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.



Existence of translation-invariant basis on $C_c(\mathbb R)$
No, there is no shift-invariant basis of $V=C\_c(\mathbb R).$ I'll use the formulation in [YCor's answer](https://mathoverflow.net/a/392049/164965), 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 $P(1)=0.$


