How to prove that $A \in B(\ell^1 , \ell^2 )$ and compute $\|A\|$ Consider the Banach spaces
$\ell^1 = \{x = (x_1, x_2,\dots) : \sum_{i=1}^{\infty}|x_i| < \infty, x_i \in C \text{ for all } i\}$,
$\ell^2 = \{x = (x_1, x_2,\dots) : \sum_{i=1}^{\infty}|x_i|^2 < \infty, x_i \in C \text{ for all } I\}$,
which are equipped with the norms $\|x\|_1 = \sum_{i=1}^{\infty}|x_i|$ and $\|x\|_2 = \sqrt{\sum_{i=1}^{\infty}|x_i|^2}$, respectively.
For each $x \in \ell^1$ define
$$A(x) = (x_1,1/2(x_1 + x_2),1/4(x_1 + x_2 + x_3),1/8(x_1 + x_2 + x_3 + x_4),\dots)$$
Prove that $A \in B(\ell^1,\ell^2)$ and compute |A|.
I have try using Jensen inequality to prove that $A \in B(\ell^1,\ell^2)$, but I have not reached any conclusions
 A: By definition, showing that $A$ is a bounded operator from $\ell^1$ to $\ell^2$ consists of finding a positive constant $C > 0$ such that the following holds:
$$ \| A(x) \|_{\ell^2} \le C\|x\|_{\ell^1}, \quad \forall x \in \ell^1. $$
Now, unravel the definition of $A$ to find that, for $x = (x_1, x_2, \ldots) \in \ell^1$,
\begin{align} 
    \| A(x) \|^2_{\ell^2} &= x_1^2 + \frac{1}{4}(x_1 + x_2)^2 + \frac{1}{16}(x_1 + x_2 + x_3)^2 + \cdots \\
    &\le (1 + \frac{1}{4} + \frac{1}{16} + \cdots)(x_1 + x_2 + x_3 + \cdots)^2 \\
    &= (1 + \frac{1}{4} + \frac{1}{16} + \cdots)\|x\|^2_{\ell^1}.
\end{align}
The conclusion follows by taking the square root on both sides (as $1 + 1/4 + 1/16 + \cdots < +\infty$).
Regarding the norm of the operator $A,$ by definition
$$ \| A \|_{B(\ell^1, \ell^2)} = \sup_{\| x \|_{\ell^1} \le 1} \| A(x) \|_{\ell^2}.$$ Now try to play around with various $x$. Bear in mind that for $x$ to be in $\ell^1$, this means that somehow the tail of the sequence has to decrease as a certain rate. Compare this rate of decay with the decay of $(1, 1/2, 1/4, \cdots)$ to understand which terms will contribute the most in the norm, and ultimately find the $x$ achieving this supremum.
