Linear space of translatable functions.

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 ${f(x+b) : binmathbb{R}}$ 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 ${P(x)e^{lambda x}: Ptext{ is a polynomial of degree} leq k}$, so condition 3 implies condition 1. $square$

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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 $lambdainmathbb{C}$; 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 $0leq kleq n-1$ 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.