While studying weakly compact operators the book proves that every compact operator is weakly compact and also that every weakly compact operator is bounded, I got it, ad they conclude those results place weak compactness between compactness and boundedness as a property for linear operators between Banach spaces.

As counterexamples they say the identity operator in $l_1$ is a bounded operator ( I can see this) but isn't weakly compact (not so sure how to prove that) and that the identity operator in $l_2$ is weakly compact (why?) but isn't bounded.

The definition for a weakly compact operator is the following:

Let X,Y be Banach spaces, we say $T:X\longrightarrow Y$ is weakly compact, if for every bounded set $B\subset X$ , $T(B)$ is relatively weakly subset of $Y$.

Any guide in how to prove this? I don't want the full answer, thank you.


First, to prove (weak) compactness of an operator $T:X \rightarrow Y$, you just need to show that $T(B_X)$ is relatively (weakly) compact in $Y$.

I will first comment on the $\ell_2$ case. I think you meant that the identity operator on $\ell_2$ is weakly compact but not compact (as it is obviously bounded). So I will show that.

As $I(B_{\ell_2}) = B_{\ell_2}$, $I$ cannot be compact since $\ell_2$ is infinite-dimensional. However, $\ell_2$ is a Hilbert space and hence reflexive, so $B_{\ell_2}$ is weakly compact by Banach-Alaoglu theorem and $I$ is weakly compact.

The $\ell_1$ case is similar. $I$ is obviously bounded but $B_{\ell_1}$ is not weakly compact as $\ell_1$ is not reflexive, therefore $I$ is not even weakly compact.

  • $\begingroup$ Yeah, the identity in $l_2$ isn't compact. Thank you now I got it. $\endgroup$ – ipreferpi Sep 27 '19 at 20:52

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