Let $c_o$ be the set of the sequences who converge to $0$. Consider the map
$$\varphi:c_o\rightarrow \mathbb{R}\ \ \text{or} \ \ \mathbb{C}$$ Such that for every $(a_j)_j \in c_o$ $$\varphi((a_j)_j)=\sum_{j=1}^{\infty} \frac{a_j}{2^j}$$
I want to compute the norm of the operator $\varphi$ and show that $||\varphi||_{op}=1$, i.e.:
$$\sup_{x\in c_o}\frac{|\varphi(x)|}{||x||_{\infty}}=1$$
I`m kinda stuck in how to proceed. Also, the exercise does not say that the norm in $c_o$ is $|| \ \ ||_{\infty}$, is it likely to be the right one?
Indeed the norm in $c_0$ is the $|\!|\cdot|\!|_{\infty}$ norm. Clearly the norm of your functional is at most $1$ because if $(a_j)\in c_0$ is an element whose norm is at most $1$, then $$\left|\sum_{j=1}^{\infty}\frac{a_j}{2^j}\right|\leq\sum_{j=1}^{\infty}\frac{1}{2^j}=1$$ to see that the norm is precisely $1$, let $\varepsilon>0$. Choose $N$ sufficiently large so that $\sum_{j=1}^N\frac{1}{2^j}>1-\varepsilon$. Let $a_0=(a_j)\in c_0$ be the element whose first $N$ entries are $1$ and all the rest are zeros. Then the norm of $\varphi(a_0)$ is greater than $1-\varepsilon$. Since $\varepsilon$ was arbitrary, this proves that the norm of $\varphi$ is not smaller than $1$.