Scale Operator $Uf(x)=f(kx)$

Question

I am looking for an operator $U$, that can do this to a function: $$Uf(x)=f(2x).$$

In particular I am happy if there is an $U$ for the general case: $Uf(x)=f(kx)$.

Does such an operator exist for all $f(x)$? What is the operator for known classes of $f(x)$?

I have so far 1 condition on $f(x)$: $$\int_{-\infty}^{\infty} f(x)^2\,dx\text{ converges.}$$

I dont really know where to look for such an operator, or what keyphrases here are.

Answer 1

Yes, such operator exists and it is linear. The answer above by Qiaochu Yuan seems to confuse function with operator. That said there is no such function $U:\mathbb{R}\to\mathbb{R}$, but there is a linear operator U as said Arturo Magidin. And on the set of analytic functions it has the following infinite (because the set of analytic functions is infinite-dementional) matrix:

$$U=\left( \begin{array}{cccccc} 1 & 0 & 0 & . & 0 & . \\ 0 & k^1 & 0 & . & 0 & . \\ 0 & 0 & k^2 & . & 0 & . \\ . & . & . & . & . & . \\ 0 & 0 & 0 & . & k^n & . \\ . & . & . & . & . & . \end{array} \right)$$

To obtain the resulting function you have to multiply this matrix by the argument function:

$f(k x)= Uf(x)= f(0)+\frac {f'(0)}{1!} k x+ \frac{f''(0)}{2!} k^2 x^2+\frac{f^{(3)}(0)}{3!} k^3 x^3+ \cdots+\frac{f^{(n)}(0)}{n!} k^n x^n +\cdots.$

The confusion with your question probably arose because you used the function composition sign $\circ$ where it should not be if you meant applying the operator.

Answer 2

Proposition: There exists no function $U : \mathbb{R} \to \mathbb{R}$ such that, for every function $f : \mathbb{R} \to \mathbb{R}$, we have the relation $(U \circ f)(x) = f(2x)$. (This is the interpretation of the question I have managed to glean from your comments.)

Proof. Let $f, g$ be two functions such that $f(1) = g(1)$ but such that $f(2) \neq g(2)$. Then $(U \circ f)(1) = (U \circ g)(1)$ but $f(2) \neq g(2)$; contradiction.

As you can see the proof is robust under many modifications of the conditions on $f$.

Answer 3

There are the special linear conformal transformations SL(2,R) associated with the differential operators

$S_{-1}f(z)=\exp(ad/dz)f(z)=f(z+a)$

$S_{0}f(z)=\exp(bzd/dz)f(z)=f(e^b z)$

$S_{1}f(z)=\exp(cz^{2}d/dz)f(z)=f(z/(1-cz))$

The $z^{m+1}d/dz$ ($m=-1,0,1$) are a representation of a subgroup of the infinite Witt Lie algebra associated with the Virasoro algebra, and their exponential maps can be used to construct Möbius, or linear fractional, transformations.

$S_{0}$ is one rep of a scaling operator that you might be looking for with $b=\ln(k)$.

For more info see my notes "Mathemagical Forests" (pages 13-15) at my little "arxiv".

Also see answers to MS Q116633.