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

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.

• 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. May 5 at 22:38
• 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. May 6 at 9:48
• 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.) May 6 at 9:55
• @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. May 6 at 22:36
• 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 May 7 at 8:25

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 $$P(1)=0.$$

• There's probably a sign typo in "$T^{-2^{-n}/4}+T^{-2^{-n}/4}$ "?
– YCor
May 7 at 10:58
• 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? May 7 at 14:18
• @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 $x \geq 1$. May 7 at 14:46
• @StefaanVaes Indeed, that's very cute. May 7 at 15:29
• @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 $f_{n-1}=(1+T^{2^{-n}})f_n.$ May 8 at 8:08

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?

• 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... May 6 at 21:54

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).

• [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. May 7 at 11:29