Determine an explicit expression for $f$. 
Let $f:\mathbb{R}\rightarrow\mathbb{R}$ a continuous function, bounded such that the space  $\mathrm{lin}\{f_k(x)=f(x+k)∣k ∈\mathbb{N}\}$ is finite-dimensional.
Determine an explicit expression for $f$.

The sequence $(f_k)$ generates a vector space of finite dimension means that the sequence $(f_k)$ admits a non-zero minimal polynomial.
Let $P_f(X) = \sum_{k=0}^n p_k X^k$ this polynomial.
We can then construct $f$ by induction: first we construct $f$ continuous (and therefore bounded) on $[0, k]$ such that $\sum_{k = 0}^d p_k f(k)=0$, then $f$ extends to $\mathbb{R}$ through the relation $$\forall x \in \mathbb{R} \sum_{k=0}^d p_kf (x+k)=0.$$
How can I continue ?
NB: Writing this exercise I've realized that he was here. Unfortunately the answer doesn't give us a explicit expression for $f$. Then please do not close my question as a duplicate.
EDIT: This function $$f_\theta : n + x \mapsto (\sin(n\theta)-\sin((n-1)\theta)) x - \sin(n\theta)$$ for  $n \in \mathbb{Z}, 0 \leq x < 1$ works.
 A: Let $f:\mathbb{R}\to\mathbb{R}$ be a bounded piecewise continuous function. For $k\in\mathbb{Z}$, we define
$f_k:\mathbb{R}\to\mathbb{R}$ by $f_k(x)=f(k+x)$, and we consider
$$
V_f=\text{Span}\{f_k:k\in\mathbb{N}\}.
$$
We want to characterize the functions $f$ for which $\dim V_f<+\infty$.

Proposition 1.
  Consider a  real polynomial $P(X)=X^d-\sum_{k=0}^{d-1}a_kX^k$ 
  having $d$ distinct complex zeros of module $1$, and consider the space of bilateral linear recurrent sequences
  $$
\cal{L}_P=\left\{(u_n)_{n\in\mathbb{Z}}:\forall\,n\in\mathbb{Z},~u_{n+d}=\sum_{k=0}^{d-1}a_ku_{n+k}
\right\}.
$$
  Let $\left(e^{(1)},e^{(2)},\ldots,e^{(d-1)}\right)$,  with $e^{(k)}=\big(e^{(k)}_n\big)_{n\in\mathbb{Z}}$, be a basis of $\mathcal{L}_P$. Also, consider $B_k:\mathbb{R}\to\mathbb{R}$ to be a piecewise  continuous $1$-periodic function, for $k\in\{0,\ldots,d-1\}$. Then,
  the function $f:\mathbb{R}\to\mathbb{R}$ defined by
  $$
f(x)=\sum_{r=0}^{d-1}B_r(x)e^{(r)}_{\lfloor x\rfloor}
$$
  is a piecewise continuous bounded function satisfying $\dim V_f<+\infty$.

Proof.
First, note that the sequence $e^{(k)}$ is a linear combination
of geometric sequences of the form $(\lambda^n)_{n\in\mathbb{Z}}$
with $|\lambda|=1$, so it is bounded. This proves that the considered function $f$ is a bounded piecewise continuous function. 
Moreover, note that
$$
\sum_{k=0}^{d-1} a_kf(x+k)=\sum_{k=0}^{d-1}B_r(x)\sum_{k=0}^{d-1}a_ke^{(r)}_{\lfloor x\rfloor+k}
=\sum_{k=0}^{d-1}B_r(x)e^{(r)}_{\lfloor x\rfloor+d}=
f(x+d)
$$
Replacing $x$ by $x+n$, we conclude that
$$f_{n+d}\in \text{Span}\{f_{k+n}:0\leq k\leq d-1\}$$
for every $n$, and this allows us to show by mathematical induction that
$$f_n\in\text{Span}\{f_k:0\leq k\leq d-1\},\quad\hbox{for every $n\geq d$.}$$  Hence $\dim V_f<+\infty$.$\qquad\square$
We will prove that the converse of the Proposition 1. is also true. Indeed, we have  the following result:

Proposition 2.
  Consider  a bounded piecewise continuous function $f:\mathbb{R}\to\mathbb{R}$ such that  $\dim V_f<+\infty$. Then,
  $f$ has the form described in   Proposition 1. 

