Norm of $f$ in a dual space space $(\ell^{\infty})^{*}$ For $x \in \ell^{\infty}$, let $f(x)= \sum_{n \in \mathbb{N}}x_{n}2^{-n}$ determine the norm of $f$ in $(\ell^{\infty})^{*}$ (the dual space of $(\ell^{\infty})$.
Notes: I think I need to related this back to the fact that $(\ell^{\infty})^{*}$ is isometrically isomorphic to $\ell^{1}$ so I know there is some map T such that the norm of T$x$ in the dual space is the norm of $x$ in $\ell^{1}$, but I'm not sure how to relate this back to find the norm of $f$.

We know that $f$ is the image of $(a_n)$, where $a_n=2^{-n}$, under the isometric embedding $i:\ell^1\to (\ell^\infty)^*$, so we simply need to compute $\|f\|_1$. This is $1$.
Alternatively, we know that $\|f\|\ge |f(1)|=1$, and then you just need an upper bound by $1$.
