Could we get rid of explicit fields derivatives in Quantum Field Theories? For instance, if we choose the following scalar field Lagrangian, which is (I hope) Lorentz-invariant,  where  $l$ is a a length scale, and with a $(-1,1,1,1)$ metric: 
$$ \mathfrak{L}(x) \sim  l^{-6}  \int d^4y \space e^{\frac{- y^2}{ l^2}} (\Phi(x + y)\Phi(x - y) - \Phi(x) \Phi(x))  $$
There are no explicit derivatives of fields in this expression. 
Of course, if we develop this expression in orders of $l$, we will find, for instance, at the first order (in $l^0$), a term proportional to:
$$\partial_i \Phi(x) \partial^i \Phi(x).$$
And, of course, we will find higher order derivatives of fields with higher order development in $l$.
But the initial expression does not involve explicit derivative of fields.
 A: Of course one can formulate derivative-free Lorentz invariant actions. But these will typically not be causal. This means that fields at spacelike distance will typically not commute, as reqired for a good interpretation of the fields at a fixed time. Hence the fields lack one of the most important requirements of a relativistic QFT, needed e.g., for good cluster separation properties.
Moreover, your integral is not even classically well-defined. Indeed, $y^2$ is unbounded below in Minkowski space, so that the exponential blows up. You'd need to choose a better behaved function of $y^2$ as density.
For the Euclidean case (definite metric, O(4) invariance rather than Lorentz invariance), your action is sensible for $l>0$. However it is questionable whether for $l>0$ there is an analytic continuation to Minkowski space, as this already fails on the classical level.
A: The energy $E$ of a field quantum contains a momentum $\vec{p}$ for all known quanta $$E^2=\vec{p}^2+m^2$$ and the momentum operator is a space derivative $-i\frac{\partial}{\partial\vec{x}}$ (I use units where $c=\hbar=1)$. It is important and unavoidable thing, so the answer is "no", any realistic QFT has to contain the field derivatives.
