For each real $t$, we define an operator $L(t)$, acting on functions from $\mathbb{R}$ to $\mathbb{R}$, by setting $(L(t)f)(x)=f(x-t)$ for all functions $f$ and all reals $x$.
I know that $L$ is representation of the additive group $\mathbb{R}$ on the space of functions (or continuous functions) defined on the real line. It restricts to a representation of $\mathbb{R}$ on the space $\mathcal{P}$ of univariate polynomials $\mathbb{R}\left[x\right]$ (which we consider as embedded into the space of functions).
Find all the invariant subspaces of polynomials of $L$ (that is, all subrepresentations of the representation $L$ on $\mathcal{P}$).
Let $V = \mathbb{R}[x]$ and let $V_n$ be its subspace of polynomials of order at most $n$. Each $V_n$ is $L(\mathbb{R})$-stable. Note that $V_n = 0$ if $n$ is negative.
Let $\Delta : V \to V$ be the linear operator $L(0) - L(1)$. Then $(\Delta f)(x) = f(x) - f(x-1)$ so $\Delta(V_n) \subseteq V_{n-1}$ and $\Delta(x^n) \equiv n x^{n-1} \mod V_{n-2}$ for all $n \geq 1$.
Let $W$ be some $L(\mathbb{R})$-stable subspace. Then $\Delta(W) \subseteq W$. Let $n$ be largest possible integer such that $W$ contains $V_n$ (if it exists). We will show that $W = V_n$ in this case.
Suppose for a contradiction that $f(x) \in W \backslash V_n$. Then $f(x) \equiv ax^m \mod V_{m-1}$ for some non-zero $a$ and some $m \geq n+1$, so
$(\Delta^{m-n-1} f)(x) = a m (m-1) \cdots (n+2) x^{n+1} + v \in W$
for some $v \in V_n$. Since $am(m-1) \cdots (n+2) \neq 0$ and $v \in V_n \subseteq W$ we get $x^{n+1}\in W$ and hence $V_{n+1} \subseteq W$. This contradicts the maximality of $n$.
Therefore either $W = V_n$ for some $n$, or no such $n$ exists in which case $W$ contains all $V_m$ and hence $W = V$.
Thus $0 = V_{-1} \subset V_0 \subset V_1 \subset \cdots \subset \bigcup V_n = V$ are the only possible $L(\mathbb{R})$-stable subspaces of $V$.