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Christian Remling
Christian Remling
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Your operator is a rank one perturbation of the multiplication operator $(Mf)(x) = (x^3/2)f(x)$$(Mf)(x) = (x/2)f(x)$, which has (purely) absolutely continuous spectrum equal to $[0,1/2]$. Since the ac spectrum is invariant under trace class perturbations (so certainly under rank one perturbations), your operator $L$ still has the same ac spectrum, so doesn't even come close to having pure point spectrum (and thus it isn't "diagonalizable," if you want to put it this way, though I personally don't think it's very good terminology).

Your operator is a rank one perturbation of the multiplication operator $(Mf)(x) = (x^3/2)f(x)$, which has (purely) absolutely continuous spectrum equal to $[0,1/2]$. Since the ac spectrum is invariant under trace class perturbations (so certainly under rank one perturbations), your operator $L$ still has the same ac spectrum, so doesn't even come close to having pure point spectrum (and thus it isn't "diagonalizable," if you want to put it this way, though I personally don't think it's very good terminology).

Your operator is a rank one perturbation of the multiplication operator $(Mf)(x) = (x/2)f(x)$, which has (purely) absolutely continuous spectrum equal to $[0,1/2]$. Since the ac spectrum is invariant under trace class perturbations (so certainly under rank one perturbations), your operator $L$ still has the same ac spectrum, so doesn't even come close to having pure point spectrum (and thus it isn't "diagonalizable," if you want to put it this way, though I personally don't think it's very good terminology).

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Christian Remling
Christian Remling
  • 24.2k
  • 2
  • 48
  • 83

Your operator is a rank one perturbation of the multiplication operator $(Mf)(x) = (x^3/2)f(x)$, which has (purely) absolutely continuous spectrum equal to $[0,1]$$[0,1/2]$. Since the ac spectrum is invariant under trace class perturbations (so certainly under rank one perturbations), your operator $L$ still has the same ac spectrum, so doesn't even come close to having pure point spectrum (and thus it isn't "diagonalizable," if you want to put it this way, though I personally don't think it's very good terminology).

Your operator is a rank one perturbation of the multiplication operator $(Mf)(x) = (x^3/2)f(x)$, which has (purely) absolutely continuous spectrum equal to $[0,1]$. Since the ac spectrum is invariant under trace class perturbations (so certainly under rank one perturbations), your operator $L$ still has the same ac spectrum, so doesn't even come close to having pure point spectrum (and thus it isn't "diagonalizable," if you want to put it this way, though I personally don't think it's very good terminology).

Your operator is a rank one perturbation of the multiplication operator $(Mf)(x) = (x^3/2)f(x)$, which has (purely) absolutely continuous spectrum equal to $[0,1/2]$. Since the ac spectrum is invariant under trace class perturbations (so certainly under rank one perturbations), your operator $L$ still has the same ac spectrum, so doesn't even come close to having pure point spectrum (and thus it isn't "diagonalizable," if you want to put it this way, though I personally don't think it's very good terminology).

Source Link
Christian Remling
Christian Remling
  • 24.2k
  • 2
  • 48
  • 83

Your operator is a rank one perturbation of the multiplication operator $(Mf)(x) = (x^3/2)f(x)$, which has (purely) absolutely continuous spectrum equal to $[0,1]$. Since the ac spectrum is invariant under trace class perturbations (so certainly under rank one perturbations), your operator $L$ still has the same ac spectrum, so doesn't even come close to having pure point spectrum (and thus it isn't "diagonalizable," if you want to put it this way, though I personally don't think it's very good terminology).