Background: I am a physicist with decent background in mathematics.
Reading the article on functional derivatives on wikipedia gives:
The functional derivative $\frac{\delta F}{\delta \rho(x)}$ of a functional $F[\rho]$ is defined by
$$\int{ \frac{\delta F}{\delta \rho(x)} \phi (x)dx}=\lim_{\epsilon \to0}\frac{F[\rho+\epsilon\phi]-F[\rho]}{\epsilon}$$
where $\phi$ is an arbitrary function.
Note that the equation above is analogous to equating two equivalent definitions of the directional derivative, for functions of $N$ variables.
Here is my question: The equation is supposed to hold for all functions $\phi$ and yield the same derivative $\frac{\delta F}{\delta \rho(x)}$. But replacing $\phi(x) \to 2\phi(x)$ doubles the LHS but not the RHS of the equation, so the equation can't hold for all $\phi$.
I posit that wikipedia isn't totally correct, and that it should only hold for normalized functions $\phi$, as in the definition of directional derivatives. Is this the case?
Providing a reference would be very useful (the wiki source didn't help).
Best Answer
No, you scaled the right hand side improperly.
If you take $f(x)=2\phi(x)$, $$2\int{ \frac{\delta F}{\delta \rho(x)} \phi (x)dx}=\lim_{\epsilon \to0}\frac{F[\rho+\epsilon 2 \phi]-F[\rho]}{\epsilon},$$ you trivially see that $$\int{ \frac{\delta F}{\delta \rho(x)} \phi (x)dx}=\lim_{\epsilon \to0}\frac{F[\rho+\epsilon 2 \phi]-F[\rho]}{2\epsilon},$$ which amounts to the original expression in the limit.
As you soundly observed, this is analogous to the directional derivative of a scalar function of vectors, and scales the same way. In fact, all functional derivative manipulations draw their intuition from your vector analog.