How does functional derivative transform under coordinate transformation

Question

The definition of functional derivative I will be using will be that described in "Using delta function as a test function" section of the following Wikipedia link: https://en.wikipedia.org/wiki/Functional_derivative#Using_the_delta_function_as_a_test_function

Instead of $\frac{\delta F}{\delta \rho(x)}$, I would like to compute $\frac{\delta F}{\delta \rho(x')}$, where x and x' are related by, say, x=ix'.

What is the relationship between $\frac{\delta F}{\delta \rho(x)}$ and $\frac{\delta F}{\delta \rho(x')}$?

Answer 1

The answer actually stays unchanged. You're still taking functional derivative with respect to the same point. If you refer back to the definition of functional derivative, such as the one in Wikipedia, you'll see that it's the coefficient that matters.

Answer 2

Wikipedia defines the functional derivative through the formula (see this section) $$\left.\frac{\mathrm{d}}{\mathrm{d}\epsilon} F[\rho + \epsilon \phi]\right|_{\epsilon=0} =: \int \frac{\delta F}{\delta \rho(x)} \phi(x) \mathrm{d}x.$$

Suppose now we want to perform a change of coordinates $x \to y$. Then we have $$\left.\frac{\mathrm{d}}{\mathrm{d}\epsilon} F[\rho + \epsilon \phi]\right|_{\epsilon=0} =: \int \frac{\delta F}{\delta \rho(x(y))} \phi(x(y)) \left|\frac{\mathrm{d}x}{\mathrm{d}y}\right| \mathrm{d}y,$$ and we see a Jacobian factor has appeared. Of course, $\phi(x(y)) = \phi(x)$, and hence we must conclude that $$\frac{\delta F}{\delta \rho(y)} = \left|\frac{\mathrm{d}x}{\mathrm{d}y}\right| \frac{\delta F}{\delta \rho(x)}.$$