Proposition 1

For an operator U on K[X] to be a composition operator it is necessary and sufficient that it commute with the derivation D on K[X].

Indeed, the Taylor formula shows that for every polynomial $f \in K[X]$ one has

if one puts $g = U(f)$ then one has

for U to be a composition operator, one must then have $UD^k = D^k U$ for every integer $k \geq 1$, and in particular $UD = DU$. Conversely, if this relation holds, it implies $UD^k = D^k U$ for every integer $k \geq 1$, by induction on $k$; the Taylor formula then shows that $g(X + Y) = U_X(f(X + Y))$.

For every polynomial $f \in K[X, Y]$ we denote by $U_0(f)$ the term independent of X in the polynomial $U_X(f)$; in particular, if $f \in K[X]$, $U_0(f)$ is the constant term in $U(f)$, and $U_0$ is a linear form on $K[X]$. For every polynomial $f \in K[X]$ let $g = U(f)$; by def. 1 of VI, p. 270

and if, in this formula, one replaces X by 0, one obtains

Thus one sees that

where $\mu_k$ is the constant term in the polynomial $U(X^k)$.

This formula shows that the $\mu_k$ determine the composition operator $U$ completely; conversely, if $(\mu_n)$ is an arbitrary sequence of elements of K then the formula (2) defines an operator $U$ which clearly commutes with D, so (VI, p. 270, prop. 1) is a composition operator. From now on we shall write (2) in the form

This formula can be interpreted in topological language as follows: if one considers the discrete topology on $K[X]$, and the topology of simple convergence on the algebra $\mathrm{End}(K[X])$ of endomorphisms of $K[X]$ (Gen. Top., X, p. 277), the series with general term $\frac{1}{k!} \mu_k D^k$ is commutatively convergent in $\mathrm{End}(K[X])$ and has sum $U$ (Gen. Top., III, p. 269).

The formula (3) shows that to every formal series $u(S) = \sum_{k=0}^{\infty} \alpha_k S^k$ in one indeterminate over K (Alg., IV. 41) one can associate the composition operator $U = \sum_{k=0}^{\infty} \alpha_k D^k$, which in future we shall denote by $u(D)$. This remark can be clarified in the following manner: