Properties of the duals of $\ell^1$ and $\ell^{\infty}$

Question

a) True or false:
(i) $(\ell^{1})^* = \ell^{\infty}$
(ii) $\ell^1 \subset (\ell^\infty)^*$
(iii) $(\ell^\infty)^* \subset \ell^1$
(iv) $(\ell^1)^{**} \subset \ell^1$

b) Give the set of dual vectors $j(x) \subset (\ell^1)'$, when $x=(1,1,0,0,0,...) \in \ell^1$.

I found this exercise from an old exam and it seems like a basic problem, one that I probably should be able to solve. However, seeing as I don't seem to completely understand duality I have been struggling with this for a while now.
As for the (i) part I am fairly certain it is true and also found some help for that elsewhere but the other parts are not clear to me at all.

Answer 1

(i) is true indeed, and is a special case of $\ell_p^* = \ell_{p/(p-1)}$ for $1\le p<\infty$. (I prefer $p$ as a subscript, where it does not get in the way of asterisks.)

(ii) is a special case of a general fact: a normed space isometrically embeds into its second dual. In a formula, $\iota:X\to X^{**}$ is an isometric embedding, where $\iota(x)$ is a functional on $X^*$ that acts by evaluating the elements of $X^*$ at $x$.

(iii) and (iv) are false. In both cases, the space on the left is the dual of $\ell^\infty$. Since $\ell^1$ is not reflexive, this dual is strictly larger than $\ell^1$. You may also want to observe that the dual of a nonseparable space is not separable.

Answer 2

What you want is just the dual of $\ell^\infty$. Infact, $(\ell^\infty)*=(\ell^1\times (c_0)^\circ)_1$. Please see the reference "Richard Holmes, Geometric functional analysis and its application" page 129