# How to prove the Identity operator in $l_1$ is bounded but no weakly compact

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