Different definitions of Functional Derivative In studying QFT and General Relativity, I came across two different definitions of Functional Derivative, and I'd like to know if they are equivalent.


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*Firstly, in Wald's book General Relativity, as well as in other GR references (Baez), the variation of the action is given in terms of a "one-parameter family of field configurations $\psi_\epsilon$" and this family can be defined as $\psi_\epsilon = \psi_0+\epsilon\phi$, where $\phi$ is an arbitrary field. Then, the following definitions are made:
$$d\psi_\epsilon/d\epsilon|_{\epsilon=0}:=\delta\psi \ \ \ \ \ \ \ \ \ \ \ \delta S:=\frac{d}{d\epsilon}S[\psi+\epsilon\phi]\Big|_{\epsilon=0}$$
where $S[\psi]$ is the functional of interest. Finally, the Functional Derivative $\delta S/\delta\psi(x)$ is defined as follows:


$$\delta S= \int \mathrm{d}^4x\frac{\delta S}{\delta\psi(x)}\phi(x) = \frac{d}{d\epsilon}S[\psi+\epsilon\phi]\Big|_{\epsilon=0}$$


*When I turn to a QFT reference, like Greiner's Field Quantization, the definition is practically the same, but the arbitrary field $\phi$ is now specified as $\delta^4(x-x')$. I understand that this specification can be interpreted as a variation at the position $x'$ alone, so that the integral can be seen as an analogous of $dS = \sum \frac{\partial S}{\partial x_i}dx_i$. It also seems important when dealing with Generating Functionals, but I haven't studied them yet, so I might be wrong.


I would like to know why is this specification ($\phi=\delta^4(x-x')$) made and if both definitions are equivalent.  
 A: Aside from the latter definition not being "mathematically sane" as md2perpe pointed out, the equivalence can be established easily as follows:
Let $\delta_{x_0}$ denote the Dirac delta distribution centered on $x_0$, eg. $\delta_{x_0}(x)=\delta(x-x_0)$.
Assume that $S$ is functionally differentiable at $\psi$. then for any variation $\phi$ we have $$ \delta S[\psi]=\int d^4x \frac{\delta S[\psi]}{\delta\psi(x)}\phi(x). $$ Since this is true for any variation (or at least those that vanish at the boundary), we can replace $\phi$ with $\delta_{x_0}$. Then we get $$ \delta S[\psi]_{\text{specific}}=\int d^4x\frac{\delta S[\psi]}{\delta\psi(x)}\delta_{x_0}(x)=\frac{\delta S[\psi]}{\delta\psi(x_0)}, $$ so using the Dirac delta distribution as a "test function" will get you the functional derivative directly, rather than indirectly.
But of course, as it has been pointed out, this is a formal manipulation and is not actually sane mathematically.
A: In most practical cases the two definitions are equivalent, but the latter definition is not mathematically sane. The functional $S$ is defined for smooth functions, and $\delta^4$ is not even a function.
