Problem: Let $l^2$ denotes the set of all sequences which is square summable, that is, \[ l^2 = \left\{ \left( x_n \right) : \sum_n \vert x_n \vert^2 < \infty \right\} . \] Let the continuous linear operator $T:l^2 \to l^2$ be defined by \[ T\left( x_1,x_2,\dots \right) = \left( 0,x_1,0,x_3,0,x_5,\dots \right) . \] Then show that $T^2 = T\circ T$ is compact whereas $T$ is not compact.
Solution: We recall that an operator on a Banach space $X$ is compact if and only if for every bounded sequence $\left( x_n \right) _{n\in \mathbb{N} }$ the image $\left( T\left( x_n \right) \right) _{n\in \mathbb{N} }$ has a convergent subsequence. Let $e_i = \left( 0,0,\dots,1,0,\dots \right) $, where $1$ is at the $i^{\text{th}}$ place. It is clear that $e_i\in l^2$. We take the sequence defined as \[ x_n = e_{2n-1},~n\in \mathbb{N} . \]
Since $\lVert e_n \rVert = 1$, the sequence is bounded. But observe that the image sequence is not Cauchy and hence it can not have a convergent subsequence. \begin{align*} T\left( x_1 \right) & = \left( 0,1,0,0,0,0,\dots, \right) = e_2 \\ T\left( x_2 \right) & = \left( 0,0,0,1,0,0,\dots, \right) = e_4 \\ \vdots \\ T\left( x_n \right) & = \left( 0,0,\ldots, 1,\ldots , \right) = e_{2n}. \end{align*} Note that \[ \lVert T\left( x_m \right) - T\left( x_n \right) \rVert = \lVert e_{2m} - e_{2n} \rVert = \sqrt{2}, \] and hence the sequence $\left( T\left( x_n \right) \right)$ is not Cauchy. Thus, $T$ is not compact. \vf \hf On the other hand \[ T^2 (x_1,x_2,x_3,\dots,) = T(0,x_1,0,x_3,0,\dots) = (0,0,0,\ldots,). \] So, $T^2 = 0$ and hence it is a compact operator.