Proof.
Let us show how to find all the building blocks of the construction. First, define $d$ by the formula:
$$
d=\min\{k\geq 1:f_k\in\text{Span}(f_0,f_1,\ldots,f_{k-1})\}.
$$
Clearly the existence of $d$ follows from the fact that $\dim V_f<+\infty$. The minimality of $d$ proves that $(f_0,f_1,\ldots,f_{d-1})$ is a basis of $V_f$, and consequently, there is a unique vector $(a_0,a_1,\ldots,a_{d-1})$ of real numbers such that 
$f_d=\sum_{k=0}^{d-1}a_kf_k$. Note that $a_0\ne 0$ because  $a_0=0$  would imply that $f_{d-1}=\sum_{k=1}^{d-1}a_kf_{k-1}$, and this  contradicts the minimality of $d$. Define the real polynomial
$P(X)=X^d-\sum_{k=0}^{d-1}a_kX^k$, and consider the space $\mathcal{L}_P$ of  bilateral linear recurrent sequences:
$$
\mathcal{L}_P=\left\{(u_n)_{n\in\mathbb{N}}:\forall\,n\in\mathbb{Z},~u_{n+d}=\sum_{k=0}^{d-1}a_ku_{n+k}
\right\}.
$$
In fact, since $a_0\ne 0$ we can go in both directions starting from the initial conditions. 
From the equality $f_d=\sum_{k=0}^{d-1}a_kf_k$, we see that
for every $x\in\mathbb{R}$ the sequence $(f_n(x))_{n\in\mathbb{Z}}$ belongs to $\mathcal{L}_P$. 
Now, since $(f_0,f_1,\ldots,f_{d-1})$ are linearly independent, there exists $(x_0,x_1,\ldots,x_{d-1})\in\mathbb{R}^d$ such that
the determinant of the $d\times d$ matrix, with $(i,j)$-entry given by $f_{i-1}(x_{i-1})$, id different from zero, (this a general property that can be proved by induction). It follows that
the sequences $\left(e^{(1)},e^{(2)},\ldots,e^{(d-1)}\right)$, with $e^{(k)}_n=f_n(x_k) $, are linearly independent, and since there are $d$ of them, they constitute a basis of the space 
$\mathcal{L}_P$. In particular, this proves that the elements of
$\mathcal{L}_P$ are bounded, (because $f$ is), and consequently,
the $d$ zeros of the characteristic polynomial $P$ in $\mathbb{C}$ are distinct and have modulus $1$.
Next, for a given $x\in\mathbb{R}$, 
The sequence $(f_n(x))_{n\in\mathbb{Z}}$ belongs to $\mathcal{L}_P$, so there are $(\lambda_0(x),\ldots,\lambda_{d-1}(x))$ such that
$$
\forall\,n\in\mathbb{Z},\quad f_n(x)=\sum_{k=0}^{d-1}\lambda_k(x)f_n(x_k)
$$
Note that $(\lambda_0(x),\ldots,\lambda_{d-1}(x))$ are obtained by solving the linear system
$$
f_j(x)=\sum_{k=0}^{d-1}\lambda_k(x)f_j(x_k),\quad j=0,\ldots,d-1
$$
So, each $\lambda_k$ is a linear combination of the functions
$(f_0,\ldots,f_{d-1})$. In particular, $\lambda_k$ is piecewise continuous. Finally
$$
f(x)=f(\lfloor x\rfloor+\{x\})=
\sum_{k=0}^{d-1}\lambda_k(\{x\})f_{\lfloor x\rfloor}(x_k)=
\sum_{r=0}^{d-1}B_k(x)e^{(k)}_{\lfloor x\rfloor}
$$
where $B_k(x)=\lambda_k(\{x\})$ which is a piecewise continuous
$1$-periodic function. This concludes our construction.$\qquad\square$
Finally, If we are interested only with continuous functions $f$ that satisfy $\dim V_f<+\infty$, we only have to assume that the $B_k$'s are continuous on $(0,1)$ and satisfy  continuity conditions at the points $  1,\ldots,d $ in order for $f$ to be continuous.
That is
$$
\sum_{k=0}^{d-1}B_k(1^-)e^{(k)}_{j-1}=
\sum_{k=0}^{d-1}B_k(0^+)e^{(k)}_{j}\quad j=1,2,\ldots,d.
$$
