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?

  • 1
    $\begingroup$ Does the the set of functions have any special property? $\endgroup$
    – Orca
    Oct 17, 2014 at 16:43
  • $\begingroup$ Not really, we can assume continuity over the entire real line. $\endgroup$
    – jII
    Oct 17, 2014 at 17:04
  • $\begingroup$ Composition operator. $\endgroup$ Oct 17, 2014 at 21:17
  • $\begingroup$ @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. $\endgroup$
    – jII
    Oct 17, 2014 at 21:27
  • $\begingroup$ You can call it a scaling operator if you want. But it is also a composition $f\circ\varphi$ where $\varphi(x)=\alpha x$. $\endgroup$ Oct 17, 2014 at 21:50

1 Answer 1


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


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