I'm reading a text on statistical field theory and, at some point, the author states the following:
"In many cases, for each field configuration $\phi$ these functionals are measures on $\Omega^{N}$, for example:
$$\frac{\delta}{\delta \phi(x)}\frac{\delta}{\delta \phi(y)}\int dz \phi^{2}(z) = 2\delta(x-y)$$
but they could be more singular objects, like derivatives of delta functions. We shall require that $(\delta Z^{N}/\delta \phi^{N})(\phi)$ exists for each $N$ and $\phi$ as signed Borel measures on $\Omega\times\cdots\times\Omega$."
Now, what does is mean? How can a functional derivative be a Borel signed measure?
Notation: Here $\Omega$ is some index set (which is a subspace of a topological space) and $\phi$ is an element of $\mathcal{C}(\Omega)$ the space of continuous functions $\phi:\Omega \to \mathbb{R}$. Furthermore, $Z=Z(\phi)$ is a functional defined on $\mathcal{C}(\Omega)$ and usually thought as an integral functional.
Best Answer
Keep in mind that functional derivatives are taken by varying the function, not its argument. So unless the functional itself involves derivatives, and integration by parts is involved in transforming the result (as in the Euler-Lagrange equations, for example) there is no reason for derivatives of the function to appear. Consider an even better case with continuous $K(x,y)$: $$Z[\phi]=\iint K(x,y)\phi(x)\phi(y)dxdy$$
Then the second variation is $2K(x,y)\delta\phi(x)\delta\phi(y)$, so the second functional derivative comes out as $2K(x,y)$. The example they give simply takes a more singular kernel $K(x,y)=\delta(x-y)$ so the functional reduces to the diagonal integration of the square: $$Z[\phi]=\iint \phi^2(x)dx.$$ If we stick some sign-changing kernel $k(x)$ into it, $$Z[\phi]=\iint k(x)\phi^2(x)dx,$$ the second functional derivative will be a signed Borel measure $2k(x)\delta(x-y)$ supported on the diagonal $x=y$. Indeed, we can take $$Z[\phi]=\iint \phi(x)\phi(y)\,\mu(dx,dy)$$ as a functional for any signed Borel measure $\mu$.