Existence of translation-invariant basis on $C_c(\mathbb R)$

Question

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.

Answer 1

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 $P(1)=0.$

Answer 2

Here are some remarks.

Let $B$ be the ring $\mathbf{C}[\mathbf{R}]=\mathbf{C}[T^r:r\in\mathbf{R}]$. Then $V=C_\mathrm{c}(\mathbf{R})$ is a $B$-module in the obvious way ($T^r$ acting as $T_r$). The question is essentially one about the understanding of $V$ as an abstract $B$-module.

A first remark is that $V$ is a torsion-free $B$-module. Indeed, if $Pf=0$ with a nonzero element of $B$, then, up to multiply $P$ by an invertible element, we can suppose $P=1-\sum_{r>0}a_rT^r$ (finitely supported sum), and hence $f=\sum_{r>0}a_rT_rf$. Looking at the support of $f$ yields a contradiction unless $f=0$. (In particular, the action of $\mathbf{R}$ on $V\smallsetminus\{0\}$ is free.)

Next, assuming its existence, consider a basis $(f_i)_{i\in I}$ of $V$ with the required invariance. The invariance (and freeness of the action) yields an action of $\mathbf{R}$ on $I$, satsifying $T_r\cdot f_i=f_{r\cdot i}$.

Partition $I$ according the the $\mathbf{R}$-action: $I=\bigsqcup_{k\in K}I_k$, and define $V_k$ as generated by $f_i$ for $i\in I_k$. Then $V=\bigoplus_{k\in K}V_k$, and $V_k$ is a $B$-submodule, generated by $f_i$ for any $i\in I_k$. Since $V$ is a torsion-free $B$-module, we deduce that $V_k$ is a free $B$-module of rank 1, and hence $V$ is a free $B$-module (choose one basis element per orbit).

Conversely, if $V$ is a free $B$-module, taking a $B$-basis and considering its translates yields an $\mathbf{R}$-basis with the required invariance.

Hence the question is equivalent to:

Is $V=C_\mathrm{c}(\mathbf{R})$ a free module over the group algebra $B=\mathbf{C}[\mathbf{R}]=\mathbf{C}[T^r:r\in\mathbf{R}]$ acting by translation?

Answer 3

Let me answer a related question by showing that the following set $W$ of (not necessarily continuous) functions $\mathbb{R} \to \mathbb{C}$ does not admit a basis stable under translations, or even under just translations by integers:

$$ W = \{f\colon \mathbb{R}\to \mathbb{C} \;|\; \forall t\in\mathbb{R}(\{k\in\mathbb{Z} \,|\, f(t+k)\neq 0\}\text{ is bounded})\} $$

(I was hoping to get the set of functions $\mathbb{R} \to \mathbb{C}$ with compact support, but then I realized it is smaller than the not-entirely-natural set $W$ above; still, I think the statement is interesting enough to be mentioned.)

Indeed, let $A := \mathbb{C}[T^{\pm 1}] = \{\sum_{k=-N}^N a_k T^k : N\in\mathbb{N}, a_k\in\mathbb{C}\}$ be the ring of Laurent polynomials (i.e., the group algebra over the infinite cyclic group with generator $T$), acting on $W$ in the obvious way ($T$ acts by translation by $1$, i.e., as $T_1$): arguing as in YCor's answer, there is a basis for $W$ which is stable under $\mathbb{Z}$-translations iff $W$ is a free $A$-module.

Now let $I = \{t \in \mathbb{R} : 0\leq t < 1\}$ (I write this to avoid the notation $[0,1\mathclose[$, which non-French people can't understand 😉). I claim that $W \cong A^I$ (direct product of $I$ copies of $A$) as an $A$-module. Indeed, given an element of $W$, say $f\colon \mathbb{R} \to \mathbb{C}$, identify it with the function $I \to A$ taking $0\leq t < 1$ to $\sum_{k\in\mathbb{Z}} f(t+k)\, T^k$: this is a bijection $W \to A^I$ (pretty much by definition of $W$), it is $\mathbb{C}$-linear and preserves the action of $T$, so it is an isomorphism of $A$-modules.

Now $A^I$ is not a free $A$-module: indeed, by theorem 3.1 of O'Neill, “When a ring is an F-ring”, J. Algebra 156 (1993) 250–258, since $A$ is not semiprimary (because its Jacobson radical is $0$ and it is not Artinian), the $A$-module $A^I$ can only be free if it is free of finite rank, which it clearly isn't (e.g., for cardinality reasons).