of differentiation $$ D(S) = \{ f\in L^2\cap AC : f'\in L^2, f(\pm 2\pi)=0 \} , \quad\quad Sf = -if' $$ is closed and has the maximal operator as its adjoint: $$ D(S^*) =\{ f\in L^2\cap AC: f'\in L^2\} , \quad\quad S^*f=-if' $$ Now since $g$ is smooth and $g>0$, your condition that $f\in AC$ is the same as asking that $gf\in AC$. This means that your $T=Sg$ is the composition of $S$ with the (bounded) operator of multiplication by $g$.

Thus $T$ is indeed closed, and $T^*=gS^*$. In particular, $D(T^*)=D(S^*)$.

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  • $\begingroup$ 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 $\lim_{x \rightarrow \pm 2 \pi} g(x)f'(x)=0$. $\endgroup$
    – Antonio Rapallino
    Commented Jan 23, 2015 at 22:56
  • $\begingroup$ 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 $gf'$? $\endgroup$
    – Antonio Rapallino
    Commented Jan 23, 2015 at 23:12
  • $\begingroup$ @AntonioRapallino: That is correct. Since $f'$ can be an arbitrary $L^2$ function here, the limit need not exist. $\endgroup$
    – Christian Remling
    Commented Jan 23, 2015 at 23:13
  • $\begingroup$ 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 $(S\circ f)^*= f \circ S^*$? $\endgroup$
    – Antonio Rapallino
    Commented Jan 24, 2015 at 22:04
  • $\begingroup$ @AntonioRapallino: This is correct. It's easy to verify, from the definition of the adjoint (use that the bounded operator is defined everywhere). $\endgroup$
    – Christian Remling
    Commented Jan 24, 2015 at 22:10

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