Operator for scaling a function?

Question

Let $\mathbb{F}$ denote the set of functions of the form $f: \mathbb{R} \to \mathbb{R}$.

I am interested to know whether there exists a well-known linear map $T_\alpha: \mathbb{F} \to \mathbb{F}$ such that $Tf(x) = f(\alpha x)$.

What would be the formal name of such an operator?

Not really, we can assume continuity over the entire real line. — jII, Oct 17 '14 at 17:04
@T.A.E. the composition operator is incorrect as it does not achieve the task at hand. Perhaps I was ambiguous, but $\alpha$ is a constant real number. — jII, Oct 17 '14 at 21:27
You can call it a scaling operator if you want. But it is also a composition $f\circ\varphi$ where $\varphi(x)=\alpha x$. — Disintegrating By Parts, Oct 17 '14 at 21:50
$a^{x \partial_x}f(x)=f(ax)$ https://math.stackexchange.com/questions/1341495/exponential-of-powers-of-the-derivative-operator — Rayamon, Nov 22 '17 at 22:40

score 0 · Answer 1 · answered Apr 30 '19 at 18:00

It is called the dilatation operator, and it is but a coordinate change of the Lagrange shift operator. Define $$ x\equiv e^y , \qquad \alpha \equiv e^\beta, $$ so that $$ T f(e^y)= f(e^{y+\beta}). $$ That is, T is a shifter, $$ T= e^{\beta \frac{d}{dy}}=e^{\ln\! \alpha ~~~x\frac{d}{dx}}, $$ so that $$ T f(x) = f(\alpha x). $$

Operator for scaling a function?

1 Answers1