This is a homework assignment.
I am not sure if my proof is enough to show that $A \in L(X,Y)$. I proved linearity of A. Boundedness is as follows.
Given: $X = l_1 \qquad Y = l_2,$
$A(x1,x2,...) = (y1,y2,...), \qquad \qquad \text{where}\quad y1=x1, \quad y_n=x_n-x_{n-1},\quad n = 2,3,...$
Boundedness of A: $\quad\exists M: ||Ax_n||_2 \leq M||x_n||_1 $
Operator is continuous iff it is bounded:
$\qquad M ||x_n||_1 \rightarrow 0\quad \Rightarrow\quad ||Ax_n||_2 \rightarrow 0$
$ M ||x_n||_1 \rightarrow 0: \quad M\left(\sum_{i=1} ^\infty \xi_i ^n \right) \rightarrow 0$
$||Ax_n||_2 \rightarrow 0: \quad||y_n||_2\rightarrow 0 \Leftrightarrow ||x_n - x_{n-1}||_2 \rightarrow 0 \Leftrightarrow \left(\sum_{i=1}^\infty|\xi_i^n - \xi_i^{n-1}|^2 \right)^{1/2} \rightarrow 0 $
$\left(\sum_{i=0}^\infty|\xi_i^n - \xi_i^{n-1}|^2 \right)^{1/2} \rightarrow 0 \Leftrightarrow \sum_{i=1}^\infty|\xi_i^n - \xi_i^{n-1}| \rightarrow 0 \Leftrightarrow \sum_{i=1}^\infty \xi_i ^n \rightarrow \sum_{i=1}^\infty\xi_i^{n-1} \Leftrightarrow x_n \rightarrow x_{n-1}$