Boundedness and norm of a sequence operator 
Let $s = \{s_{n}\}_{n=1}^{\infty}$ be a fixed and bounded sequence of real numbers, i.e. $s \in (\ell^{\infty},\|\cdot\|_{\infty})$.  Consider the operator $T_{s} : \ell^{2} \to \ell^{2}$ defined by
  $$ 
    T_{s}(\{a_{n}\}) := \{s_{n}a_{n}\}, \phantom{..} \forall \{a_{n}\} \in \ell^{2}, n \in \mathbb{N}.
$$
  Show that $T_{s}$ is a bounded linear operator and find $\|T_{s}\|$.

I am seeking hints or tips. Thank you.
 A: Linearity is quite trivial to show. So we only focus on proving boundedness.
Recalling the definition, we only need to show that for all $x\in\ell^2$ s.t. $\|x\|=1$ ($\ell^2$ norm), there exists $M>0$ s.t. $\|T_s x\|\le M$. Then by your definition of $T$, we have that
$$\|T_s x\|=\|\{s_n x_n\}\|=\left(\sum (s_n x_n)^2\right)^\frac12\le \left(\sum \|s\|^2_{\infty}(x_n)^2\right)^\frac12=\|s\|_{\infty}\left(\sum (x_n)^2\right)^\frac12=\|s\|_{\infty}\cdot 1.$$

EDIT We can also show that $\|s\|_\infty$ is exactly the value of $\|T_s\|$:

1). if $\|s\|_\infty$ is attained, i.e., exists some $k$ such that $|s_k|=\|s\|_\infty$. Then $x_n:=\chi_{\{k\}}(n)$ obviously satisfy that $\|x\|=1$ and $\|Tx\|=\sqrt {s_k^2}=\|s\|_\infty.$
2). otherwise, $s_n<\|s\|_\infty$ but there exists a subsequence of $s$, say $\{s_{N,k}\}_{k\in\Bbb N}$ s.t. $s_{N(k)}\to \|s\|_\infty$ as $k\to\infty$. Let $\{(x_m)\}\subset\ell^2$ s.t. $(x_m):=\{\chi_{\{N(m)\}}(n)\}_{n\in\Bbb N}$, then 
$$\|T_sx_m\|=\sqrt {s^2_{N(m)}}\to\|s\|_\infty,\quad\text{as}\,\, m\to\infty.$$
Either case, we have that
$$\|T_s\|:=\sup_{\|x\|=1}\|T_sx\|\ge \|s\|_\infty.$$
But $\|s\|_\infty$ as previously illustrated is also an upper bound for $\|T_s\|$, it follows that $\|T_s\|= \|s\|_\infty$.
