Action of the Product of Two Linear Functionals on a Polynomial I am looking for help with the following problem.  Here we denote the action of a linear functional on a polynomial by 
$$\langle L\mid p(x)\rangle$$
Suppose that there are two linear functionals $L_f$ and $L_g$ and their action on a polynomial is defined as
$$\langle L_f\mid p(x)\rangle=\int_{-\infty}^\infty f(x)p(x)\,dx$$
$$\langle L_g\mid p(x)\rangle=\int_{-\infty}^\infty g(x)p(x)\,dx$$
And so I'm assuming then that the linear functional is simply
$$L_f=L(f(x))=\int_{-\infty}^\infty f(x) \, dx$$
I am looking to find the following action:  $\langle L_fL_g\mid p(x)\rangle$.  Its been a while since I've done the appropriate linear algebra and need some help.  I recall that in the language of linear transformations, linear functionals are simply transformations onto the base field.  But I'm confused about the notation $L_fL_g$.  Would it simply be the product
$$L_fL_g=L(f(x))L(g(x))=\int_{-\infty}^\infty f(x) \, dx\int_{-\infty}^\infty g(x) \, dx$$
which i don't think is correct.  Or is it interpreted as
$$L_f L_g=L_{fg}=\int_{-\infty}^\infty f(x)g(x) \, dx$$
which, again, I don't think is correct.  
 A: The take away here is that the product of two functionals is a functional, where we think of product not in the operator sense, but as the product of the images of a functional. Functionals map vector spaces to the field on which they are based, and the product of two elements of a field is well defined. However, it’s important to note that the product doesn’t give a linear functional.
The big hint here is that your professor said to think of convolution. Just like the Fourier transform of the convolution of a pair of functions is the product of the Fourier transforms, the same holds true here.
Let $f$ and $g$ be the symbol for the functionals $L_f$ and $L_g$ as you described. Let $p$ be a polynomial.
Then $$(L_f \cdot L_g) p := L_f p \cdot L_g p = \int_{-\infty}^\infty f(x)p(x) dx \cdot \int_{-\infty}^\infty g(y)p(y) dy.$$
If we assume that $f$ and $g$ are nice enough that we can combine the two integrals into one double integral we find:
$$\int_{-\infty}^\infty f(x)p(x) dx \cdot \int_{-\infty}^\infty g(y)p(y) dy = \int_{-\infty}^\infty  \int_{-\infty}^\infty f(x)p(x) \cdot g(y)p(y) dx dy$$
By a change of variables:
$$ = \int_{-\infty}^{\infty} \int_{-\infty}^\infty f(t-s)g(s)(p(t-s)p(s)) ds dt = \int_{-\infty}^\infty (f \cdot p) \star (g \cdot p)(t) dt.$$
And that is our new functional. Note that it can't be written as $L_h$ for some function $h$, since it is a nonlinear functional.
