Editing Functional derivative (section)

==Using the delta function as a test function==
In physics, it is common to use the [[Dirac delta function]] <math>\delta(x-y)</math> in place of a generic test function <math>\phi(x)</math>, for yielding the functional derivative at the point <math>y</math> (this is a point of the whole functional derivative as a [[partial derivative]] is a component of the gradient):<ref>{{harvp|Greiner|Reinhardt|1996|p=37}}</ref>
<math display="block">\frac{\delta F[\rho(x)]}{\delta \rho(y)}=\lim_{\varepsilon\to 0}\frac{F[\rho(x)+\varepsilon\delta(x-y)]-F[\rho(x)]}{\varepsilon}.</math>

This works in cases when <math>F[\rho(x)+\varepsilon f(x)]</math> formally can be expanded as a series (or at least up to first order) in <math>\varepsilon</math>. The formula is however not mathematically rigorous, since <math>F[\rho(x)+\varepsilon\delta(x-y)]</math> is usually not even defined.

The definition given in a previous section is based on a relationship that holds for all test functions <math>\phi(x)</math>, so one might think that it should hold also when <math>\phi(x)</math> is chosen to be a specific function such as the [[Dirac delta function|delta function]]. However, the latter is not a valid test function (it is not even a proper function).

In the definition, the functional derivative describes how the functional <math>F[\rho(x)]</math> changes as a result of a small change in the entire function <math>\rho(x)</math>. The particular form of the change in <math>\rho(x)</math> is not specified, but it should stretch over the whole interval on which <math>x</math> is defined. Employing the particular form of the perturbation given by the delta function has the meaning that <math>\rho(x)</math> is varied only in the point <math>y</math>. Except for this point, there is no variation in <math>\rho(x)</math>.