Let $\phi:c_0\to\mathbb{K}$ given by $\phi\left((a_j)_{j=1}^\infty\right)=\sum_{j=1}^\infty\frac{a_j}{2^j}$. In this exercise $$c_0:=\{(\xi_j)_{j=1}^\infty\subset\mathbb{K} : (\xi_j)_{j=1}^\infty\hspace{.1cm}\mbox{ converges to zero}\},$$
and $\mathbb{K}$ it represents $\mathbb{R}$ or $\mathbb{C}$.
One of my goals is to show that $\phi\in(c_0)'$, where $(c_0)'$ is the dual of $c_0$. For this I must show that $\phi$ is bounded, therefore it will be continuous and therefore $\phi$ will be in $(c_0)'$.
Here are my concerns:

*

*In these cases I must prove that the operator is well defined. This is one of the points that I have not resolved yet and in which I have doubts about how to approach it.


*I estimate that $\|\phi\|=1$. One of the inequalities I must test is obvious (this is $\|\phi\|\geq 1$), which I intend to demonstrate as follows: consider the succession $(\xi_j)_{j=1}^\infty$ defined by $$\xi_j =
\left\{
\begin{array}{ll}
1, & \mbox{si } j<N  \\
0, & \mbox{si } j\geq N,
\end{array}
\right.$$
for some $N\in\mathbb{N}$. Then $$\|\phi\| \geq |\phi(\xi_j)|=\left|\sum_{j=1}^\infty \frac{\xi_j}{2^j}\right|=\left|\sum_{j=1}^{N-1} \frac{1}{2^j}\right|\geq 1.$$
It would remain to prove that $\|\phi\|\leq 1$, in which I request a suggestion.


*Finally, my last question and the one I really need more help with is how to prove that there is no $ x\in c_0$ such that $ \|x\| \leq 1$ and $\|\phi\|=|\phi(x)|$.
Thanks for your attention. Greetings.
 A: I assume that $c_0$ is being given the supremum norm, that is
$$
\|x\|:=\sup\{|x_n|:n\in\mathbb N\}
$$
First, we check that $\phi$ is well-defined, i.e. that the series $\sum_n \frac{x_n}{2^n}$ converges whenever $x\in c_0$. To see this, note that, if $x\in c_0$, then in particular $\|x\|<\infty$.  For each $n$, $|x_n|\le \|x\|$, so that
$$
\sum_n \left|\frac{x_n}{2^n}\right|\le \|x\|\sum_n \frac{1}{2^n}<\infty
$$
Remark: $\|x\|<\infty$ is all that is needed for this to work.

Now, we check that $\|\phi\|=1$. Fix $n\in \mathbb{N}$. Let $y\in c_0$ be defined by $y_k=1$ for $k\le n$ and $y_k=0$ for $k>n$. Obviously, $\|y\|=1$. Then
$$
\|\phi\|\ge |\phi(y)|=\sum_{k=1}^n\frac{1}{2^k}
$$
Therefore
$$
\|\phi\|\ge\sum_{k=1}^n\frac{1}{2^k}
$$
for every $n$. letting $n\to\infty$, we get that
$$
\|\phi\|\ge \sum_{k=1}^\infty\frac{1}{2^k}=1
$$
so that $\|\phi\|\ge 1$.
To see that $\|\phi\|\le 1$, fix $x\in c_0$ with $\|x\|\le 1$.  Then $|x_n|\le 1$ for every $n$, and so
$$
|\phi(x)|=\left|\sum_{n=1}^\infty \frac{x_n}{2^n}\right|\le \sum_{n=1}^\infty\left|\frac{x_n}{2^n}\right|\le\sum_{n=1}^\infty \frac{1}{2^n}=1
$$
Thus $|\phi(x)|\le 1$ for every $x\in c_0$ with $\|x\|\le 1$. This implies that $\|\phi\|\le 1$
