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Functionla Analysis separability

  1. Apr 13, 2014 #1
    Let T: l^2 -> l^2 be bounded linear operators. K=L(l^2,l^2) be the space of T, Prove that K=L(l^2,l^2) is not separable

    I know that if a space contains an uncountable number of non intersecting open balls then it is not separable. But how can I apply this statement here ( I mean how to construct such open balls) And are there any easier way to do it ???
     
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  3. Apr 13, 2014 #2

    dextercioby

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    I don't know a direct proof of the result, but I have to ask which topology you're considering on K.
     
  4. Apr 13, 2014 #3

    micromass

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    A diagonal operator has the form

    [tex]T(x_1,x_2,x_3,...) = (\alpha_1 x_1, \alpha_2 x_2, \alpha_3 x_3, ...)[/tex]

    where ##\alpha_k\in \mathbb{C}##.

    It can easily be checked that ##T## is bounded if and only if the sequence ##(\alpha_k)_k## is bounded. Furthermore, we have ##\|T\|= \|(\alpha_k)_k\|_\infty##. This yields an isometric embedding ##\ell^\infty\rightarrow \mathcal{B}(\ell^2)##. Thus it suffices to show ##\ell^\infty## is not separable, which is well-known.
     
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