# Under what circumstances is the functional derivative (of an action functional) an actual function?

In general, for a functional $F[\phi]$, the functional derivative is

$$\frac{\delta F[\phi]}{\delta \phi} [f(x)] = \lim_{\varepsilon \to 0} \frac{F[\phi + \varepsilon f ] - F[\phi]}{\varepsilon}$$

which is itself a distribution. But for quite a wide class of action functionals, the functional derivative will be as well a smooth function

$$\frac{\delta S[\phi]}{\delta \phi} [f(x)] = \int \frac{\delta S[\phi]}{\delta \phi}(x) f(x) dx$$

As far as I know, the only exception arises for the Polyakov-type actions, in which case the stress-energy tensor will involve Dirac distributions. This is because the action is linear in the metric components, and $\delta g(x) / \delta g(y) \approx \delta(x-y)$. This isn't quite limited to linear field components since the same is true for the Nambu-Goto action and other actions of the same form, which has the action $\approx \sqrt{\phi}$.

Is there a general theorem to predict when a functional derivative will be a function or not? I'd say it's probably true for any polynomial term $\phi^n(x)$, $n > 1$, or at least $\phi \partial \phi$, but is there any more general theorem regarding this?

• Comments to the post (v2): 1. Smooth local Lagrangian densities $\Rightarrow$ functional derivatives are a smooth function. 2. Hence the Polyakov action does not have a distributional stress-energy-momentum tensor. – Qmechanic Jun 16 '17 at 20:15