# Operator for scaling a function?

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?

• Does the the set of functions have any special property?
– Orca
Oct 17, 2014 at 16:43
• Not really, we can assume continuity over the entire real line.
– jII
Oct 17, 2014 at 17:04
• Composition operator. Oct 17, 2014 at 21:17
• @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, 2014 at 21:27
• You can call it a scaling operator if you want. But it is also a composition $f\circ\varphi$ where $\varphi(x)=\alpha x$. Oct 17, 2014 at 21:50

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 $$T f(x) = f(\alpha x).$$