Consider the $\mathbb{R}$-vector space of sufficiently nice real-valued functions on the unit square $I^2$, where "sufficiently nice" could be taken to mean any one of a number of things - say continuous for now.

In analogy with matrix multiplication, we can define the product of two such functions $F$ and $G$ as

$$(F\times G)(i,j) = \int_0^1F(i,t)G(t,j)dt.$$

We can check immediately that this operation is associative (the proof is exactly the same):

$$((F\times G)\times H)(i,j) = \int_0^1(F\times G)(i,t)H(t,j)dt$$ $$=\int_0^1\left(\int_0^1F(i,s)G(s,t)ds \right)H(t,j)dt$$ $$=\int_0^1F(i,s)\left(\int_0^1 G(s,t)H(t,j)dt \right)ds$$ $$=\int_0^1F(i,s)(G\times H)(s,j)ds = (F\times (G\times H))(i,j)$$

Also, $\times$ is obviously bilinear with respect to usual addition of real-valued functions, and hence defines a ring structure on $C(I^2)$ which is considerably different from the usual ring structure (but addition is the same).

Extending the matrix analogy, we see that each $F$ also defines a linear operator $C(I) \to C(I)$ in the usual way, as

$$F(f)(i)=\int_0^1 F(i,t)f(t) dt$$

for each $f : I \to \mathbb{R}$.

Also, all of this actually generalizes usual matrix multiplication if we subdivide the square $I^2$ into a bunch of small rectangles and let $F$ be constant on each subrectangle, being more or less careful on boundaries.

The only candidate for a unit element for $\times$ is the distribution which has a weight $1$ dirac delta on the diagonal, and is $0$ everywhere else (in other words, the product of the dirac delta with the Kronecker delta! :))

Now my question is: what is this? Is it of any interest, or a mere curiosity? For example, could a notion of "determinant" be assigned to these objects?

Integral Operator on $L^2$ spaces. Voting to close. – Pietro Majer Sep 4 '11 at 7:14