# Definition of functional derivative

In this book,
A derivative of a function is defined as follows: $$\frac{df}{dx} = \lim_{\epsilon \to 0} \frac{f(x+\epsilon) - f(x)}{\epsilon}.$$

And define a functional derivative of a functional $F[f]$ as follows: $$\frac{\delta f}{\delta f(x)} = \lim_{\epsilon \to 0} \frac{F[f(x^\prime) + \epsilon\delta(x-x^\prime)] - F[f(x^\prime)]}{\epsilon}.$$

I don't understand why change of functional F is $\epsilon\delta(x-x^\prime)$. Why not define $$\frac{\delta f}{\delta f(x)} = \lim_{\epsilon \to 0} \frac{F[f(x^\prime) + \epsilon] - F[f(x^\prime)]}{\epsilon}~?$$ What is the meaning of $\epsilon\delta(x-x^\prime)$?

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• Would Mathematics be a better home for this question? – Qmechanic Mar 25 '17 at 9:38
• The correct mathematical dedefinition of "functional" derivative is given by either the Fréchet or Gâteaux derivative. – yuggib Mar 25 '17 at 12:50

$$$\frac{\delta F}{\delta f(x)} = \lim_{\epsilon \to 0} \frac{F[f(x^\prime) + \epsilon] - F[f(x^\prime)]}{\epsilon}$$$
would be the normal derivative of $F()$ at the point $f(x')$. The point of a functional derivative is that $\delta(x-x^\prime)$ is an unknown function that goes to zero at the boundaries. In contrast $\epsilon$ is just a number. So we are determining an extremum of $F(f(x))$ by allowing small changes in $f(x)$ , not in $x$.
$F()$ is not a 'normal' function ('normal' being for example : $\mathbb{R} \mapsto \mathbb{R}$ ). Instead of a number, $F()$ maps a function (e.g. $f(x)$ ) to a number. A trivial example would be : $F(f(x))=\int_{a}^{b} f(x) dx$.
A functional derivative normally gives you as output not a function, but a differential equation with which you can determine $f(x)$. So you're determining for which function $f(x)$ , $F(f())$ will be an extremum.