# finite dimensional vector spaces of functions left invariant by translation

Let $E$ be a finite dimensional vector space of functions $\mathbb{R} \rightarrow \mathbb{R}$ such that $\forall f \in E, \forall t \in \mathbb{R}, x \mapsto f(x-t) \in E$.

Example of such spaces include polynomials of any degree, exponentials, sine and cosine, and of course the linear span of these spaces.

For instance, functions of the form $a_0 + a_1 x + a_2 x^2 + a_3 e^{-\frac{x}{3}} + a_4 \cos 2 x + a_5 \sin 2 x$ live in a 6 dimensional subspace of real functions of real variables.

Are there any other such spaces, or do sine and cosine, exponential, and polynomials cover it?

• Is your space of functions $\mathbb R\to\mathbb R$ of continuous functions? Polynomials, sines and exponentials really cannot span every such function, e.g. $|x|$ or "stranger" functions like $\chi_{\mathbb Q}$, the Dirichlet function. – yellowquark Apr 16 '15 at 19:11
• Also, something like $e^t\sin t$ isn't in your 6-dimensional space. – yellowquark Apr 16 '15 at 19:17
• I never claimed that those spaces spanned the entire space, this isn't the purpose. – Arthur B. Apr 16 '15 at 19:32

## 1 Answer

Let $T_a$ denote translation by $a$, so $(T_af)(x)=f(x+a)$. Consider the complexification $E\otimes\mathbb C$. Since $T_aT_b=T_{a+b}=T_bT_a$, we find simultaneous eigenspaces of all $T_a$. Let $W$ be such a simultaneous eigenspace and let $\lambda(a)$ be the eigenvalue of $T_a|_W$. Then $$\tag1\lambda(a+b)=\lambda(a)\lambda(b)$$ and $\lambda(0)=1$. We verify that $f\in W$ implies $f(x)=f(0)\cdot \lambda(x)$, so that $\dim W=1$. If we require that all functions are continuous, it is well-known that the only solutions to $(1)$ are exponential functions. Thus we have finitely many $\lambda_j\in\mathbb C$ such that the functions $x\mapsto e^{\lambda_jx}$ span the eigenspaces of each $T_a$. If $a_0$ is not an integer multiple of any of the $\frac{2\pi i}{\lambda_j-\lambda_k}$, then all eigenspaces of $T_{a_0}$ as well of $T_{a_0/m}$, $m\in\mathbb N$, are onedimensional because the eigenvalues $e^{\lambda_ja_0}$ (resp. $e^{\lambda_ja_0/m}$) differ. Since this excludes only countably many values, such $a_0$ can certainly be picked. Then the $T_{a_o/m}$ also share their generalized eigenspaces. Let $U_j$ the $j$th generalized eigenspace (i.e., belonging to $e^{\lambda_ja_0/m}$ for $T_{a_0/m}$) and let $d_j=\dim U_j$. Let $V_j=\{\,e^{-\lambda_jx}\cdot f\mid f\in U_j\,\}$. Then on $V_j$, $T_{a_0/m}-\operatorname{Id}$ is nilpotent and from the theory of iterated differencex, we know that each $v\in V_j$ is a polynomial function of degree $\le d_j$ when restricted to the set $\frac{a_0}m\mathbb Z$. Since this polynomial is already determined by its values at $x=0,a_0,\ldots, d_ja_0$, we conclude that each $v\in V_j$ is a polynomial fucntion also on $a_0\mathbb Q$. By continuity, $v$ is a polynomial on all of $\mathbb R$.

Therefore, $$\tag2U_j=\{\,p(x)e^{\lambda_j x}\mid p\in\mathbb C[x], \deg p\le d_j\,\}.$$ since $E\otimes \mathbb C=\bigoplus_j U_j$, ell elements of $E\otimes C$ can be written as sums of polynomials times exponentials. The elements of $E$ are precisely those of the form $\frac12(f+\overline f)$ with $f\in E\otimes \mathbb C$, which according to $(2)$ means precisely that $E$ is spanned by functions of the form $e^{\lambda x}x^k\sin \omega x$ and $e^{\lambda x}x^k\cos \omega x$.

• Assuming $f$ is differentiable, can we show every such subspace is spanned by $\{ e^{\lambda x} x^k \sin x, e^{\lambda x} x^k \cos x\}$ ? – Arthur B. Apr 16 '15 at 20:21
• @ArthurB.I rewrote completely, now it is an answer :) – Hagen von Eitzen Apr 17 '15 at 8:23