This question is Old but I reckon its still worth answering since I can't find any other related questions. We had this question in a Hilbert space course I am currently doing at Uni. It is in a slightly different form but it should be more or less the same for this question.
Proposition.
let $(a_n)_{n \in \mathbb{N}}\subset \mathbb{C}$ be a fixed bounded sequence. Then the operator
\begin{align*}
T:&\ell_2(\mathbb{N},\mathbb{C}) \rightarrow \ell_2(\mathbb{N},\mathbb{C})\\
&(x_n) \mapsto(a_nx_n).
\end{align*}
Is a bounded linear operator and $||T||_{\text{op}}=\sup_{n \in \mathbb{N}}|a_n|$.
Proof:
Throughout this proof we will endow $\ell_2$ with the usual 2-norm.
We start with Linearity.
let $\lambda \in \mathbb{C}$ and $(x_n)_{n \in \mathbb{N}},(y_n)_{n \in \mathbb{N}}\in \ell_2(\mathbb{N},\mathbb{C})$ then we compute;
\begin{align*}
T(\{\lambda x_n+y_n\}_{n \in \mathbb{N}})&= \{a_n(\lambda x_n+y_n)\}_{n \in \mathbb{N}} & \text{[Definition of $T$]}\\
&=\{a_n\lambda x_n+a_ny_n\}_{n \in \mathbb{N}}\\
&=\{\lambda a_n x_n\}_{n \in \mathbb{N}}+\{a_ny_n\}_{n \in \mathbb{N}} & \text{[Point wise addition]}\\
&=\lambda \{a_n x_n\}_{n \in \mathbb{N}}+\{a_ny_n\}_{n \in \mathbb{N}}\\
& = \lambda T(\{x_n\}_{n \in \mathbb{N}})+T(\{y_n\}_{n \in \mathbb{N}}).
\end{align*}
Thus we can see that $T$ is linear.
Next we show that $T$ is bounded.
Let $(x_n)_{n \in \mathbb{N}},\in \ell_2(\mathbb{N},\mathbb{C})$ be a arbitrary sequence. Then we compute;
\begin{align*}
||T(x_n)||_2 = ||(a_nx_n)||_2 = \sqrt{\sum_{i=1}^{\infty}|a_nx_n|^2}.
\end{align*}
Now The boundedness of $(a_n)$ becomes crucial. Since $(a_n)$ is bounded we can find $\sup_{n \in \mathbb{N}}|a_n|$ (if we knew more about $(a_n)$ we could find this exactly). Now let $M = \sup_{n \in \mathbb{N}}|a_n|$, then we have
\begin{align*}
||T(x_n)||_2 = \sqrt{\sum_{i=1}^{\infty}|a_nx_n|^2} \leq. \sqrt{\sum_{i=1}^{\infty}|Mx_n|^2} = M\sqrt{\sum_{i=1}^{\infty}|x_n|^2} = M||(x_n)||_2.
\end{align*}
Since $(x_n)$ was arbitrary, this is true for all sequences in $\ell_2$. Consequently $T$ is bounded.
Now finally we find $||T||_{\text{op}}$:
Recall that $||T||_op$ is the minimum of the bounds of $T$.
To show that $||T||_{\text{op}}=\sup_{n \in \mathbb{N}}|a_n|$ we need to show the following;
If $M \geq 0$ is a finite constant such that
\begin{align*}
||T(x_n)||_2 \leq M||(x_n)||_2\ \ \ \ \ \ \ \ (1)
\end{align*}
for all $(x_n)\in \ell_2$ then $\sup_{n \in \mathbb{N}}|a_n| \leq M$. So we fix $M \geq 0$ that satisfies $(1)$. Now for all $(x_n) \in \ell_2$ we have
\begin{align*}
||T(x_n)||_2 = \sqrt{\sum_{i=1}^{\infty}|a_nx_n|^2} \leq M\sqrt{\sum_{i=1}^{\infty}|x_n|^2} = M||(x_n)||_2.
\end{align*}
Now this holds in particular when $||(x_n)||_2 =1$. For every $n \in \mathbb(N)$ we define the sequence
\begin{align*}
x_n &= (x_{n,k})_{k \in \mathbb{N}}\\
x_{n,k} &:= \delta_{k,n}.
\end{align*}
where $\delta_{k,n}$ is the Kronecker delta. Then we have
\begin{align*}
||x_n||_2 = \sqrt{\sum_{i=1}^{\infty}|x_{n,i}|^2}=\sqrt{|x_{n,n}|^2} = 1
\end{align*}
thus $x_n \in \ell_2$. Then we have
\begin{align*}
Tx_n &= (a_k x_{n,k})_{k \in \mathbb{N}}\\
Tx_{n,k} &:= a_k \delta_{k,n}.
\end{align*}
Now we also have
\begin{align*}
||Tx_n||_2 = \sqrt{\sum_{i=1}^{\infty}|a_kx_{n,k}|^2}=\sqrt{|a_n|^2} =|a_n|.
\end{align*}
then by the way we chose $M$ we have
\begin{align*}
||Tx_n||_2 \leq M||x_n||_2 \implies |a_n| \leq M.
\end{align*}
However remember that there was nothing special about $n$ so that $|a_n| \leq M$ for all $n \in \mathbb{N}$ and in particular this means $\sup_{n \in \mathbb{N}}|a_n| \leq M$, which proves the result.
Hint: Let $(x_n)=\frac 1 N(1,1,..,1,0,0,...)$ where there are $N$ $1$'s. Then $\|A(x_n)\|=\frac 1 N \sum\limits_{k=1}^{\infty } \frac {a^{k}} {k!}(N-k) =\sum\limits_{k=1}^{\infty }\frac {a^{k}} {k!}(\frac {N-k} N )$. An Application of DCT shows that $\|A(x_n)\| \to e^{a}$ as $N \to \infty$. Hence $\|A\|=e^{a}$.
Best Answer
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 $1$.