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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.

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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 $\mu$.

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  • $\begingroup$ @Canifold thanks so much for your answer! I'm a bit confused about these afirmatives about these functions being Borel signed measures. When you say $2 k(x)\delta(x-y)$ is a signed Borel measure, you mean that you can define a signed measure $\nu(A) = \int_{A}2k(x)\delta(x-y)d\mu$ with respect to $\mu$ so that our functional $Z[\phi]$ is nothing more than the integration with respect to these measure $\nu$? $\endgroup$
    – IamWill
    Dec 12 '19 at 11:56
  • $\begingroup$ @Willy.K Yes, it is supported on the diagonal, and your $\mu$ is just the 1D Lebesgue measure there. $\endgroup$
    – Conifold
    Dec 12 '19 at 11:59
  • $\begingroup$ @Canifold Oh, now it becomes clear. So, when the author says "we require that $(\delta Z^{N}/\delta \phi)(\phi)$ exists for each $N$ and $\phi$ as signed Borel measures" he is basically saying that the $N$-th functional derivative must be Borel integrable? This would be enough to define a signed Borel measure as I did before right? $\endgroup$
    – IamWill
    Dec 12 '19 at 12:04
  • $\begingroup$ No, only a function can be integrable. He means that the variation can be computed by integration over some Borel measure, i.e. as an integral over a measure defined on Borel subsets. $\endgroup$
    – Conifold
    Dec 12 '19 at 12:10
  • $\begingroup$ Oh, right. So I should not evaluate at $\phi$ right? $\endgroup$
    – IamWill
    Dec 12 '19 at 12:17

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