Integral operator

Remarks

Example: on $X = Y = [0, 1]$ with Lebesgue measure, the kernel $K(x, y) = 1$ for $y < x$ and $K(x, y) = 0$ for $y \ge x$ gives the Volterra operator $(Vf)(x) = \int_0^x f(y)\, dy$. It is defined for every integrable $f$, because $\int_0^1 \abs{K(x, y) f(y)}\, dy \le \int_0^1 \abs{f}$. Non-example: the identity map of $L^2([0,1])$ is linear but is not an integral operator with a function kernel. Such a kernel would give $\int_A K(x, y)\, dy = 1$ for almost every $x \in A$ and $0$ for almost every $x \notin A$, for every interval $A$. This would force the measure $K(x, y)\, dy$ to be a point mass at $x$, which is impossible. If $K$ is in $L^2(X \times Y)$, the operator maps the Lebesgue space $L^2(Y)$ to $L^2(X)$ and is bounded with norm at most $\norm{K}_{L^2}$. Such operators are called Hilbert–Schmidt integral operators, and they are compact.