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What are the functions $f$ so that a set $\{a \cdot f(x+b) : a \in \mathbb{R}, b \in \mathbb{R}\}$ is a finite dimensional linear vector space ?

Is there a complete characterization of such functions?

$e^{c x}$ (where $c$ is some constant) is a good example of a base of one dimensional space.
$sin (c x), cos(c x)$ seems to be two dimensional example.

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Please rephrase the title or question if you know proper mathematical names. –  Koszmarny Aug 19 '10 at 15:26
Unless I misunderstand your notation, $f(x) = e^x$ is a function such that the set you want is a one-dimensional vector space. And thus trivially it has a finite-dimensional dense subspace. –  Willie Wong Aug 19 '10 at 16:01
I'm trying to expand my understanding of Fourier transform. –  Koszmarny Aug 19 '10 at 16:19

3 Answers 3

up vote 6 down vote accepted

The question, as stated, is about the set of multiples of translates, but from the example quoted, $\sin x,$ I suspect that OP really meant the span.

Theorem Let $f$ be a continuous complex-valued function on $\mathbb{R}.$ Then the following conditions are equivalent:

  1. The translates $\{f(x+b) : b\in\mathbb{R}\}$ span a finite-dimensional vector space;

  2. $f$ satisfies a homogeneous constant coefficient linear differential equation;

  3. $f$ is a finite linear combination of functions $f_{k,\lambda}(x)=x^k e^{\lambda x}.$

Proof. If $f$ is assumed infinitely differentiable then all derivatives of $f$ belong to the $\mathbb{R}$-span of translates of $f.$ Thus condition 1 implies that $f$ and its derivatives of order up to $n$ are linearly dependent over $\mathbb{R},$ which is condition 2. The smoothness assumption may be removed by using the Fourier or Laplace transform.

The equivalence of conditions 2 and 3 is a basic fact of ODEs. Finally, a direct computation shows that $f_{k,\lambda}(x)$ spans the $(k+1)$-dimensional vector space $\{P(x)e^{\lambda x}: P\text{ is a polynomial of degree} \leq k\}$, so condition 3 implies condition 1. $\square$

Condition 1 – 3 have the following representation-theoretic interpretation. The additive group of $\mathbb{R}$ acts on itself by the right multiplication. This gives rise to a linear representation of $\mathbb{R}$ on the functions on $\mathbb{R}$ via translations called the right regular representation, and condition 1 states that $f$ belongs to a finite-dimensional subrepresentation $V$. Finite-dimensionality of $V$ implies that $V$ contains an irreducible subrepresentation $W$, which must be one-dimensional (Schur's lemma), hence $W$ is spanned by a character of $\mathbb{R}.$ All continuous characters are the exponential functions $e^{\lambda x}$ for various $\lambda\in\mathbb{C}$; however, using a Hamel basis of $\mathbb{R},$ it is easy to see that there are uncountably many others.

Condition 2 is the Lie algebra analogue of condition 1: viewing $\mathbb{R}$ as a one-dimensional Lie group, the content of Lie's theorem is that its finite-dimensional (continuous) representations correspond (by differentiation and exponentiation) to f.d. representations of the abelian one-dimensional Lie algebra, i.e. to a single linear transformation on $V.$ The span $V_{n,\lambda}$ of the functions $f_{k,\lambda}$ with $0\leq k\leq n-1$ from condition 3 is an $n$-dimensional indecomposable representation of $\mathbb{R},$ whose infinitesimal version is a Jordan block of order $n$ with $\lambda$ on the diagonal. Moreover, any subrepresentation isomorphic to $V_{n,\lambda},$ i.e. corresponding to the same Jordan block, must be $V_{n,\lambda}$ itself.

