Compact Operators

We shall use the notion of sequential compactness in a separable Hilbert space. An example would be the closed unit ball $\bar{B}$ in \H, which is not compact unless \H is finite-dimensional. If \H is finite-dimensional, then $\bar{B}$ is compact by the Hein-Borel theorem. Otherwise, \H is infinite-dimensional. Suppose $\{e_n{\}}_{n=1}^{\infty }$ is an orthonormal basis, then no subsequence of the sequence $\{e_n{\}}_{n=1}^{\infty }$ will converge to any point in $\bar{B}$, because $\Vert e_n-e_m\Vert =\sqrt{2}$ for all $n\ne m$.

Compact Operator

Suppose $V$ and $W$ are normed spaces. An operator $T:V\to W$ is compact if all bounded sequences $\{x_n\}$ in $V$ have a subsequence $\{x_{k_n}\}$ such that $\{T(x_{k_n})\}$ converges in $W$. Equivalently, $T$ is compact if the closure of the image of the unit ball in $V$ under $T$ is compact in $W$.

e.g.

• $T$ is compact if $\mathrm{rank}T$ (i.e. dimension of the image) is finite.
• Hilbert-Schmidt operators are compact. (See the corollary)

Proposition

Composition of a compact operator of bounded operator is compact.

Theorem

Suppose T\colon \H\to\H is a bounded linear operator on Hilbert space \H, then

• If $S$ is a compact operator on \H, then $ST$ and $TS$ are also compact.
• If $\{T_n{\}}_{n=1}^{\infty }$ is a family of compact linear operators with $\Vert T_n-T\Vert \to 0$ as $n$ tends to infinity, then $T$ is compact.
• Conversely, if $T$ is compact, there is a sequence $\{T_n\}$ of operators of finite rank such that $\Vert T_n-T\Vert \to 0$.
• $T$ is compact if and only if $T^{\ast }$ is compact.