Finding the norm of $(x_n) \mapsto \sum x_{n}2^{-n}$ in $(\ell^\infty)^*$

Question

The question says: for $x \in \ell^{\infty}$, let $f(x)=\sum_{n \in \mathbb{N}}x_{n}2^{-n}$. Determine the norm of $f$ in $({\ell^{\infty}})^{*}$.

I have tried to use the definition of the dual space norm. I have written down that

$$\|f\|= \sup\left\{\frac{|\sum_{n \in \mathbb{N}}x_{n}2^{-n}|}{\sup\{x_n\}}\right\},$$ where the outer supremum is taken over all $x$ in $\ell^{\infty} \setminus \{0\}$.

I just don't know where to go from here.

Answer 1

First of all, for all $x\in\ell^{\infty}$ we have:

$|f(x)|=|\sum_{n=1}^\infty x_n2^{-n}|\leq\sum_{n=1}^\infty |x_n|2^{-n}\leq\sum_{n=1}^\infty ||x||_{\infty}2^{-n}=||x||_{\infty}\sum_{n=1}^\infty 2^{-n}=||x||_{\infty}$

And so $||f||\leq 1$. Also, let $y=(1,1,1,1,...)$. Then $||y||_{\infty}=1$, and so:

$||f||\geq\frac{|f(y)|}{||y||_{\infty}}=|f(y)|=\sum_{n=1}^\infty 2^{-n}=1$

So the norm of $f$ is $1$.