Is there an example of a non compact linear operator $T$ from a Banach space $X$ to itself such that $T^2$ is compact? Of course the converse is true, as $T^2$ is compact if $T$ is. Here $T^2$ means $T$ composite with $T$.

  • $\begingroup$ please provide more concrete information, i.e. an example or an actual problem where this would be an issue. $\endgroup$ May 20, 2015 at 1:55
  • $\begingroup$ @JonasMeyer sorry for the mistake. $\endgroup$ May 20, 2015 at 2:00
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    $\begingroup$ Try for $T^2=0$. $\endgroup$ May 20, 2015 at 2:05
  • $\begingroup$ @JonasMeyer I just could not find such an example on an infinite dimensional space $\endgroup$ May 20, 2015 at 2:15
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    $\begingroup$ You can find an example of $T^2=0$ on a two dimensional space, and extend the idea to operators on $Y\oplus Y$. $\endgroup$ May 20, 2015 at 2:41

1 Answer 1


Let our Banach space be $\ell^1$, which is the set of all $a \in \mathbb{R}^{\mathbb{N}}$ such that $\sum_n |a_n|<\infty$. The norm is the $L^1$ norm.

For $i \in \mathbb{N}$, let $e_i \in \ell^1$ denote the $i^{th}$ standard basis vector, i.e, the sequence whose $i^{th}$ index is $1$, but all other indices are $0$.

Consider the linear map $T: \ell^1 \to \ell^1$ which sends $e_i \mapsto e_{i+1}$ when $i$ is odd, and sends $e_i \mapsto 0$ when $i$ is even. Equivalently, you can say $T(a_1,a_2,a_3,...) = (0,a_1,0,a_3,...)$. Notice that $T$ is bounded since $\|Tv\|_1 \leq \|v\|_1$ for all $v \in \ell^1$.

Then $T^2 = 0$, so $T^2$ is compact.

However, $T$ itself is not compact. Indeed, the sequence $T(e_{2i+1})=e_{2i+2}$ has no Cauchy subsequence in $\ell^1$, since $\|e_{2i+2}-e_{2j+2}\|_1 = 2$ whenever $i \neq j$.


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