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Consider the linear operator $T: \ell_2\to \ell_2$ defined by for $x=x=(\xi_1,\xi_2,\dots)\in \ell_2$ $$ Tx=(0,\xi_1, 0, \xi_3, 0, \xi_5, 0,\dots, ) $$ Clearly, $T$ is bounded linear operator on $\ell_2$.

I try to prove that $T$ is not compact but $T^2$ is compact operator.

If we take $x_1=(1,0,0,\dots), x_3=(0,0,1,0,0,\dots),\dots, x_{2n-1}=(0,\dots, 0, 1,0,\dots)$ where $1$ appears on the $2n-1$-th coordinate. Then $\{x_{2n-1}\}\in \ell_2$ and is a bounded sequence in $\ell_2$.

But I am stuck on how to show that $\{T x_{2n-1}\}$ does not has a convergent subsequence...

I can show that it is not Cauchy: for $n\neq m$, $$ \|Tx_{2n-1}-Tx_{2m-1}\|=\sqrt{2} $$

Moreover, $T^2x=0$. Because $rank(T^2)=0$, $T^2$ is then compact.

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  • $\begingroup$ What you're showing is much stronger than "the sequence is not Cauchy"! you're saying that the distance between any two distinct terms of the sequence is constant. In particular no subsequence is Cauchy and thus there are no convergent subsequences. $\endgroup$
    – Michh
    Commented Dec 4, 2023 at 0:31
  • $\begingroup$ What I don't understand is why $T^2x = 0$ rather than $T^2x = Tx$? $\endgroup$
    – Michh
    Commented Dec 4, 2023 at 0:33
  • $\begingroup$ @Michh But one sequence $\{x_k\}$ is not Cauchy, it could have a convergent subsequence ,right? $\endgroup$
    – Hermi
    Commented Dec 4, 2023 at 0:35
  • $\begingroup$ @Michh That because $T^2x=T(0,\xi_1,0,\xi_3,\dots)=(0,0,\dots)$. $\endgroup$
    – Hermi
    Commented Dec 4, 2023 at 0:36
  • $\begingroup$ Read my comment carefully. If you take any subsequence it will not be Cauchy because of the equality you wrote $\endgroup$
    – Michh
    Commented Dec 4, 2023 at 0:38

2 Answers 2

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Assume by contradiction that there is a convergent subsequence $Tx_{n_i}$. In particular this subsequence is Cauchy. Then for $\epsilon = 1$, there exists $i_0$ such that for $i,j\geq i_0$, $\lVert Tx_{n_j}- Tx_{n_i}\rVert \leq 1$. This contradicts your equality. Thus there is no convergent subsequence.

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  • $\begingroup$ So to say one operator $T$ is not compact, we just need to find one bounded sequence $\{x_k\}$ so that $\{Tx_k\}$ does not have a convergent subsequence? $\endgroup$
    – Hermi
    Commented Dec 4, 2023 at 0:53
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    $\begingroup$ Yes this is how you negate "$T$ is compact". @Hermi $\endgroup$
    – Michh
    Commented Dec 4, 2023 at 0:55
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Consider the operator $$ Sx=(\xi_2, \xi_3, \xi_4,\dots, ) $$ Then $$STx=(\xi_1, 0, \xi_3, 0, \xi_5, 0,\dots, ) $$ The operator $ST$ is thus the orthogonal projection onto infinite dimensional subspace consisting of sequences which vanish at coordinates with even indices. Clearly any infinite dimensional projection is not a compact operator.

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  • $\begingroup$ For $x=(\xi_1,\xi_2,\xi_3,\dots)$, $STx=S(0,\xi_1,0,\xi_3,0,\xi_5,\dots)=(\xi_1, 0, x_3,\dots)$, right? Why do we have $ST=I$? $\endgroup$
    – Hermi
    Commented Dec 4, 2023 at 4:54
  • $\begingroup$ The operator $ST$ restricted to $\ell^2(2\mathbb{N}-1),$ i.e. sequences vanishing at even coordinates, acts as identity. $\endgroup$ Commented Dec 4, 2023 at 7:10

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