TITLE: Existence of translation-invariant basis on $C_c(\mathbb R)$
QUESTION [17 upvotes]: Consider the space $C_c(\mathbb R)$ of complex-valued continuous functions of compact support. This is a vector space over $\mathbb C$, and I am not considering any topology, so the question is algebraic.
A set $A \subseteq C_c(\mathbb R)$ is called shift-invariant if for all $f \in A$ and $t \in \mathbb R$ we also have $T_t f \in A$, where $T_tf(x)=f(x-t)$.
I have the following question:
Question Does there exist a shift-invariant basis $B$ for $C_c(\mathbb R)$?
It is a standard application of Zorn's lemma that there exists a subset $A \subseteq C_c(\mathbb R)$ which is a maximal linearly independent shift-invariant subset, but I do not think that $A$ is always a basis. Such a set $A$ has the property that for all $g \notin \mbox{Span}(A)$ the elements $\{ T_t g : t \in \mathbb R \}$ are linearly independent and none belong to $\mbox{Span}(A)$, but unfortunately this does not seem enough to show that $A \cup \{ T_t g : t \in \mathbb R \}$ is linearly independent.

REPLY [11 votes]: No, there is no shift-invariant basis of $V=C_c(\mathbb R).$ I'll use the formulation in YCor's answer, so we need to show that $V$ is not a free $B$-module where $B=\mathbb C[T^r:r\in \mathbb R],$ with $T^r$ acting as the translation $T_r.$
Let $f(x)=\max(0,1-|x|).$ Suppose there is a $B$-module basis $\{v_i\}$ for $V.$ Then we could write $f$ as a finite sum $\sum P_i v_i$ with $P_i\in B.$ By integrating both sides, we have $P_i(1)\neq 0$ for some $i$ (we can evaluate elements of $B$ at $T=1$ meaningfully - just sum the coefficients).
If we define $f_n(x)=f(2^nx)$ then $f_n=\tfrac12(T^{-2^{-n}/4}+T^{2^{-n}/4})^2 f_{n+1}.$ This means $f,$ and hence $P_i,$ are divisible by $1+T^{2^{-n}}$ for all $n\geq 1.$
This is a contradiction:
Lemma. Assume $P\in B$ is divisible by $1+T^{2^{-n}}$ for all $n\geq 1.$ Then $P(1)= 0.$
Proof: Write $P$ as a finite sum $\sum r a_rT_r.$ Consider a coset $C\in \mathbb R/\mathbb Z[1/2].$ Each polynomial $\sum_{r\in C} a_rT^r$ must still be divisible by $1+T^{2^{-n}}$ for all $n\geq 1.$ So we can reduce to the case where only one of these cosets has non-zero coefficients $a_r.$ This means that $P$ can be written in the form $\sum_m a_m T^{m2^{-n}+r}$ for some integer $n$ and some real $r,$ and where $m$ ranges over a finite set of integers. By applying a shift we can furthermore assume $m$ ranges over non-negative integers, and $r=0.$ We have then reduced to the case where $P$ is a polynomial in $T^{2^{-n}}.$
By injectivity of the unit circle $S^1,$ we can pick a group homomorphism $\phi:\mathbb R\to S^1$ sending $2^{-n-1}$ to $-1.$
This gives a ring automorphism $\sigma$ of $B$ taking $T^r$ to $\phi(r)T^r.$ By assumption $P$ is divisible by $1+T^{2^{-n-1}},$ so $P=\sigma(P)$ is also divisible by $1-T^{2^{-n-1}},$ giving $P(1)=0.$