Invariant Subspaces of $C_c^\infty(\mathbb{R})$

Question

Does there exist a nontrivial subspace of $C_c^\infty(\mathbb{R})$ that is invariant under (horizontal) translation (i.e. any element of this subspace must also have its translates in the subspace)? I thought the answer is yes, but after trying examples such as the bounded functions of $C_c^\infty(\mathbb{R})$ (which yields the trivial subspace of itself) and analytic functions of $C_c^\infty(\mathbb{R})$ (which appears to yield the trivial zero subspace), I am not so sure.

Answer 1

I think your idea to consider analytic functions could work.

Let $f \in C^\infty(\mathbb{R},\mathbb{R})$ be a function which is nowhere real analytic (see https://en.wikipedia.org/wiki/Non-analytic_smooth_function#A_smooth_function_which_is_nowhere_real_analytic). Let $$ g(x):=\exp(-\frac{1}{1-x^2}) ~ (x \in (-1,1)) \quad g(x)=0 ~ (|x|\ge 1). $$ Then $g \in C^\infty_c(\mathbb{R})$ and $g$ is real analytic in each $x \in \mathbb{R}\setminus \{-1,1\}$. Set $$ U= {\rm span} \{x\mapsto g(x+c): c \in \mathbb{R}\}. $$ Each $h \in U$ is of the form $$ h(x)=\sum_{k=1}^n \alpha_k g(x+c_k), $$ hence the set of points where $h$ is not real analytic is finite. Now $gf \in C^\infty_c(\mathbb{R})\setminus U$ since the set of points where $gf$ is not real analytic contains $(-1,1)$.

Answer 2

Take any nonzero $\psi\in C^\infty_c(\mathbb R).$ Then $\int_{-\infty}^{\infty} \psi'(x)\,dx=0$ and the same is valid for any rescaling and translation of $\psi,$ as well as for any finite sum of such.

Now let $X$ be the linear space generated by all translations and rescalings of $\psi'.$ Then, for all $\varphi\in X$ one has $\int_{-\infty}^{\infty} \varphi(x) \, dx = 0$ so $X$ is a nontrivial subspace of $C^\infty_c(\mathbb R).$

Easier Just take $X = \{ \varphi\in C^\infty_c(\mathbb R) \mid \int_{-\infty}^{\infty} \varphi(x) \, dx = 0 \}.$