I have a question. Given the Hilbertspace $H=l^2(\Bbb{N})$ with multiplicationoperator $T_f$, with $f:\Bbb{N}\rightarrow\Bbb{C}$ and $T_f\psi=f\psi$. I want to prove the following statement: Suppose only that $f$ takes only real values. Show that $T_f$ is self adjoint on the domain $D(T_f)=${$\psi\in l^2{\Bbb{N}}:f\psi\in l^2(\Bbb{N})$}.

I think the argumentation is very logical, but follows me is it not enought to say that $T_f^*=T_{\bar{f}}=T_f$ is $f$ is real valued. But wat is the argument for the claim?

Thank you :)

  • 1
    $\begingroup$ You know the formal definition of the adjoint $T^*$, right? Use it to show that $T_f$ is its own adjoint. $\endgroup$ – Nate Eldredge Jan 22 '13 at 17:31
  • $\begingroup$ This is stated already (but this is not enought,true?) $\endgroup$ – Unital Jan 22 '13 at 17:33
  • $\begingroup$ The hard part is showing that $T_f^* = T_{\bar{f}}$. Why not try writing out a proof of this fact, and ask about any parts where you are unsure? $\endgroup$ – Nate Eldredge Jan 22 '13 at 18:03

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