Struggling with the "Umbral Algebra"

Question

I am reading a lot of papers and material that deal with "Umbral Calculus" and "Umbral Algebra" and am finding myself very confused and am hoping I can get a little more clarity. For those willing to help i should let you know that most, if not all, the material i will be referencing comes from a publication by Steven Roman and Gian-Carlo Rota called "The Umbral Calculus". Section three deals with "The Umbral Algebra"

Let $P$ be a commutative algebra of all polynomials in a single variable $x$ with coefficients in a field $K$ of characteristic zero and let $P^*$ be the vector space of all linear functionals on $P$. Denote the action of a linear functional on a polynomial by

$$\langle L\mid p(x)\rangle$$

From here $P^*$ is made into an algebra by allowing the product of two linear functionals $L$ and $M$ to be defined as

$$\langle LM\mid x^n\rangle=\sum_{k=0}^n\binom{n}{k}\langle L \mid x^k\rangle \langle M\mid x^{n-k}\rangle$$

Now that the product is defined, it is mentioned that this umbral algebra is related to the algebra of real functions under convolution. Let $f$ and $g$ be functions such that

$$\int_{-\infty}^\infty f(x)x^n dx \text{ and }\int_{-\infty}^\infty g(x)x^n dx$$

are defined for all integers $n\ge 0$. This is where I get confused pretty hardcore. The paper now defines linear functionals $L_f$ and $L_g$ by

$$\langle L_f \mid p(x)\rangle=\int_{-\infty}^{\infty}f(x)p(x)dx$$ $$\langle L_g \mid p(x)\rangle=\int_{-\infty}^{\infty}g(x)p(x)dx$$

Then it states the product $L_fL_g$ is the linear functional

$$\langle L_f L_g \mid p(x)\rangle=\int_{-\infty}^{\infty}h(x)p(x)dx$$

where

$$h(x)=\int_{-\infty}^{infty}f(t)g(x-t)dt$$

Here is my first question: Under the assumption that $\langle L\mid p(x)\rangle$ was defined as "the action of a linear functional on a polynomial", is it also true that the action of a linear functional on a polynomial is itself a linear functional? This would seem true by definition alone, but verification helps.

Here is my second question: obviously $p(x)=x^n$ is a polynomial in $x$ How was it determined that convolution was the necessary definition of product that allowed this algebra to be defined? The paper makes it seem obvious but I am one that has struggled with the dual space. My major confusion comes from the definition of an action. Is it true the action of a linear functional on a polynomial is itself a linear functional? If so, why name it differenlty? Why call it "the action"?

Finally, as a comment, i have struggled with the concepts of abstract algebra and am aware of the topic of group actions. Are these topics related and can the reader point me in a direction of books/papers that makes these ideas more obvious?

Answer 1

I don't really understand your questions but I can try and clarify what is going on. The polynomials $P = K[x]$ form a commutative algebra and a vector space over $K$ whose basis is given by the monomials $(x^i)_{i=0}^{\infty}$. The elements in the dual space $P^{*}$ are linear functionals $L \colon K[x] \rightarrow K$ and so they "eat" polynomials and return scalars in a linear way. Instead of writing $L(p)$ for the result of applying $L$ to the polynomial $p \in K[x]$, your text uses the notation

$$ \left< L \, | \, p(x) \right> = L(p(x)) $$

and reasonably calls it "the action" of $L$ on $p$. Since $L$ is linear, it is uniquely determine by its action on the basis elements $x^i$ of $P$ and so given two linear functionals $L,M$ one can define a new linear functional $LM$ implicitly by requiring that

$$ (LM)(x^n) = \sum_{k=0}^n {n \choose k} L \left(x^k \right) M \left( x^{n-k} \right). $$

Now, assume $K = \mathbb{R}$. Given $f \colon \mathbb{R} \rightarrow \mathbb{R}$, we can try and define a linear functional $L_f$ on $P$ by setting

$$ (L_f)(p) := \int_{-\infty}^{\infty} f(x) p(x) \, dx.$$

For this to be well-defined, we will need to make sure that the integral on the right exists for all polynomials $p \in P$. Now, given $f,g$ for which $L_{f},L_{g}$ is well-defined, we can ask whether the multiplication $L_{f}L_{g}$ is of the form $L_{h}$ for some $h \colon \mathbb{R} \rightarrow \mathbb{R}$. It turns out that $M_{f}M_{g} = M_{f \star g}$ when $f \star g$ is the convolution of $f$ and $g$ (assuming it exists). This relates the product of the linear functionals $M_f,M_g$ with the product of the functions $f,g$ in the sense of convolution.