Pincherle derivative

In mathematics, the Pincherle derivative^[1] $T'$ of a linear operator $T:\mathbb {K} [x]\to \mathbb {K} [x]$ on the vector space of polynomials in the variable x over a field $\mathbb {K}$ is the commutator of $T$ with the multiplication by x in the algebra of endomorphisms $\operatorname {End} (\mathbb {K} [x])$ . That is, $T'$ is another linear operator $T':\mathbb {K} [x]\to \mathbb {K} [x]$

T':=[T,x]=Tx-xT=-\operatorname {ad} (x)T,\,

(for the origin of the $\operatorname {ad}$ notation, see the article on the adjoint representation) so that

T'\{p(x)\}=T\{xp(x)\}-xT\{p(x)\}\qquad \forall p(x)\in \mathbb {K} [x].

This concept is named after the Italian mathematician Salvatore Pincherle (1853–1936).

Properties[edit]

The Pincherle derivative, like any commutator, is a derivation, meaning it satisfies the sum and products rules: given two linear operators $S$ and $T$ belonging to $\operatorname {End} \left(\mathbb {K} [x]\right),$

$(T+S)^{\prime }=T^{\prime }+S^{\prime }$ ;
$(TS)^{\prime }=T^{\prime }\!S+TS^{\prime }$ where $TS=T\circ S$ is the composition of operators.

One also has $[T,S]^{\prime }=[T^{\prime },S]+[T,S^{\prime }]$ where $[T,S]=TS-ST$ is the usual Lie bracket, which follows from the Jacobi identity.

The usual derivative, D = d/dx, is an operator on polynomials. By straightforward computation, its Pincherle derivative is

D'=\left({d \over {dx}}\right)'=\operatorname {Id} _{\mathbb {K} [x]}=1.

This formula generalizes to

(D^{n})'=\left({{d^{n}} \over {dx^{n}}}\right)'=nD^{n-1},

by induction. This proves that the Pincherle derivative of a differential operator

\partial =\sum a_{n}{{d^{n}} \over {dx^{n}}}=\sum a_{n}D^{n}

is also a differential operator, so that the Pincherle derivative is a derivation of $\operatorname {Diff} (\mathbb {K} [x])$ .

When $\mathbb {K}$ has characteristic zero, the shift operator

S_{h}(f)(x)=f(x+h)\,

can be written as

S_{h}=\sum _{n\geq 0}{{h^{n}} \over {n!}}D^{n}

by the Taylor formula. Its Pincherle derivative is then

S_{h}'=\sum _{n\geq 1}{{h^{n}} \over {(n-1)!}}D^{n-1}=h\cdot S_{h}.

In other words, the shift operators are eigenvectors of the Pincherle derivative, whose spectrum is the whole space of scalars $\mathbb {K}$ .

If T is shift-equivariant, that is, if T commutes with S_h or $[T,S_{h}]=0$ , then we also have $[T',S_{h}]=0$ , so that $T'$ is also shift-equivariant and for the same shift $h$ .