# kronecker product of integral operator

I'm wondering whether can we define an explicit Kronecker product of Hilbert–Schmidt integral operators. $T(f)=\int k(x,y)f(y)dy$.

Since integral operator is an extension of matrix operation, I'm curious whether there's a corresponding concept of Kronecker product.

Let's say for example that the domain you are looking at is the unit interval $[0,1]$, then these operators are linear maps from $L^2([0,1])$ to itself. Hence the tensor product of two such maps should be a linear map from $L^2([0,1]) \otimes L^2([0,1])$ to itself. Explicitly the map just sends $f_1 \otimes f_2$ to $T_1(f_1) \otimes T_2(f_2)$.
Now let's identify $L^2([0,1]) \otimes L^2([0,1])$ with $L^2([0,1]^2)$ in the natural way. That is we send $f_1 \otimes f_2$ to the function $f_1(x)f_2(y)$ on $[0,1]^2$, the span of such functions is dense in $L^2([0,1]^2)$ so this identification makes sense when we use the typical completed tensor product for Hilbert spaces. Now by working it out explicity for functions $f(x,y) = f_1(x)f_2(y)$ and extending linearly we get that our map is just: $$T_1 \otimes T_2(f)(x,y) = \int \int k_1(x,t)k_2(y,z)f(x,y)dtdz$$ which we can recognize as another Hilbert-Schmidt integral operator, but this time it's on $L^2([0,1]^2)$.