$T$ is a linear operator Define $T: l^2 \mapsto l^2$ by $(Tx)_i = \frac{x_i}i \; \forall i=1,\ldots,n\ldots$. Prove that $T$ is a linear operator with $\|T\|=1$.
 A: Linearity :
1. $T((x_i + y_i))= {(x_i+y_i)\over i}={x_i\over i}+{y_i\over i}$.
2. $T((\alpha x_i))={\alpha x_i\over i}=\alpha{x_i\over i}$
Norm : 
Let $||(x_i)||=\sqrt{\sum {x_i^2 }}=1$.
$||(T(x_i))||=\sqrt{\sum {{x_i^2}\over {i^2} }}$
Observe that for each $i$, ${x_i^2\over i^2}<= x_i^2 \implies \sum {{x_i^2}\over {i^2} }<=\sum {x_i^2} = 1$. Therefore $||T||<=1$.  Now consider the sequence $(1,0,0,...)$.$||T((1,0,0,..))||=1$. Therefore $||T||=1$.
A: Well, it's easy to see $T$ is additive, viz. if $(x_i), (y_i) \in l^2$, then
$(x_i) + (y_i) = (x_i + y_i), \tag{1}$
so that
$T((x_i) + (y_i)) = T(x_i + y_i) = (\dfrac{x_i + y_i}{i}) = (\dfrac{x_i}{i}) + (\dfrac{y_i}{i}) = T(x_i) + T(y_i); \tag{2}$
furthermore, for any scalar $\alpha$,
$\alpha(x_i) = (\alpha x_i), \tag{3}$
whence
$T(\alpha (x_i)) = T(\alpha x_i) = (\dfrac{\alpha x_i}{i}) = \alpha (\dfrac{x_i}{i}) = \alpha T(x_i); \tag{4}$
from (2) and (4) we see that
$T(\alpha (x_i) + \beta (y_i)) = T(\alpha (x_i))+ T(\beta (y_i)) = \alpha T(x_i) + \beta T(y_i); \tag{5}$
by (5), $T$ is linear.  As for $\Vert T \Vert$, we have
$\Vert T(x_i) \Vert^2 = \langle T(x_i), T(x_i) \rangle = \langle (\dfrac{x_i}{i}), (\dfrac{x_i}{i}) \rangle = \sum_1^\infty \dfrac{x_i^2}{i^2} \le \sum_1^\infty x_i^2 = \Vert (x_i) \Vert^2, \tag{6}$
whence
$\Vert T(x_i) \Vert \le \Vert (x_i) \Vert \tag{7}$
for all $(x_i) \in l^2$; thus
$\Vert T \Vert \le 1. \tag{8}$
May we in fact have $\Vert T \Vert < 1$?  The answer is clearly, "We may not!", since taking
$(z_i) = (\delta_{1i}) \tag{9}$
yields
$T(z_i) = (z_i) \tag{10}$
with $\Vert (z_i) \Vert = 1$,
and so
$1 = \Vert (z_i) \Vert = \Vert T(z_i) \Vert \le \Vert T \Vert \Vert (z_i) \Vert = \Vert T \Vert; \tag{11}$
it follows that
$\Vert T \Vert = 1. \tag{12}$
QED.
Note:  Wrote this one up last night and fell asleep before posting; when I awoke, the fine answer of Srinivas K (and Plus One, incidentally) was already there; but I decided to post anyway, though it seems some things can only be done one way, in the hope that an even slightly different perspective may clarify further for some.  Fiat Lux! End of Note.
