Linear space of translatable functions. 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.
 A: 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:


*

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

*$f$ satisfies a homogeneous constant coefficient linear differential equation; 

*$f$ 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.  
A: 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
A: 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.
