# compact operator

Let $X$ and $Y$ be two Banach spaces. A (completely continuous operator) is a linear operator $T\colon X\to Y$ that maps the unit ball in $X$ to a set in $Y$ with compact closure. It can be shown that a compact operator is necessarily a bounded operator.

The set of all compact operators on $X$, commonly denoted by $\mathbb{K}(X)$, is a closed two-sided ideal of the set of all bounded operators on $X$, $\mathbb{B}(X)$.

Any bounded operator which is the norm limit of a sequence of finite rank operators is compact. In the case of Hilbert spaces, the converse is also true. That is, any compact operator on a Hilbert space is a norm limit of finite rank operators.

###### Example 1 (Integral operators)

Let $k(x,y)$, with $x,y\in[0,1]$, be a continuous function. The operator defined by

 $(T\psi)(x)=\int_{0}^{1}k(x,y)\psi(y)\,\mathrm{d}y,\qquad\psi\in C([0,1])$

is compact.

Title compact operator CompactOperator 2013-03-22 14:26:59 2013-03-22 14:26:59 mhale (572) mhale (572) 8 mhale (572) Definition msc 46B99 completely continuous