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Suppose we work in $H=l^2(\Bbb{N})$ and suppose the multiplication operator $T_f$ such that $T_f\psi=f\psi$ and $f:\Bbb{N}\rightarrow \Bbb{C}$. We denote by $B_1(H)$ the trace class of operators.

Question: I want to find a sufficient and necessary condition for $f$, such that $T_f\in B_1(H)$.

Can someone help me with this question? (This a question from a exam for Introduction to Functional analysis.)

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1 Answer 1

The adjoint is given by $T_{\bar f}$ hence $(T_f^*T_f)^{1/2}\psi=|f|\psi$. By definition, $T_f$ is trace-class if and only if the sequence $\{f(n)\}$ is summable, i.e. in $\ell^1$.

Multiplication operators are interesting in the following sense. With the notations of the OP,

  • $T_f$ is bounded if and only if the sequence $f$ is bounded;
  • $T_f$ is compact if and only if $f(n)\to 0$;
  • $T_f$ is trace-class if and only if $f\in\ell^1$;
  • $T_f$ is Hilbert-Schmidt if and only if $f\in\ell^2$.
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Very good little list of characterizations. (Strange that this is not better-known...) –  paul garrett Dec 29 '12 at 0:44
    
Thank you for the help. Is the following true?! $T_{\bar{f}}$ is the adjoint operator because: $<fx,y>=f<x,y>=f\bar{<y,x>}=<f\bar{y},\bar{x}>=<x,\bar{f}y>$ ? What are the eigenvalues of the operator? –  Mathematic Research Dec 29 '12 at 19:50
    
How do you put $f$ outside the inner product? Maybe it's more simple written them as sums. The $f_j$ are eigenvalues with $e_j$ as eigenvector. Do you think they are the only ones? –  Davide Giraudo Dec 29 '12 at 22:12

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