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

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.

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Strange question... If $x$ is defined on the whole real line, then $Uf(x):=f(kx)$ is a perfectly good definition of a linear operator (and it preserves the condition that the integral converges). – Hans Lundmark Jan 11 '11 at 16:06
@kakemonsteret: that is an explicit expression for U. I don't understand. – Qiaochu Yuan Jan 11 '11 at 16:15
@kakemonsteret: What you give is an "explicit description". If you are looking for some formula $U(w) =$(expression of $w$) that you can just plug in $f(x)$ into and the result will be $f(2x)$, you are not going to find it in general, or for any but the most restrictive class of functions; the expression you give is perfectly fine way of defining it, though. – Arturo Magidin Jan 11 '11 at 16:20
@kakemonsteret: Formulas are finite strings of characters. There are a countable infinity of them. Real functions are sets of pairs $(x,y), x,y \in \mathbb{R}$ in which there is only one pair for a given $x$. If you insist the domain be all of $\mathbb{R}$, you must have a pair for each $x \in \mathbb{R}$. There are $|\mathbb{R}|^{|\mathbb{R}|}$ of them-a few more than formulas. – Ross Millikan Jan 11 '11 at 17:03
@kakemonstert: a formula is a finite sequence of strings from a valid alphabet; there are only countably many. A function, say, from $\mathbb{R}$ to $\mathbb{R}$, is any subset of $\mathbb{R}\times\mathbb{R}$ that satisfies a certain set of conditions. There are uncountably many of them (in fact, even more than real numbers). So, most definitely no, they are not the same thing and they are not equal in numbers. There are far more functions than formulas. – Arturo Magidin Jan 11 '11 at 17:04

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.

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You should add the caveat that this explicit form works only if $kx$ is in the radius of convergence of the power series about $0$. – Zhen Lin Jan 12 '11 at 8:12
It it considered to have this matrix everywhere. It is just not interpreted as a matrix multiplied by Taylor series where the series does not converge. – Anixx Jan 12 '11 at 8:19
-1. I am sorry but this answer is pretty nonsensical. Firstly, by far most functions from $\mathbb{R}$ to $\mathbb{R}$ are not differentiable, let alone infinitely differentiable. So your expression almost never makes sense, which begs the question what "the operator has the matrix" is supposed to mean. Secondly, it was the OP who was confusing functions $\mathbb{R}\rightarrow \mathbb{R}$ with operators on the space of such functions (which are also functions by the way). He was using the notation for the former, but seems to have intended the latter. – Alex B. Jan 12 '11 at 9:26
His actual intended question, as I understand it, "does there exist an operator on the space of functions that is right composition with multiplication by $k$" is trivial, as was repeatedly pointed out to him in the comments (to no avail, alas): simply define the operator to be just that. – Alex B. Jan 12 '11 at 9:29
Yea I mean operator, someone edited my question and put in that composition sign – user1708 Jan 12 '11 at 11:07

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 $f$.

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While there is no $U\colon\mathbb{R}\to\mathbb{R}$, there are of course functions $U_k\colon\mathbb{R}^{\mathbb{R}}\to\mathbb{R}^{\mathbb{R}}$ for which $(U_k(f))(x) = f(kx)$ for all $x$; the problem is that the OP seems to want such $U_k$ to be "expressible" in some simple formulaic way. E.g., if you restrict yourself to affine maps $f(x) = ax+b$, then $U(g) = 2g-g(0)$ would (presumably) be such a formula. But for any sufficiently large class of functions, there won't be a "formula" like that. – Arturo Magidin Jan 11 '11 at 21:10
@Arturo: then the OP should clarify what kind of simple formulaic ways are being allowed here. I don't see what sense in which U(f(x)) = f(2x) isn't already simple and formulaic. – Qiaochu Yuan Jan 11 '11 at 21:15
I fully agree with you, as I agreed in the comments before. (-: – Arturo Magidin Jan 11 '11 at 21:20
@ Arturo Magidin, in you example of formula for $U(f)$ for $f(x)=ax+b$ you used right composition, so I fail to see how this formula is more simple or formalistic than the definition: $U[f(x)]=f(kx)$ – Anixx Jan 12 '11 at 10:34
If $U$ is allowed to be a linear operator, then $U=\exp\left(\ln(k)x\frac d{dx}\right)$ does the trick, see also joriki's answer to my very similar question – Tobias Kienzler Apr 3 '12 at 11:52

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.

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For restricted domain f, of course. – Tom Copeland Apr 6 '12 at 0:27