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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.

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  • $\begingroup$ Do you know the answer if you drop the compact support condition? I imagine periodic functions mess things up then but I don't see a clear argument. $\endgroup$
    – Wojowu
    May 5 at 22:38
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    $\begingroup$ In other words, you're asking whether the continuous functions with compact support are a permutation representation of the group of translations of the line. Clearly there can't be a purely formal answer to this (e.g., “just apply Zorn's lemma”) because not every group representation is a permutation representation. $\endgroup$
    – Gro-Tsen
    May 6 at 9:48
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    $\begingroup$ It might also be worth making the utterly trivial remark that if we consider the vector space $\mathbb{R}^{(\mathbb{Z})}$ of $\mathbb{Z}$-indexed sequences of real numbers with finite support, it has a translation-invariant basis, namely that consisting of the sequences $e_n$ where $e_n$ is the sequence whose $n$-th term is $1$ and all other terms are $0$. (Utterly trivial, but suggests that a positive answer to the question is not implausible.) $\endgroup$
    – Gro-Tsen
    May 6 at 9:55
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    $\begingroup$ @Wojowu Without the compact support condition, the following argument was provided on Twitter: first try to decompose $1$ on the basis and translate to conclude that some nonzero constant must be part of the basis. Then try to decompose $x$ on the basis and translate to get a contradiction. $\endgroup$
    – Gro-Tsen
    May 6 at 22:36
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    $\begingroup$ A related fact: the space of bounded functions $f : \mathbb{R} \to \mathbb{R}$ with bounded support does not have a translation-invariant basis. (There is no continuity assumption here.) This was Advanced Problem #6278 in the American Mathematical Monthly (1981), see doi.org/10.2307/2321767 $\endgroup$ May 7 at 8:25
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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 $P(1)=0.$

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  • $\begingroup$ There's probably a sign typo in "$T^{-2^{-n}/4}+T^{-2^{-n}/4}$ "? $\endgroup$
    – YCor
    May 7 at 10:58
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    $\begingroup$ Nice. This shows that there is even not a basis that is only $\mathbb{Z}[\frac 1 2]$-invariant. What a $\mathbb{Z}$-invariant basis as in Gro-Tsen's answer? $\endgroup$ May 7 at 14:18
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    $\begingroup$ @Mikael de la Salle: I think that there is a $\mathbb{Z}$-invariant basis, namely the function $g(x) = \max(0,1-2|x|)$ together with any vector space basis for the vector space of continuous functions $f$ that are zero at $x \leq 0$ and at $x \geq 1$. $\endgroup$ May 7 at 14:46
  • $\begingroup$ @StefaanVaes Indeed, that's very cute. $\endgroup$ May 7 at 15:29
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    $\begingroup$ @NickS: I don't think that fixed point equation is quite right - there is a dilation involved. I think the self-similarity is a bit of a red herring; my $f$ is just a particular concrete function that is in the image of each operator $1+T^{2^{-n}}.$ For example you could instead use $f_n=g*\chi_{[0,2^{-n}]}$ where $*$ is convolution, $g$ is any nice function, and $\chi_{[0,2^{-n}]}$ is the indicator function of $[0,2^{-n}]$; then $f_{n-1}=(1+T^{2^{-n}})f_n.$ $\endgroup$
    – Harry West
    May 8 at 8:08
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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?

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  • $\begingroup$ This ring $B$ is pretty horrendous (it's an inductive limit of rings isomorphic to $\mathbb C[t_1,t_1^{-1},t_2,t_2^{-1},...t_n,t_n^{-1}]$ where $n$ is arbitrarily large). And the space $C_c(\mathbb R)$ is a pretty horrendous $B$-module. I would be very surprised if that $B$-module turned up to be free... $\endgroup$ May 6 at 21:54
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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).

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    $\begingroup$ [off-topic chatter] "I write this to avoid the notation [0,1[, which non-French people can't understand" : I'd like to disagree, but I'm French… :) Looks pretty obvious what is meant, though, isn't it? I always use the notation in my papers and never got any remark about that. $\endgroup$ May 7 at 11:29

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