I am trying to solve an exercise from An Introduction to Banach Space Theory by R. E. Megginson, but I am a little bit stuck. It is Exercise 1.114:
1.114. Characterize the elements of $c_0^*$ that are norm-attaining. Conclude that the norm-attaining functionals form a dense subset of $c_0^*$.
In this context, $c_0$ is the sequence space of bounded functions that converges to $0$, viewed as a Banach space with the supremum norm. $c_0^*$ stands for the continuous dual of $c_0$, i.e., linear and bounded functionals over the field $K$, which is either $\mathbb{R}$ or $\mathbb{C}$.
Previously on the text, it is shown that there exists elements on $c_0^*$ that are not norm-attaining (in particular, $c_0$ is not reflexive.) Additionally, we already know that $l_1$ is isometrically isomorphic to $c_0^*$ as Banach spaces. This isomorphism is given by $$T:l_1\longrightarrow c_0^*,\qquad T(x)=\varphi : c_0\longrightarrow K, \quad x=(x_n)\in l_1,$$ where $$\varphi(y)= \sum_n x_ny_n,\qquad y=(y_n)\in c_0.$$ Thus, we can identify any element $\varphi\in c_0^*$ with a sequence $(x_n)\in l_1$; we write $\varphi=(x_n)$ and $c_0^*=l_1$ with a slightly abuse of notation.
My attempt: In the Example 1.10.4 it is proved that the functional $\varphi=(x_n)\in l_1$, with $x_n=2^{-n}$, is not norm-attaining. Indeed, $||\varphi||_{op}=\sum_n2^{-n}=1$, but for any $(\alpha_n)\in B_{c_0}$ (the unitary ball of $c_0$) we have $$ |\varphi((\alpha_n))| = |\sum_n 2^{-n}\alpha_n| \leq \sum_n2^{-n}|\alpha_n|<\sum_n 2^{-n}=1.\qquad\qquad(1)$$ So $\varphi$ is not attaining its norm.
From this example, my guess is that $\varphi =(x_n)$ is norm-attaing if and only if $||\varphi||_{op}>1$.
Attempt of Proof. If $||\varphi||_{op}\leq 1$, a computation similar to that in (1) implies that $\varphi$ is not attaining its norm. Conversely, and here is where I am stuck, given $\varphi\in c_0^*$, I would like to exhibit a sequence $(\alpha_n)\in B_{c_0}$ such that $||\varphi||_{op}=\varphi((\alpha_n))$. I would be grateful if any of you provide me a hint or a direction to build such a sequence $(\alpha_n)\in c_0$.
I do not have any progress with the second part of the exercise.
Thank you all.
$\varphi=(x_n)$ is norm attaining if and only if $x_n=0$ for all $n$ sufficiently large.
Proof: If $x_n=0$ for all $n$ sufficiently large then norm of $\varphi$ is attained at $(\alpha_n)$ where $\alpha_n= \frac {\overline {x_n}} {|x_n|}$ when $x_n \neq 0$ and $0$ when $x_n=0$.
Suppose norm is attained. Then $\sum |x_n|=\sum x_ny_n$ for some $(y_n) \in c_0$ with $|y_n| \le 1$ for all $n$. But then, $\sum |x_n|=\sum x_ny_n\le \sum |x_n||y_n|\le \sum |x_n|$. You will get strict inequality (which is a contradiction) unless $|y_n|=1$ for any $n$ with $x_n \ne 0$. But remember that we need $y_n \to 0$. Hence, this cannot happen unless $x_n=0$ for all $n$ sufficiently large.