Find $\|T\|_{\ell^\infty \rightarrow \ell^2}$ Let $(v_n)_{n \geq 1} \in \ell ^2$ be a fixed bounded sequence of real numbers. Define a mapping $T$ on $\ell^\infty$ using the formula $$T(a_1, a_2,\ldots)=(v_1a_1,v_2a_2,\ldots), \qquad x=(a_1,a_2,\ldots) \in \ell^\infty$$
We can use the fact that, $Tx \in \ell^2 $ for every $x \in \ell^{\infty}$ and $T: (\ell^\infty,\|\cdot \|_\infty) \rightarrow (\ell^2,\|\cdot \|_2)$ is a linear operator and $\|Tx\|_2 \leq \|(v_n)\|_2 \cdot \|x\|_\infty$ for all $x \in \ell^\infty$

$$||T||_{\ell^{\infty} \rightarrow \ell^2}= \sup _{*}||Tx||_{\ell^2}$$
Where $*$ can be $||x||_{\ell^{\infty}}$ $\leq 1$ or $<1$ or $=1$ according to my notes. There is also the infimum definition of it. I don't know which one is best to use but I also have no idea how to evaluate it because of stuff like $||x||_{\ell^{\infty}}$, I have no idea what the subscript would mean because it is a set - not a number (or infinity).
 A: Since $\|T\|=\sup\{\|Tx\|_2:~\|x\|_\infty\leq 1\}$, by the above inequality we get $\|T\|\leq \|(v_k)\|_2.$ On the other hand, note that for $x=(1,1,\ldots)\in \ell_\infty$, we have $\|x\|_\infty=1$ and $Tx=(v_1,v_2,v_3,\ldots)$, so $\|Tx\|_2^2=\sum_k v_k^2\leq \|T\|^2$. Therefore $\displaystyle \|T\|=\Big(\sum_k v_k^2\Big)^{1/2}=\|(v_k)\|_2$.
EDIT.  Most of these exercises, that ask you to find the norm of a bounded linear operator $T$, consist of two parts. The first is to prove that $T$ is indeed a bounded linear operator, by finding $M>0$ such that 
$\|Tx\|_Y\leq M\|x\|_X$ for all $x\in X.$ This implies that $\|T\|=\sup\{\|Tx\|_Y:~\|x\|_X\leq 1\}\leq M$. Now comes the second part: if your inequalities are "good" enough, then $M$ will be the norm of $T$. To prove this you need to show that $\|T\|\geq M.$ And we come to the most "difficult" part of the whole task: you have to find an element $x_0\in X$ of norm $\|x_0\|_X\leq 1$ (and when we say find, we mean to really find!) such that $\|Tx_0\|_Y\geq M$. Then $\|T\|\geq \|Tx_0\|_Y\geq M.$ (note: such an element does not always exists, but in most exercises you can hope to find one).
