# Chain rule with functional derivative?

I posted the same question on math exchange but no answer yet, so I post it also here:

"I'd like to make the functional derivative of the functional $$S[\phi(x)]$$ with respect to the Fourier transform $$\widetilde{\phi}(p)$$ such that

$$\phi(x)=\int\frac{d^{d}p}{(2\pi)^{d/2}}e^{ip\cdot x}\widetilde{\phi}(p)$$

Can I use the following chain rule

$$\frac{\delta S}{\delta\widetilde{\phi}(p)}[\phi]=∫{d^{d}x}\frac{\delta\phi(x)}{\delta{\widetilde{\phi}}(p)}\frac{\delta S}{\delta\phi(x)}[\phi]=∫ \frac{d^{d}x}{(2\pi)^{d/2}}e^{ip\cdot x}\frac{\delta S}{\delta\phi(x)}[\phi]$$

or not?"

I'd give you my answer considering a functional

$$S = \int_x L(\phi'(x), \phi(x), x) \, dx \ .$$

Since I didn't know the answer (and maybe still don't it, so watch out!), after the evaluation of the function $$\phi(x)$$ and its derivative with the transform,

$$\phi(x) = \int_k \hat{\phi}(k) e^{ikx} dk \quad, \quad \phi'(x) = i \int_k k \hat{\phi}(k) e^{ikx} dk$$

I'd start from the definition of the variation of a functional. I'd first evaluate the functional $$S\left[\hat{\phi}(k) + \varepsilon\hat{w}(k) \right]$$ and expand it at the first order in $$\varepsilon$$ (with $$\hat{w}(k)$$ the function usually indicated with $$\delta \hat{\phi}(k)$$ to remember that it's the variation of function $$\hat{\phi}(k)$$)

\begin{aligned} S\left[ \hat{\phi}(k) + \varepsilon\hat{w}(k)\right] & = \int_x L\left(i \int_k k [ \hat{\phi}(k) +\varepsilon \hat{w}(k) ] e^{ikx} dk, \int_k [ \hat{\phi}(k) + \varepsilon \hat{w}(k) ] e^{ikx} dk, x \right) dx \\ & = \int_x \left\{ L\left( i\int_k k \hat{\phi}(k)e^{ikx} , \int_k \hat{\phi}(k)e^{ikx}, x \right) +\frac{\partial L}{\partial \phi'} i \varepsilon \int_k k \, \hat{w}(k) e^{ikx} dk +\frac{\partial L}{\partial \phi} \varepsilon \int_k \hat{w}(k) e^{ikx} dk \right\} dx +o (\varepsilon) \ , \end{aligned}

and subtract the non-variated functional to the the variation,

\begin{aligned} \nabla_{\hat{\phi}} S & = \lim_{\varepsilon \rightarrow 0} \frac{1}{\varepsilon} \left\{ S\left[ \hat{\phi}(k) + \varepsilon\hat{w}(k)\right] - S\left[ \hat{\phi}(k) \right] \right\} = \\ & = \int_x \left\{ \frac{\partial L}{\partial \phi'} i \varepsilon \int_k \hat{w}(k) e^{ikx} dk +\frac{\partial L}{\partial \phi} \varepsilon \int_k \hat{w}(k) e^{ikx} dk \right\} dx = \\ & = \int_k \underbrace{\int_x \left\{ i k \frac{\partial L}{\partial \phi'} +\frac{\partial L}{\partial \phi} \right\} e^{ikx} dx}_{=: \nabla_{\hat{\phi}(k) }S} \, \hat{w}(k) dk \ , \end{aligned}

with the last expression recast with the definition of the sensitivity w.r.t. the function $$\hat{\phi}(k)$$.

• So the answer coincide, since this question raised when I was studying Polchinski eq. in LPA so I should have specified that S does not depend on the derivatives. Thank you btw Commented May 31 at 16:27
• You're welcome. Glad to hear that it's been useful (and I'll imply I managed to provide a meaningful answer) Commented May 31 at 16:31