The norm of linear functional $x\mapsto \sum_{n=1}^{\infty} \frac{x_n}{2^n}$ on $c_0$

Question

Consider the mapping $\phi :c_0 \to \mathbb{R}$ defined by $\sum_{n=1}^{\infty} \frac{x_n}{2^n}$. Compute $\|\phi\|$ Does there exist a $x \in c_0$ such that $\|x\|=1$ and $\|\phi\|=|\phi(x)|$

Clearly $\|\phi\|=\sup_{\|x\|=1} |\phi(x)|=\sup_{\|x\|=1} \sum_{n=1}^{\infty} \frac{x_n}{2^n} \leq \sup_{\|x\|=1} \sum_{n=1}^{\infty} \frac{1}{2^n}=1$

Now from above we have that $\|\phi\|\leq 1$ and as $x \in c_0$ we can choose $x_n=1$ for n as large as we want and get |$\phi(x)|$ as close as we want to $1$ . But then since $\lim_{n \to \infty} x_n=0$ , it is impossible that

$\|x\|=1$ and $\|\phi\|=|\phi(x)|$

Question:Does my argument make sense and if yes how could I write it down more rigorously?

Edit : I am sorry, I was sloppy. $c_0=\{\text{Space of all sequences whose limit goes to zero}\}$ and I am using the $\infty- norm$

Answer 1

Your argument is right. To express it more rigorously, you can do something like this: for every $\|x\|<1$, $$ \sum_{n=1}^{\infty} \frac{x_n}{2^n} < \sup_{\|x\|=1} \sum_{n=1}^{\infty} \frac{1}{2^n}=1$$ where the inequality is strict because $x_n\le 1$ for all $n$ and $x_n<1$ for some $n$.

On the other hand, the vector $x = (1,\dots,1,0,\dots)$ with $N$ ones has $ \sum_{n=1}^{\infty} \frac{x_n}{2^n} = 1-2^{-N}$, which can be arbitrarily close to zero.

Together, the above paragraphs show that the norm is $1$ and it is not attained.