TITLE: A question about the vector space spanned by shifts of a given function
QUESTION [10 upvotes]: Let $E$ be the space of continuous real functions and $f\in E$
Let $T_t$ denote the shift operator: $T_t(f)(x)=f(t+x)$
Let $T(f)$ be the linear span of the set $\{T_t(f) \;|\; t\in \mathbb R\}$
Suppose $T(f)$ is finite-dimensional.
Prove that $f$ is differentiable and that $f'\in T(f)$

I have very few clues about what should be done to solve this problem.
Here's my work so far.
Let $T_{a_1}(f),\ldots,T_{a_n}(f)$ be a basis of $T(f)$.
Let $x\in \mathbb R$. There exists $\lambda_1,\ldots,\lambda_n$ such that $T_{x}(f)=\sum_{k=1}^n \lambda_i T_{a_i}(f)$
Since $\displaystyle \frac{f(x+h)-f(x)}{h}=\frac{T_{x}(f)(h)-T_{x}(f)(0)}{h}=\sum_{k=1}^n \lambda_i \frac{T_{a_i}(f)(h)-T_{a_i}(f)(0)}{h}$, it suffices to prove that $f$ is differentiable at $ a_1,\ldots,a_n$, but that's still difficult.

REPLY [2 votes]: With a little use of Functional Analysis.
First observe that $T(\,f)$ is a closed subspace of $C(\mathbb R)$, equipped with the topology of locally uniform convergence, as it is finite-dimensional.
Let $j(x)$ be a non-negative $C^\infty$ function supported in $\big[-\frac{1}{2},\frac{1}{2}\big]$, with $\int_{-1/2}^{1/2}j(x)\,dx=1$. Then the convolution $j*f$ is equal to
$$
j*f=\lim_{n\to\infty}\sum_{k=1}^nj(t_{k,n})T_{-t_{k,n}}f,
$$
where $t_{k,n}=-\frac12+\frac{k}{n}$, and hence $j*f\in T(\,f)$. Let $$\hat T(\,f)=T(\,f)\cap C^\infty(\mathbb R).$$
Now if we set $j_\varepsilon(x)=\frac{1}{\varepsilon}j\big(\frac{x}{\varepsilon}\big)$, then $j_\varepsilon*f\to f$, locally uniformly, as $\varepsilon\to 0$. Thus every element of $T(f)$ lies in the closure of $\hat T(\,f)$. But as $\hat T(\,f)$ is finite dimensional, it is closed, and hence $\hat T(\,f)=T(\,f)$. Likewise the limit $f'=\lim_{h\to 0}\frac{1}{h}(T_hf-f)$ is locally uniform,  as $f\in C^\infty$ and as $\frac{1}{h}(T_hf-f)\in T(f)$ so is $f'$.