linear translationally invariant operator represented as a stencil. Let $L$ be a linear translationally invariant operator $L:\{f:\mathbb{Z} \rightarrow \mathbb{C}\} \rightarrow \{f:\mathbb{Z} \rightarrow \mathbb{C}\}$, i.e. $$LT=TL,$$ where $T:\{f:\mathbb{Z} \rightarrow \mathbb{C}\} \rightarrow \{f:\mathbb{Z} \rightarrow \mathbb{C}\}$ is the translation operator defined by $(Tf)(x) := f(x+1)$ for every $f$.
I'd like to see a proof for that $L$ can be represented by some stencil, i.e., there exist numbers $\alpha_k \in \mathbb{C}$, such that 
$$(Lf)(x)= \sum_k \alpha_k f(x+k) $$
is true for every $f$ and $x$. I guess a proof could be an elegant one-liner, but somehow I don't see how to do that.
Edit: I think that this statement is very similar to the theorem that the translation invariant operators on $L^2$ are exactly the multiplier operators.
 A: The following "proof" has a couple of errors:
Let $P=[0,2\pi]$ and $F$ be the Fourier transformation, i.e.,
$$F:\ell^2(\mathbb{Z}) = \{f:\mathbb{Z} \rightarrow \mathbb{C} : \sum_{k\in\mathbb{Z}}|f(k)|^2 < \infty\} \rightarrow \{\hat{f}(k):P\rightarrow \mathbb{C} :\int_P |\hat{f}(k)|^2 dk < \infty \}\}$$ with 
$$(Ff)(k) = \hat{f}(k) = \sum_{x\in\mathbb{Z}} f(x) e^{-i k x}$$
and
$$(F^{-1}\hat{f})(x) = f(x) = \frac{1}{|P|}\int_P \hat{f}(k) e^{i k x}.$$
The eigenfunctions of $T$ are $e^{i k \cdot}$ since $Te^{ik x} = e^{ik(x+1)} = e^{ik} e^{ikx}$ for every $x$. Since $TL=LT$ the eigenfunctions of $L$ are the same, i.e., there exist numbers $m_k\in\mathbb{C}$, such that $Le^{ikx}=m_ke^{ikx}$ for every $x$.
We additionaly have that \begin{align}\widehat{Tf}(k) &= (F f(\cdot + 1) )(k)\\ &= 
\sum_{x\in\mathbb{Z}} f(x+1) e^{-i k x} \\ &= \sum_{x\in\mathbb{Z}} f(x) e^{-i k (x-1)}  =  e^{ik} \hat{f}(k).\end{align}
Using $f(x)=\frac{1}{|P|}\int_P \hat{f}(k) e^{i k x} dk$ we finally find
\begin{align}
(Lf)(x) &=  \frac{1}{|P|}\int_P m_k \hat{f}(k)  e^{i k x} dk \\
&=   \frac{1}{|P|}\int_P m_k \hat{f}(k)  e^{i k x} dk \\
&=   \frac{1}{|P|}\int_P m_k \widehat{T^x f}(k)  dk \\
&=   \frac{1}{|P|}\int_P m_k (FT^x f(\cdot))(k)  dk \\
&=   \frac{1}{|P|}\int_P m_k (F f(\cdot + x))(k)  dk \\
&=   \frac{1}{|P|}\int_P m_k \sum_{\ell\in\mathbb{Z}} f(x+\ell) e^{-i k \ell}  dk \\
&= \sum_{\ell\in\mathbb{Z}} f(x+\ell)( \frac{1}{|P|}\int_P m_k  e^{-i k \ell}  dk )\\
&:= \sum_{\ell\in\mathbb{Z}} f(x+\ell) \alpha_k.
\end{align}
