This is probably a minor point, but I have seen in some QFT texts the Euler-Lagrange equation for a scalar field, [tex] \partial_{\mu} \left(\frac{\delta \cal{L}}{\delta (\partial_{\mu}\phi)}\right) - \frac{\delta \cal L}{\delta \phi }=0 [/tex] i.e. [itex] \cal L [/itex] is treated like a functional (seen from the [itex] \delta [/itex] symbol). But why would it be a functional? Functonals map functions into numbers, and in our case [itex] \cal L [/itex] is a function of the fields (and their derivatives).
If you want to be completely rigorous, the action is the true functional. The variational derivatives of the Lagrangian (density) should be considered distributions.
But why should there be a functional derivative of [itex] \cal L [/itex]? we have [itex] \cal L [/itex] which is a function of [itex] (\phi, \partial_\mu \phi) [/itex] and we differentiate (as a function) with respect to [itex] \partial_\mu \phi [/itex]
It's common to abuse the notation and use ##\delta## for these derivatives in order to distinguish them from the coordinate derivatives ##\partial_\mu##.