# Does a compact operator always have a kernel?

I am sorry if this question is stupid..... I raise it when I read Lax's book Functional Analysis. We know that some integral operators are compact, for example an integral operator from $L^2[Y]$ to $L^2[X]$ defined by $$(Kf)(x) = \int{K(x,y)f(y)dy}$$ is compact if $X$ and $Y$ are compact space and the kernel is square integrable. But how about the converse question? Does any compact operator from $L^2[Y]$ to $L^2[X]$ always has a kernel $K(x,y)\in{L^2}$? If not, under what conditions this compact operator have such kernel?

Moreover, how about other spaces, such as $C(X)$, $L^p$ space? (For genreal $L^p$ space, I am not sure this question is meaningful, since the above integral may diverge....)

Can anyone give me a hint or any references? Thank you!

If you insist that $K$ must be square-integrable, then even on $L_2$ this is not true. Indeed, every such integral operator must be Hilbert–Schmidt but there are plenty of compact operators which are not Hilbert–Schmidt. (For instance any diagonal operator with respect to a fixed orthonormal basis whose eigenvalues are not (square) summable but tend to 0.) See a related question here.