Continuous integral kernels give rise to completely continuous integral operators Let $X$ be a compact space, and $K : X\times X\rightarrow \mathbb{C}$ a continuous function. Then how can I see that the integral operator $T_K : L^2(X)\rightarrow L^2(X)$ given by
$$T_Kf(x) := \int_X K(x,y)f(y)dy$$
is "completely continuous"?
Also, what is the proper definition of "completely continuous"? (I've been taking it to mean a compact operator".
This is a footnote in Gelfand-Piatetskii-Shapiro's book Representation theory and automorphic functions.
 A: From Encyclopedia of Mathmatics:

Definition (Complete continuity of an operator). A bounded linear operator $f$, acting from a  Banach space  $X$ into another space $Y$, that transforms weakly-convergent sequences in $X$ to norm-convergent sequences in $Y$. Equivalently, an operator $f$ is completely-continuous if it maps every relatively weakly compact subset of $X$ into a relatively compact subset of $Y$.

We have to prove that if a sequence $\left(f_n\right)_{n\geqslant 1}$ converges weakly to $0$ in $\mathbb L^2$, then the sequence $\left(\left\lVert T_Kf_n\right\rVert_2\right)_{n\geqslant 1}$ converges to $0$. This can be done using the dominated converge theorem:


*

*weak convergence ensures that for any $x$, the sequence $\left(\left(T_Kf_n\right)\left(x\right)\right)_{n\geqslant 1}$ converges to $0$;

*the combination of Hölder's inequality with boundedness of a weakly convergent sequence gives a dominating function.


However, complete continuity is not equivalent to compactness in general:

It is easy to see that every compact operator is completely continuous, however the converse is false.  For example, recall that the Banach space $X=l_1$ has the Schur Property, that is weak sequential and norm sequential convergence coincide.  It follows that the identity operator from $X$ to $X$ is completely-continuous, but it is not compact since $X$ is  infinite-dimensional.  If $X$ is reflexive, then every completely-continuous operator is compact, so the two classes of operators do coincide in that case.  The term "completely-continuous operator" originally meant what we now call "compact operator", which has sometimes resulted in confusion.