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Since OP's literal formulation implies that $f$ satisfies condition 1, one easily concludes that the requisite $f$ must be a constant multiple of $e^{\lambda x}$. –  Victor Protsak Aug 19 '10 at 19:46
I thought at first, after seeing the $sin(x)$ example, that span was what was meant. But now I am not too sure after thinking about it for a bit. The identity $a \sin(x) + b \cos(x) = \sqrt{a^2 + b^2}\sin(x + c)$ where $\sin c = b / \sqrt{a^2 + b^2}$ (assuming $a$ positive) implies that for the case of $\sin$, the span and the set are equal. –  Willie Wong Aug 19 '10 at 20:04
Willie, you need both $\sin x$ and $\cos x$ to make it translation invariant. Put differently, $\sin x$ is a linear combination of $e^{ix}$ and $e^{-ix}$ with non-zero coefficients, so the space spanned by its translates is 2-dimensional. –  Victor Protsak Aug 19 '10 at 20:48
I agree that the vector space is spanned by sin(x) and cos(x), but since cos(x) is just a translate of sin(x) anyway, isn't the vector space generated just by the translations and dilations of sin(x) only, as the OP intended? (I have a slight suspicion that this discussion between us is just about semantics.) In other words, I can write $V = \{ a \sin(x) + b \cos(x) : a,b\in R\}$ or $V = \{ a \sin(x + b): a,b\in R\}$ and refer to the same thing? –  Willie Wong Aug 19 '10 at 22:14
You are absolutely right! I am sorry, I didn't think through your first comment. So a real-valued continuous function $f$ of real argument has the property that its dilated translates form a f.d. vector space if and only if $f(x)=e^{\lambda x}(A\cos\omega x+B\sin\omega x).$ –  Victor Protsak Aug 20 '10 at 2:08

Not a complete answer but I think this might be a good start, at least in the special case when $f$ is a smooth function.

Since you require that the set is a vector space, the ratio $h^{-1}(f(x+h)-f(x))$ must belong to it and hence must be of the form $a_h f(x+b_h)$ for some $a_h,b_h$ depending on $h$. Since we are assuming that $f(x)$ is differentiable, the limit of $a_h f(x+b_h)$ as $h\to0$ exists and is precisely $f'(x)$, for every $x$. Take subsequences so that $a_h$ and $b_h$ have a (possibly infinite) limit. Now there are several cases to consider, depending on the combination of limits we get. Let us restrict to the case $a_h\to a$, $b_h\to b$ for some finite reals $a,b$ (I told you this is just a start). Then the function $f$ must satisfy the delay ODE

$$f'=a f(x+b).$$

When $b=0$ you get your exponentials. When $b=\pi/2$ you get $\sin$ and $\cos$. For other values of $b$: this is a well studied class of equations and it's easy to find pointers (keyword: delay ODE). Or solve by hands...

EDIT: this argument does not use the assumption that the space is finite-dimensional :) Let's say it is a nice complement to the much more polished answer by Victor

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So for $f(x)=x^2$ one has $$f(x+b)=(1-b^2)f(x)+\frac{b^2+b}{2}f(x+1)+\frac{b^2-b}{2}f(x-1)$$ And similarly, $x^k$ (or any polynomial of degree $k$) gives a dimension $k+1$ example. The same goes for $x^ke^x$.

edit As pointed out, these are examples where the span of all the $af(x+b)$ has finite dimension, however it is strictly larger than that set.

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I think you have the relation the other way around? For $f(x) = x^2$, let $a_1 = 1, b_1 = 0, a_2 = -1, b_2 = 1$, then $f_1(x) = a_1f(x + b_1) = x^2$ and $f_2(x) = a_2 f(x + b_2) = -(x+1)^2 = - x^2 - 2x - 1$. $f_1$ and $f_2$ are in the set defined by the OP. But $g = f_1 + f_2 = - 2x - 1$ is not in the span of rescalings and translations of $f(x)$. In fact, by modifying the argument I just wrote down, you can see that for $f(x)$ polynomial the resulting set is not a vector space. –  Willie Wong Aug 19 '10 at 19:15

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