Compact operator

Remarks

Equivalent form: $T$ is compact if and only if for every bounded sequence $(x_n)$ in $X$, the sequence $(Tx_n)$ has a convergent subsequence in $Y$. A compact operator is bounded, since a compact set is bounded.

Example: on $\ell^2$ let $T$ be the diagonal operator $T(x_1, x_2, \ldots) = (x_1, x_2/2, x_3/3, \ldots)$. The finite-rank operators $T_N$ that keep the first $N$ entries of $Tx$ and set the rest to $0$ satisfy $\norm{T - T_N} \le 1/(N+1)$. A norm limit of finite-rank operators into a Banach space is compact, because the image of the unit ball is then totally bounded; so $T$ is compact.

Non-example: the identity of $\ell^2$ is not compact. The basis vectors $e_n$ lie in the unit ball, and $\norm{e_n - e_m} = \sqrt{2}$ for $n \ne m$, so $(e_n)$ has no convergent subsequence.

The set of compact operators on a Banach space is a closed two-sided ideal of the algebra of bounded operators, the ideal of compact operators in the Hilbert space case.